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Talk:Martingale (betting system)

Blackjack
I am confused and would appreciate any insight into how a game like blackjack, where sometimes the odds pay more than 1:1 affect this system? Eventually you will get blackjack, which pays 3:2 which should increase the winning chances right? Of course, given unlimited time and limited money you will mathematically still eventually lose everything. —Preceding unsigned comment added by 190.22.94.61 (talk) 13:54, 20 June 2008 (UTC)

 * Given 3:2 odds, it's NOT true that unlimited time and limited money means you will eventually lose everything. The biased random walk (which is what you're talking about) has a positive chance of going to infinity and never reaching zero. 142.1.133.165 (talk) 16:02, 13 March 2013 (UTC)

Hmm.. How can the player lose $160 in any session if the starting bid is $10? See, first he bets $10, then $20, then $40, then $80. The sum of his bets is $150. So the given calculations do not look right. Can anyone explain and/or correct the article??? Second point: why does it say that on average each winning session gives $4.30? Each winning session gives the initial bet, so that is $10 in this scenario. So why isn't it as follows: for 100 sessions (96 winning, 4 loosing), the player: - wins 96 times of $10 = $960 - looses 4 times of $150 = $600 In the end he gains $360 dollars. 23/Aug/2006

 * There is poor math here everywhere. He has $160 bankroll. So if he starts at $10 his bets are $10 $20 $40 $80 totaling $150. He will loose 4 bets in a row 20/38^4 = 7.67% of the time. The other 92.3% of the time he will win $10. So all up he ends up with average of 0.923*10 - 0.0767*150 = -$2.275 lost each time he uses the progression. It does not make moeny.--155.144.251.120 22:15, 21 December 2006 (UTC)

 * Can someone check the math from...
 * "In 150 turns, there is a 73.5% chance that you will lose 6 times in a row on the pass line.
 * In 150 turns, there is a 70.7% chance that you will lose 6 times in a row at coin tossing."
 * I recalculated with the method they used, and got 70.74% for PASS LINE, and 68.18% for COIN TOSSING. 206.213.43.100 (talk) 21:56, 7 May 2012 (UTC)

I believe that if used with craps and you bet on the field starting with a bet of 5$ and doubling up until you win, that eventually you will turn a profit. You see there are 36 possible combinations of dice, 17 of which win you money and 19 of which where you lose money. But snake eyes and 12s gives you three times the pay out so you really have a 19/36 chance rather than 17/36. Sure you could get a very unlucky streak but the odds are in your favor to win. TrueSilver 21:46, 12 August 2006 (UTC)
 * You don't know the rules for craps. If you really had 19/36 chance to win then the player has the edge in the game and not the house. Progression betting would be mute because you could use the kelly criterion to win even more money. The payouts you have said don't exist.--155.144.251.120 22:15, 21 December 2006 (UTC)

Is this encyclopedic? I'm not sure of exactly how the Wikipedia stands on howtos... Ed Cormany 03:19 18 Jul 2003 (UTC)
 * It needs re-writting because it is rather poor. --155.144.251.120 22:15, 21 December 2006 (UTC)

Question: who invented the Martingale system, and when? Martin

Is there any reason why this should not be merged into the main Martingale article? --Henrygb 17:09, 3 Apr 2005 (UTC)

This seems pretty POV to me.. Like warning to some gambling addicts that this will not work. This is one of the best betting strategies on roulette and works pretty good if you find a high limit table somewhere..

 * I don't see how this is any more POV than stating that perpetual motion machines are not physically possible. You can't ever have positive expected value on an unbiased roulette table.

 * You're right. It's a statement of fact. But it's best just to let the idiots who think they can win at roulette using Martingale (or anything) find it out the hard way. 82.13.187.127 08:48, 27 December 2005 (UTC)

But does the strategy really require that the gambler has an infinite wealth? I would easily try this out once as soon as I have a companion that could lend me any amount of money for a very short period of time without interest. I would pay him back everything within a minute or so—guaranteed. It's risk free for him and it's risk free for me—and yet I know I will win the amount I want. Am I really an idiot then? INic 02:09, 1 February 2006 (UTC)

 * In what way is it risk-free? It is perfectly clear that there is *always* a chance of catastrophic loss. With any real amount of money, you have a chance to lose it all. That is *not* risk-free.

 * No the only catastroph that can happen is if I lose all money I have and have to stop playing. But this never happens here because I can always get a cheque with the amount I need written on it so that I can continue to play. And I know also that I wil win pretty soon. I will never experience 100 losses in a row for example. INic 02:08, 11 February 2006 (UTC)

 * No the only catastroph that can happen is if I lose all money I have and have to stop playing. Yes, indeed.  Ponder that for a while.
 * But this never happens here because I can always get a cheque with the amount I need written on it so that I can continue to play. Your credit rating must be astronomic.
 * I will never experience 100 losses in a row for example. And even in the unlikely chance that you do, provided your initial bet was $1, you will only lose $1,267,650,600,228,229,401,496,703,205,375.  And it's so much more likely that you'll win a single paltry buck, after all. Aragorn2 11:49, 4 May 2006 (UTC)

 * If you play this strategy long-term, you are *mathematically guaranteed* to lose all your money, since you will eventually hit a large enough losing streak to clean you out. With "infinite wealth", you can keep doubling your bet each time and you will eventually hit a win and recoup your losses; therefore with infite wealth you win on average. But without infinite wealth, you will eventually lose.

 * I will play it only once as I said. INic 02:08, 11 February 2006 (UTC)

 * Note: No, even with infinite wealth you will on average have a loss equal to the house edge. Given infinite players or infinite hands/spins/trials there exists several people who will die without ever having seen a win.  Yes this is unlikely but it makes the math correct and if we're playing infinites let's play infinites.  — Preceding unsigned comment added by AddBlue (talk • contribs) 07:18, 20 February 2012 (UTC)

 * Incedentally, why would the money have to be borrowed? Surely you have your own which you can gamble with? Or would you just like to ensure that if you do lose, somebody else pays?--Rhebus 11:00, 6 February 2006 (UTC)

 * The money have to be borrowed because it could happen that I have to play with a cheque representing more money than there is in the world, or even more gold than is available in the universe. But that doesn't matter because I will pay him back within a minute. INic 02:08, 11 February 2006 (UTC)

 * Casinos tend to have maximum stakes. Mikekelly 00:01, 24 February 2006 (UTC)

 * Correct, but we're not investigating a real gambling situation here but rather a thought experiment. My question is if I'm really an idiot if I want to play given that the conditions I specify are fulfilled? I would in fact play and I don't think I'm an idiot. INic 01:33, 28 February 2006 (UTC)

 * If you can find someone who will act as casino bunghole royale and allow you to wager an unlimited stake without having to pay up until after the game is over then you are sure to win. So in a thought experiment where there is no maximum stake you would not be an idiot for using this strategy. Your original question was "does this strategy really require infinite wealth to pull off"? The answer is yes. If you do not have some infinite source of wealth to wager with then this is a losing system. Mikekelly 09:18, 28 February 2006 (UTC)

 * But I know from the outset that I always will pay back the amounts that I borrow within minutes. The guy lending me money does this risk-free. This means that he easily can lend me more money than what he actually has. If the casino accepts cheques for example I can literally gamble without limit and without risk of losing. In practice I know, in addition, that I will never have to double my stake more than say 30 times. This ensures that the gamble will be not only finite in time but very short. So where is the supposedly required infinitude in wealth? INic 00:00, 19 March 2006 (UTC)

 * Your "guy lending me unbounded amounts of money" is your infinitude of wealth. Mikekelly 11:53, 20 March 2006 (UTC)

 * But those cheques aren't worth more than the paper they are written on. As we know that no one ever will go to the bank with our cheques to cash them in we don't have to have any money at all on our bank account. We can write any amount on them risk free anyway. In fact, it's the very property that we can write any amount that guarantees that we can write any amount. If there is a limit we can't write anything without losing in the long run, at least in theory. It could still be so improbable that we would lose that we could safely ignore that case. INic 02:06, 11 May 2006 (UTC)

 * When you find a casino that will let you use a billion dollar cheque to wager with, unverified, let me know. 160.39.190.180 02:06, 25 June 2006 (UTC)

 * I find your argument very strange, iNic. You are saying that you don't need infinite wealth to execute the strategy, because you are able to write infinite checks that don't actually represent real wealth. I think the problem with this discussion is jumping back and forth between 'in practice' and 'in theory'. In practice, no casino would allow you to wager more than some secured amount that they can reasonably expect to collect from you. You would always be limited by your credit, even if the table did not have explicit limits. In theory, you have constructed a game in which you are writing down numbers that don't represent real wealth, and the imaginary casino accepts those numbers, even though they don't represent actual wealth. So yes, in that constructed scenario, you don't need infinite wealth. You don't even need any money to play at all, you can just write numbers on a piece of paper to play. But that's not an interesting scenario. It doesn't prove anything. It's like saying that you can prove that everything is free as long as you stipulate that you infinite credit. Tristanreid 22:01, 28 July 2006 (UTC)

 * This page takes a mathematical look at the system and comes up with some conclusions summarized in a few tables, it might be helpful for the article or just for this debate. --BigCow 23:20, 28 July 2006 (UTC)

 * I agree that we need to clearly distinguish between 'in practice' and 'in theory' here. But not even in theory we need infinite wealth to use the strategy as you claim, all we need is unlimited wealth. Anhyway, two things happen to the situation when we go from 'theory' to 'practice,' that actually cancel out. In practice there are always limits to wealth of course, but in practice there are also always limits to a losing sequence with a 50% chance of losing every time. The problem with your reasoning is that you take the first practical limit into account but not the second. In addition I've added an extra detail that can be done in practice, the possibility to temporarily fool the casino. That can in practice be accomplished in many ways. I could have hacked the security system so that my visa card always said yes for example. The point is that I know that this strategy is risk free. I will never be charged any astronomical sum of money as I know for sure I will win in the end. Remember I will use this strategy in practice only ONCE. And of course i can use real money as my first stake. INic 11:34, 25 August 2006 (UTC)

 * But Unlimited and Infinite are the same. Infinite doesn't mean it is Hitchhiker's Guide Infinity; it means without bound, bigger than whatever you have. How much money do you need? 2N, where N is the amount you just lost. That's without bound, and that IS infinity. GumbyProf: "I'm about ideas, but I'm not always about good ideas." 05:00, 17 September 2006 (UTC)

 * Not necessarily. Infinity comes in many different flavors and one major divide is between actual infinities and merely potential infinity (that even most finitists accept). What we need here is clearly not any actual infinity. In fact, we don't even need potential infinity. The reason is that in practice I know, for sure, that I will never lose more than say 100 times in a row. It will simply never happen. And $$2^{100}$$ is very far from infinity of any flavor, actual or not. INic 02:01, 19 September 2006 (UTC)

 * Oh, COME ON! "It will simply never happen," yet the probability, $$2^{-100}$$, is the inverse of something which is "very far from infinity of any flavor?"  I would think that, even in imprecise terms, something "will simply never happen" would have an inverse which would be close to infinity.

 * No, this is in fact not the case. We can be absolutely certain that events with very small probabilities never will happen anywhere in our universe. This observation is used in statistical physics for example. This, of course, doesn't mean that these events have infinitely small probabilities in any mathematical sense of the word. For the mathematician any finite number is as far from infinity as the number one. INic 02:00, 27 September 2006 (UTC)

 * Actually, if something has any small but finite probability, it *will* happen somewhere, sometime. Since you use statistical physics as your example, the general rule "entropy tends to increase" is in general true but entropy can and does decrease by chance; just not very often. --Rhebus 11:42, 16 November 2006 (UTC)

 * Not at all. There is a positive probability that any broken thing (an egg, a cup, a window, or whatever) will spontaneously go back to once piece again. Yet, it will never ever happen. Fluctuations in entropy increase are not improbable at all—on the contrary. It follows from Boltzmanns statistical interpretation of Clausius classical law. iNic 00:41, 7 January 2007 (UTC)

I believe that this betting strategy is a sure method of not losing money and possibly winning money, just not very much relative to what you've already got. You don't need complicated stat equations to prove to yourself that this does indeed work. There are basically two main factors in determing how much you'll win.

1. You can choose when you stop playing. Therefore, if you won or break even, you can stop.

2. It depends on how much money you start with.

Let's say, I go to the casino with $2500 and I play roulette

 * NO! Even if you play a French roulette wheel with close to true odds (you play black or red, which is an even-odds bet that gives the house around a 1% advantage), where each play costs one cent and you promise to stop playing the instant you attain $2500.01, using a Martingale (or any other) betting system the odds are against you. The reason is that in this situation there are two cases: 1) You win on one of the first 16 plays, earning a total of one cent and stopping, or 2) You lose 17 games in a row, and you have not enough money left for the necessary $1,310.72 for the next bet.  Case 1) is FAR more likely than case 2), but since we know the house has an advantage, the expected winnings are negative, if only slightly.

 * This also applies to everything INic has been saying. The fact is that things with very small probabilities CAN happen in reality.  A fair roulette wheel in Brazil once spun red 32 times in a row.  That was at a probability of 0.000,000,023%, but it happened anyways.  There is also no reason entropy cannot decrease given enough time.  For systems larger than a few picograms, the second law of thermodynamics is typically true to within a few parts per million, but notice that it is not EXACTLY perfect ALL of the time.  There is no arbitrary barrier the universe sets on low probability events.  In direct response to INic, you cannot be absolutely certain very small (but finite) probability events will never occur in the universe.  Therefore, in the fictional scenario where one has a very large but finite amount of money that is sufficient to cover any losses up to Graham's number of dollars, playing the same game I outlined above with the same stakes and betting strategy will find the same result the player with just $2,500 would find, except that his losses would be exceedingly unlikely (on the order of the inverse of f, where f is A(A(A(...(1)))...), where there are many A's (perhaps A(10000)) and A(x) is the Ackerman function Ackerman(x,x)), while the potential loss would be proportionately greater (Graham's number).  Eebster the Great (talk) 02:04, 28 September 2008 (UTC)

 * But this never happens here because I can always get a cheque with the amount I need written on it so that I can continue to play.  No, you can't always get a cheque with the amount needed. No real person has an infinite amount of credit available to them. And what idiot would lend you money to invest in a scheme with _negative_ value. No matter how you slice it, the value of a dollar invested at the roulette wheel is around 97 cents. Ordinary Person (talk) 23:57, 18 April 2013 (UTC)

 * A nit, but "No matter how you slice it" is going too far. Yes, Martingale does not change EV. But, other Roulette techniques can.Objective3000 (talk) 14:11, 19 April 2013 (UTC)

 * True. Ordinary Person (talk) 11:37, 24 April 2013 (UTC)

A more simple analysis
Roulette is a game of pure chance - there is no skill - every number has an equal chance of coming up but the payouts are made at under the odds. Roulette is a NEGATIVE EXPECTANCY GAME - every time you bet on a number you are betting into a negative return in the long run. NO method works. In practice casinos couldnt care less about Martingale or any other theory.

 * It seems people here simply don't get that. THere are no more red/black squares appearing after spins, so no matter how much you bet or when you bet it, the chance of red/blaack spinning up does not change. It doesn't help that the article is poorly written either. There are mathematical proofs on the merge page that show why it won't work for gambling, there is also the Kelly Criterion which shows that the optimum bet for a game with a house edge is negetive (ie be the house or don't bet at all). These are 100% mathematical proved and verified. Of course Martingale will always be popular and has been since medieval times. --155.144.251.120 22:15, 21 December 2006 (UTC)

merger?
It is proposed to merge this "with" martingale (probability theory). It would be more plausible to merge this into that article. There is a "mergeinto" template for that purpose. Michael Hardy 19:46, 13 September 2006 (UTC)

 * There are three Martingale articles. I'd only merge this one if the third one was merged too.  Otherwise, since this is so wildly different than the probability article it should also be kept separate.  Anyway, if the paradix one is merged too, that seems best. 2005 22:51, 13 September 2006 (UTC)

 * Support This article could be a footnote of the probability article. Then, you could take out all the uncited stuff and make it into a useful side comment. GumbyProf: "I'm about ideas, but I'm not always about good ideas." 05:00, 17 September 2006 (UTC)

 * I'd support merging all the articles. They're all on the Martingale system, and merging would make the information easier to access. IMHO. 71.89.61.239 02:15, 20 October 2006 (UTC)

 * Support merge. It can and should be added to the Martingale as it only an implementation of this theory. JeffyP 20:27, 29 October 2006 (UTC)

 * Support merge. You can always redirect hits to this page back to the main one. And I agree, this is an implementation of the theory. Obscurans 20:59, 3 January 2007 (UTC)

I'd keep it as is. I found this article in searching this exact topic. I had not heard of the name of the theory, only the method of essentially doubling one's bet upon sequential losses. Had it been merged with the other topic, I likely would not have found it, much less realized the correlation between the two.

 * Keep I agree with his keep. Many people do not know it as the 'martingale' but as the 'double up' system or progression betting etc.--155.144.251.120 22:15, 21 December 2006 (UTC)

 * Keep I agree that it's a good thing to have two different articles, even though they could cross-reference each other better. As I understand it, readers here fall into two quite different groups (mathematicians and gamblers) and consequently they expect two quite different accounts of the matter. iNic 00:18, 7 January 2007 (UTC)

 * Keep The Martingale (betting system) article is written mostly in standard English, and is an easy article to read for those simply interested in the progression betting or "double-up" gambling system. The Martingale (probability theory) article contains much that is written as statistics and probability equations, and is unintelligible to many readers without advanced study of such systems. StavinChain 17:58, 25 May 2007 (UTC)

I agree that the analysis is completely incorrect - so incorrect that it should be removed until it is re-written correctly.
 * It is not incorrect, it simply does not explain exactally what a session is, but obviously it needs re-writting.--155.144.251.120 22:15, 21 December 2006 (UTC)

Nick Leeson
Would it make sense to add a bit here about Nick Leeson, who destroyed the Barings' Bank with what was in effect a martingale series of bets of the Nikkei index? --Christofurio 04:25, 4 January 2007 (UTC)

 * I just read that article but couldn't find any support for a connection to martingales. iNic 00:18, 7 January 2007 (UTC)

 * There are some similarities in that it appears he kept betting more and more (with riskier bets) in an attempt to recover the money he had lost, I presuming hoping he would eventually make it all back. Also see Barings Bank. Nil Einne 01:36, 25 April 2007 (UTC)

 * A closer example would be the collapse of LTCM, which made "sure thing" bets with small profits, but after a streak of bad luck, was forced to come up with too much capital and went bust. They claimed that had there not been a margin call, they would have made it all back in a few months, a frequent martingale claim.  —Preceding unsigned comment added by 132.228.195.207 (talk) 21:40, 31 March 2009 (UTC)

re-written
I just re-wrote most of the article and put in correct math. Feel free to improve it.

On an unrelated note, does anyone know the origin of the term "martingale", and how it's related to this betting system? --Spoon! 10:58, 22 January 2007 (UTC)

Formula still looks wrong to me
The math here still looks incorrect. It claims that the expected profit is (...), but does not take into account the fact that one could win x times in a row - I.E., the math is basically as follows: P(neverWinning) * amountLost + P(winningAtLeastOnce) * amountWon, but the amountWon increases if the player has won more than once and this is not taken into account. The formula only assumes that the player wins once and stops playing.

As an example, note tha the current formula shows the correct payoff if there are consistent losses on all x plays, but does not show the correct payoff if there are consistent gains on all x plays. The correct way to show the expected payoff of a martingale involves combinatorics and the series of corresponding payoffs and probabilities. Luckily, the series can be reduced to a closed-form solution. "I have a wonderful proof of this fact which the margin is too small to contain." (I hope to have more time to update in the next few days)

We should also show a graph that illustrates the Martingale payoff. —The preceding unsigned comment was added by 66.65.160.193 (talk) 02:21, 25 April 2007 (UTC).

 * I think you have totally missunderstood the formula. It applies generally. Having a combination of the payoffs and probabilities is impossible, because it will be infinite if the player has infinite money. That is why the formula includes how much they have. Martingale is very simple, you are trying to win 1 unit, risking all of your money to do it. It doesnt matter how many times you do this.--58.108.249.77 09:24, 6 June 2007 (UTC)

Example
The example is misleading. It is comparing a loss per round with a loss per roll and indicating that there is a difference in the edge. Martingale makes no difference to edge. Objective3000 (talk) 12:02, 18 April 2008 (UTC)

POV
Added for the obvious: an article shouldn't be calling anyone 'foolish', etc. —Preceding unsigned comment added by 219.73.23.140 (talk) 15:10, 24 May 2008 (UTC)

Silly mathematicians
It's unlikely that you'll lose any money by withdrawing it at profit at some point if you have a lot of money and play with smaller bets. This is just stupid absolutistic idealistic analysis of a situation where you play for an infinite amount of time. Well, duh, Einstein, how is that possible in real life? Gambling is by definition not risk-free. This is an alright strategy :) -- nlitement [talk]  21:45, 30 May 2008 (UTC)

 * Well, yes, it is true that this strategy usually results in positive payouts if one plays for brief periods of time and only leaves immediately after a win. However, this strategy sometimes leads to catastrophic losses, and the net effect of these catastrophic losses outweighs the net effect of the meager winnings assuming a house edge.  This doesn't need to be idealized to infinite time played, but merely shows that the Martingale system doesn't improve your pot odds.  So obviously this strategy is fine if one assumes that a high probability of winning a low amount of money inherently outweighs a low probability of losing a proportionately high amount of money -- in fact, there might be cases one could contrive where this would be true (for example, if one needs just $1 more to pay off their $1,000,000 debt, and if that dollar is not earned the whole of their $1,000,000,000 estate will be lost).  However, most of the time the exact opposite is true (if one takes a realistic utility function into account, for example) to a high degree, making the Martingale system WORSE than a system where one bets the same amount on every play.  And no matter what, in terms of mathematical expectancy, the Martingale system changes nothing.  Eebster the Great (talk) 02:12, 28 September 2008 (UTC)

Modified Martingale..?
I was wondering if there is a modified martingale system that would let you gain on a bet by more than doubling the new bet after a failed bet. For instance betting $1, $2, then $5, $10 then $25 and so on so that a person would come out ahead once (or if) he finally hit on a bet. It would have similar risks and would risk the catastrophic failure point quicker, but adds the possibility of reward rather than just breaking even. —Preceding unsigned comment added by Mzalar (talk • contribs) 19:51, 2 October 2008 (UTC)
 * No. Objective3000 (talk) 19:39, 3 October 2008 (UTC)
 * Yes there is. The reason the doubling up principle doesn't work is because you only win your initial wager on each losing streak. But when you finally hit a losing streak you can not maintain with your bankroll you lose your initial wager an exponential amount of times. Your winnings need to also increase exponentially to make up for such large losses and in this way multiple losing streaks will actually earn you more than what you lose. Do the math, it doesn't lie. Don't be fooled that nothing will work. The casinos know this very well so they have to institute an effective limit to minimise this while still maintaining one large enough to accomodate a wide enough range of low to high stakes gamers. --41.26.3.13 (talk) 03:21, 12 May 2009 (UTC)
 * Please keep this nonsense out of here. There are plenty of gambling forums where you can post phony gambling systems.Objective3000 (talk) 11:54, 12 May 2009 (UTC)
 * Please read the question again. He never asked for a guarantee but for the possibility of a reward. The universe is not the static place it was once thought to be. Recent discoveries have led many to believe there's always some order within and some have even gone as far as to mention that every atom may be governed by one universal mathematic formula. This is not the place to discuss POV arguments of what's phony and what is not. Nobody ever gave anybody a guarantee but I do think my last sentence summed it up nicely, if it wasn't possible there would be no need for casinos to have such overly restrictive betting ranges. That is all I'm going to say to someone not interested in learning something new, it is certainly no hair of my back what you believe, I am comfortable where I am and I guess you must also be comfortable where you are. For anybody who is interested in learning they should note that all computer generated random numbers are by nature not truely random and only pseudo random by a varying degree. --41.26.3.13 (talk) 15:26, 12 May 2009 (UTC) —Preceding unsigned comment added by 41.27.127.26 (talk)
 * Table limits have nothing to do with progression systems. There are many reasons that casinos assign different limit ranges to different tables:
 * •	Chip trays have a fixed amount of space. It makes no sense to populate every tray with every size chip.
 * •	High limit players often do not like to play with low limit players and low limit players often do not like to play with high limit players. So, they are segregated to a degree.
 * •	The high limit tables are watched more carefully.
 * •	High limit tables are less likely to be manned by less experienced dealers.
 * •	Winning hand payoffs are slower if players are betting many different chip sizes.
 * •	High limit tables are kept less crowded to provide more comfort to higher limit players and to speed the play of larger bettors.
 * •	The limits also prevent ridiculously large bet changes, which could be of value to an advantage player.
 * If you wish to discuss gambling systems, there are many forums available.Objective3000 (talk) 15:47, 12 May 2009 (UTC)

the article doesn't state that if you have an INFINITE bankroll you can guarantee yourself money AND keep your bankroll!
Let's say you have an infinite bankroll and want to guarantee yourself $5,000,000,000. That's five billion dollars.

You can follow these simple steps
 * 1) walk out of the casino into a bank
 * 2) open a bank account.  In the step below I propose a method for getting $5 billion into that checking account (if you have an infinite bankroll), WITHOUT affecting your infinite bankroll!
 * 3) write yourself a check for $5b
 * 4) deposit it into the new account
 * 5) The bank will deduct the five billino from your bankroll, which will continue to have infinite money in it.
 * 6) Therefore, you have just gotten a guaranteed five billion dollars without affecting your bankroll!!! (if you have an infinite bankroll).

Can anyone see any problems with this theory???

 * You're right, but there's nothing special about this betting strategy that allows this. Simply betting a dollar on a (fair) coin flip enough times will ensure that you will eventually win some fixed arbitrary amount. This phenomenon is called recursion, and is fairly common. 142.1.133.165 (talk) 16:07, 13 March 2013 (UTC)

 * Try it and come back and tell us how well it worked.Objective3000 (talk) 00:40, 9 February 2009 (UTC)

 * they said I need two proofs of ID which I didn't have :( —Preceding unsigned comment added by 82.120.236.246 (talk) 11:10, 9 February 2009 (UTC)

I was thinking about the same thing; I think that if one did indeed have infinite available cash and no table cap you could always be 'up' if following a martingale strategy. You would then just have to stop at an arbitrary point, say when you are in profit by $1,000,000 or something - and this could take a very very long time. However I can't prove that this is true mathematically, is anyone here an expert who can tell me if I'm wrong?144.173.5.196 (talk) 13:14, 11 November 2009 (UTC)

 * You're wrong.Objective3000 (talk) 17:50, 11 November 2009 (UTC)

 * Objective3000 is right, you're wrong. Assuming an infinite number of players, several of them will have losing streaks longer than their life spans.  This makes up for the winning players and brings the average win/loss exactly inline with the house edge of the bet in question. AddBlue (talk) 07:24, 20 February 2012 (UTC)

Mathematical question
Could someone explain me how to get the value for probability of 6 concesutive losses within e.g. 60, 152, 250 spins (like it is in section Alternate Mathematical analysis of a single round)? If you were able to give me some general formula for it I would be very thankful. Or just link me some site with explanation how to count it. I found it hard to deduce some formula on my own. Thank you very much.--Bab dz (talk) 20:14, 5 May 2009 (UTC)

Smaller bets = much higher win chance?
I acctually thought of this theory without any help when i was 12 years old, was planning on trying it out today then looked it up and it seems to be v well known.

Anyway, why is everyone using examples of loosing 6 times in a row. taking up the example of having $6,300 why not bet smaller stakes, say, starting at $10 then you have a much greater chance to win? (10 / 20 / 40 / 80 / 160 / 320 / 640 / 1280 / 2560/ 5120) - 10 losses in a row to loose! Now i havn't thought about this alot but the only reason i can think for not doing this is you will be winning tiny stakes :S

Very much appreciating replies ^^ —Preceding unsigned comment added by 212.239.170.3 (talk) 18:06, 20 May 2009 (UTC)
 * What's the use of a great chance of winning, if you just win tiny amounts, along with staggering losses when you eventually lose? --76.173.203.58 (talk) 04:34, 22 May 2009 (UTC)

No: With lots of small bets, you will over time approach closer and closer to an outcome reflecting the real odds (which of course are against you). Conversely, if you make one really big bet (red/black for the sake of argument), your chances of winning are APPROXIMATELY even, which is important because it is an up or down result (100% or 0%). It's still stupid to bet against the house, of course, but the odds do not become so decisive (to the house's advantage, of course) until you make lots of bets.83.5.167.4 (talk) 09:13, 3 July 2009 (UTC)

Unlimited bankroll
The system not only requires the player to have an unlimited bankroll, it also requires the casino to have unlimited solvency so it can keep paying off possible wins as the stakes increase.83.5.167.4 (talk) 09:26, 3 July 2009 (UTC)

Flawed "Intuitive analysis"
The article currently has a section called "Intuitive analysis", which says the following: "Since expectation is linear, the expected value of a series of bets is just the sum of the expected value of each bet. Since in such games of chance the bets are independent, the expectation of all bets is going to be the same, regardless of whether you previously won or lost. In most casino games, the expected value of any individual bet is negative, so the sum of lots of negative numbers is also always going to be negative."

This reasoning, "intuitive" though it might be, is actually incorrect unless the stopping time has finite expectation. I removed this with a reason in the edit summary, but User:Objective3000 undid it without one ("revert"). Oh well. Here's a more detailed explanation. Although when $$\operatorname{E}\left(X\right) \le 0$$ it is true that $$\operatorname{E}\left(\sum_{i=1}^n{X_i}\right) \le 0$$ for any fixed n, it is not true that $$\operatorname{E}\left(\sum_{i=1}^T{X_i}\right) \le 0$$ for an arbitrary stopping time $$T$$ (indeed, that was the entire reasoning behind the martingale betting system). For that to be true (e.g. Wald's equation) you need $$\operatorname{E}(T)$$ to be finite. That is why we have the conditions in the optional stopping theorems — we need a finite lifetime and a limit on bets. Let's remove this misleading reasoning, please. Shreevatsa (talk) 14:50, 16 November 2009 (UTC)

 * The stopping time DOES have a bounded expected value, with the natural assumption that the outcomes are iid. In this case, the stopping time is a geometric random variable, which has a mean. I'm going to change the intuitive description to account for this. 142.1.133.165 (talk) 16:12, 13 March 2013 (UTC)

 * So, you are saying that the analysis is flawed because it fails in a situation that is imposssible.:) We DO have a finite lifetime.Objective3000 (talk) 18:40, 16 November 2009 (UTC)

 * Yes. The conclusion is correct (with a finite1 lifetime, no betting strategy can work), but the reasoning given to arrive at it is utterly wrong. It should be removed. Shreevatsa (talk) 19:02, 16 November 2009 (UTC)
 * 1. Just a clarification: the lifetime in any play is finite (almost surely); by "finite lifetime" above I mean either "bounded lifetime", or "lifetime whose expected value is finite, and with limits on the bets". Shreevatsa (talk) 19:19, 16 November 2009 (UTC)

 * Sorry, but it seems to me that you have not shown that the reasoning is flawed. You have only shown that it would be flawed in circumstances that do not exist. That is, circumstances irrelevant to the article. Many mathemetic proofs break down when you bring infinity into the picture. But, infinity is not in this picture.Objective3000 (talk) 20:23, 16 November 2009 (UTC)

 * Ok, since it seems you're not carefully reading what I wrote, and to avoid going in circles, let's take it step-by-step: "Since expectation is linear, the expected value of a series of bets is just the sum of the expected value of each bet." Can you show this is true, even for a number of bets that is not fixed? Shreevatsa (talk) 20:28, 16 November 2009 (UTC)

 * First, if you're going to be nasty, you can talk to yourself. I carefully read and responded to what you wrote. I did not respond to what you did not write. Second, you are the one claiming the article is flawed. The proof you provided was not relevant to the article. A reason should be given for removal. I'll let someone else respond to your next response.Objective3000 (talk) 21:22, 16 November 2009 (UTC)

 * I'm sorry that came off as nasty. The clarification earlier was specifically intended to address "bringing infinity into the picture" — the number of plays can be always finite, and yet have infinite expectation (this is exactly what happens with the most common form of this betting strategy). The very first statement of the paragraph is wrong (and the comments I made were meant to demonstrate this; I realise now that they weren't successful). The fact that a proof is wrong (e.g. there's no source backing it up) is certainly sufficient reason for removal; the burden of proof is always on the claims being made! Anyway, how about the following? I'll rewrite that section, explaining the mathematics as clearly as I can, and then we can discuss before blindly reverting. Shreevatsa (talk) 23:00, 16 November 2009 (UTC)

 * Thank you. I didn't write the original text. I think the general concept here (on WP) is to accept original text that has stood for a time without a good reason to revert. I also think we would welcome a rewrite. But, within the constraints of a pure betting strategy (part of the definition of Martingale precludes infinity), it will be difficult to disprove the statement. I don't see your problem with the statement within the constraints of the article. There is no question that stopping points can affect strategies that take into affect actual changes in odds (often to the negative). Do you believe that stopping points can affect pure progression systems that ignore changing odds according to changing conditions (e.g. changing deck contents) in a finite world?Objective3000 (talk) 01:11, 17 November 2009 (UTC)

 * Yes! That's one of the surprising counter-intuitive things I learned. (I'm referring solely to "the expected value of a series of bets is just the sum of the expected value of each bet" as a statement.) I'll try to write this up (or find a source where it's written up) more clearly soon. In the meantime, to see where the reasonable-seeming argument goes wrong, here is a similar question: suppose $$E(X_1)$$ and $$E(X_2)$$ (the returns from two bets you placed, say) are both at most $$1$$. Is it true that $$E(\max(X_1, X_1+X_2))$$ (the expected value of the highest point you reached) is at most 2? :-) (A few days ago, I had almost convinced myself it was true, but one can construct simple examples where it's not.) Shreevatsa (talk) 02:32, 17 November 2009 (UTC)

 * Sorry, but stopping points have no effect on EV. As the text states, the expected value of a series of bets is just the sum of the expected value of each bet.Objective3000 (talk) 12:14, 17 November 2009 (UTC)

 * No. :-) That's the reasonable-at-first-glance naive argument, but the martingale betting system is a counterexample. The point is that the "series of bets" is not fixed, so in fact the statement doesn't even make sense, let alone be incorrect. Either formally write down what you mean so we can figure out where the communication gap is, or find a source that says so. Shreevatsa (talk) 15:19, 17 November 2009 (UTC)

 * You can call me naive all you wish. But what you are claiming has been debunked time and again. I suggest you post to an advantage player forum.Objective3000 (talk) 15:38, 17 November 2009 (UTC)

 * I'm sorry; I have no idea what that is. (I understand "forum", and from context it seems to be something related to gambling, which does not interest me.) My interest is only in fixing the incorrect mathematics. I find it hard to believe how my "claim" (which is merely pointing out at a sentence does not make much sense) has been debunked anywhere; perhaps you are thinking of something else? Here's a very simple concrete example of what I meant: Let $$X_1$$ (the result of the first bet) be $$-3$$ with probability $$1/2$$ and $$+1$$ with probability $$1/2$$, so that $$\operatorname{E}(X_1) = -1$$, and let $$X_2$$ (the result of the second bet, if any) be an independent identically distributed random variable. Now let $$T$$ (the stopping time) be as follows: you stop after the first bet if $$X_1$$ was positive, else you stop after the second. Then yes, $$\operatorname{E}(X_1) = -1$$ and $$\operatorname{E}(X_1+X_2) = -2$$, but $$\operatorname{E}(\sum_{i=1}^{T}X_i) = (1/2)1 + (1/4)(-2) + (1/4)(-6) = -3/2$$. So the "expected value of the series of bets" is -3/2, which is neither the expected value of one bet, nor the expected value of two bets. The sentence does not make any sense! (It works only if you separately condition on possible values $$T$$ can take, but that is not talking of the "expected value of the series of bets" any more). More generally, consider stopping times like "the tenth time the coin lands heads" or "the first time there are six heads in a row", or with the $$X_i$$'s not IID. What does the statement mean now? Shreevatsa (talk) 18:44, 17 November 2009 (UTC)

 * I don't know where you got your second forumla, but it's incorrect. It looks like you are assuming you will win the third play if you lose the first two. The expected value is the number of plays times -1.Objective3000 (talk) 19:19, 17 November 2009 (UTC)

 * There's no third play, at most two. Here's the explanation: with probability 1/2, $$X_1 = 1$$ and you stop ($$T=1$$). This is the first term in the formula. With the remaining probability (1/2), $$X_1 = -3$$, and you bet again ($$T=2$$). Now there are two equally probable possibilities, both with probability 1/2 of this current probability (so, 1/4): either $$X_2 = 1$$, in which case the return is $$-3 + 1 = -2$$ (the second term), or $$X_2 = -3$$, in which case the return is $$-3 - 3 = -6$$ (the third term). So the expected value from this betting "strategy" is -3/2. The broader point is that a statement like "The expected value is the number of plays times -1" does not make sense when the number of plays is not fixed (here, it can be either 1 or 2, and for the other stopping times mentioned in the previous comment, it may be any number). Shreevatsa (talk) 22:08, 17 November 2009 (UTC)

 * Expected value is based upon the original bets. If you always bet twice, there are two bets. -2/2=-1. If you use your stopping example, you bet on average 1.5 times. -1.5/1.5=-1. In either case, the EV is -1.Objective3000 (talk) 22:24, 17 November 2009 (UTC)

 * Great! Finally we have reached the content of my very first comment. :-) In saying that if you "bet on average 1.5 times", then the expected reward is -1.5, you are essentially using Wald's equation. This equation is not valid when the "average" number of times you bet is not finite: for example, let $$c=6/\pi^2$$, and suppose that with probability $$c$$ you bet just once, with probability $$c/4$$ you bet twice, with probability $$c/9$$ you bet thrice, and so on (with probability $$c/n^2$$ you bet $$n$$ times). Then, you always bet a finite number of times, but the "average" number of times you bet is not finite, so the equation is not valid. This is precisely the flaw in the betting strategy described here, but instead of explaining the actual reason it has an "intuitive analysis" that is wrong. Shreevatsa (talk) 22:49, 17 November 2009 (UTC)

 * There is an intuitive analysis AND a math analysis. They are both useful as this is an encyclopedia and not a math text.Objective3000 (talk) 22:59, 17 November 2009 (UTC)

 * But the "intuitive" analysis is in fact wrong, which is why it does not belong on an enclycopedia (at least in its current form). I do not begrudge you your statements like "stopping points have no effect on EV" and "don't know where you got your second forumla, but it's incorrect", but I think we can agree it has been a waste of time for both of us. To make a constructive suggestion: given the original incident of being reverted without an explanation, I am reluctant to waste effort, but if I am allowed to rewrite the section, I could retain the current "intuitive" claims and re-work them into some form that is actually correct. How about that? Shreevatsa (talk) 23:34, 17 November 2009 (UTC)

 * Anyone is welcome to try to rewrite anything here. But, the current statement is valid in my mind and, in fact, a long-held truism. I would view its deletion as a negative. Martingale theories have existed for centuries. To look at another encyclopedia, in one of its rare moments of levity, the Enc. Britannica says that every day someone reinvents Martingale. I think that's why it is important to provide an intuitive analysis as well as a math analysis. I also stand by my statement that stopping points have no effect on EV, as has been proved many times.

 * I would also add that we are all volunteers. Please review the manner in which you engage other volunteers. And please excuse me for saying that.Objective3000 (talk) 00:49, 18 November 2009 (UTC)

 * Hello, I had to take a long wikibreak without finishing this conversation. Indeed we are all volunteers, and I appreciate your efforts at Wikipedia. Your last paragraph suggests that I hurt your feelings in some way, which is unfortunate. Doubtless some of it is simply a communication gap (for example, I called an argument naïve, which is common in mathematics and not an insult (e.g. naive set theory and naive algorithm), but you seemed to take it as a personal remark). Anyway, apologies.

 * Returning to the mathematics, I wish to reassert that the intuitive explanation on this page is, in the absence of further qualification, wrong. Otherwise, it would not have taken until the 1960s to prove this theorem (by Doob), which as you observed, had been of interest for centuries. Specifically, the "intuitive argument", if true, should apply to any distribution. But taking a particular distribution in which the stopping time happens to have infinite expectation, it actually fails: the martingale strategy can win. (Of course, this case requires unbounded wealth as well, so it is not possible in the real world.) The correct proof, therefore, cannot arise from carrying the "intuitive argument" through, but on somehow modelling mathematically the constraint of finiteness that exists in the real world. (Hence the conditions in the optional stopping theorem.)
 * As Lenstra said, "The problem with wrong proofs to correct statements is that it is hard to give a counterexample." :-) That's what is happening here. The statement is correct (the martingale strategy does not work in the real world) but the purported proof, seductive though it is, is not strictly correct. I have tried to the best of my ability to explain why the intuitive explanation is wrong (it leads to conclusions that are not valid without further assumptions, realistic though they are), but if we still disagree, I don't know what to do. Shreevatsa (talk) 18:45, 7 January 2010 (UTC)

 * Update: I made an edit which I hope should be satisfactory (except for language issues: feel free to rewrite). I tried to clarify that the "intuitive analysis" is only valid under a certain assumption (valid in the real world) and that the theorem also holds under different (possibly weaker) conditions in which the "intuitive analysis" may not be valid. Shreevatsa (talk) 18:59, 7 January 2010 (UTC)
 * No you did not "hurt my feelings" and that is a ridiculous statement. The article is not wrong in the real world. We live in the real world. If your problem with the article is that it is not true in a non-real world, well EVERYTHING can be true and false in non-real worlds. Should we specify in every article that it is only true in the real world?Objective3000 (talk) 19:31, 7 January 2010 (UTC)

Odds for extended streaks in Martingale
so the odds for a 10x streak of coin flips would be 1023 to 1 against, I believe but with ENOUGH flips the odds of having ten consecutive heads are even, and at some yet larger number of flips the odds of NOT having ten consecutive heads is 1023 to 1 against. am i thinkin about this the right way? is that why cant think of it as not playing the static 50/50 odds but playing the odds for streaks.. because the odds are dynamic based on number of case sets? id love any clarification. —Preceding unsigned comment added by 98.212.231.88 (talk) 16:07, 23 February 2010 (UTC)

Some more mathematical material
I have just added some new text, probably too wordy, under "mathematical analysis". It's my first logged-in Wikipedia edit, and a bit of an experiment to see if I can do it right. I think there is some duplication of material already present in the article, but I preferred not to change anything written by others at my current level of experience.

My goal was to provide a more mathematical discussion of the "certain to win eventually" property at a reasonably elementary level, and to show its inapplicability to the real world in a different light than just negative expectation under bounds on time or money (which is also true, of course). —Preceding unsigned comment added by Styrofoams (talk • contribs) 02:17, 19 March 2010 (UTC)

з
if p=1/k and the bets increase a(n+1)=(k+1)*a(n), then the win with p=1/k will lead to getting 1 initial bet. Станислав Крымский (talk) 16:32, 11 July 2012 (UTC)

expected value of the stopped martingale
In the introduction of the article it says the gambler's expected value does indeed remain zero.... But I think the expected value of the stopped martingale (the martingale stopped at the stopping time defining the martingale strategy) is not zero but one. My source for this is http://mitschriebwiki.nomeata.de/Stochastik2.pdf p.77.

Example of a "generalisation"
Arbitrarily large, not infinite wealth: w Probability of winning: p=4/5 Return on winning: r=3/4 Multiplication factor: f=999.....99999 First bet: x[0]=1; Next bet: x[n]=p*x[n-1]

For a general formula: Given w[0], p[n], r[n] and r[n]<p[n] one may use Numerical methods to compute f[n] and x[n] along with w[n]; risk may also arise from the error resulted from the approximation discrepancy between rational and non rational, but real numbers.

Someone might be interested in correcting what appears on the Roulette article. 79.117.44.240 (talk) 16:17, 10 December 2013 (UTC)

Alternative mathematical analysis
I think the math in that section is incorrect, indeed, betting for one colour (either red or black) gives you a 47.37% (18/38) chance of winning (or 52.63% of losing depending on how you see it), however the writer of that section exponentially expands it to 6 saying that the chances of losing 6 spins in a row increase by 2.12% in the first 6 spins and subsequently to 50.3% after 73 spins and then to 91.1% after 250 spins.

I myself have tried spinning the roulette 250 times and more, and if those calculations stated above were true, I would have a 91.1% chance of losing 6 times in a row, which didn't happen, at most I lost 5 spins in a row a number of times but never reached the 6 fails mark, and even if it did happen if would be merely coincidental.

Besides if those calculations were correct, and the chances of losing 6 times in a row increased by the number of spins I play, what would happen if I stopped every once in a while and started from 0 all over again? my chances of winning would come back to the original percentage? I don't think so, the chances of you losing 6 times in a row are exactly the same in the first spins as they are 500 spins later, that is 52.63%. Supaman89 (talk) 21:33, 5 May 2014 (UTC)

Diagnostics
For something to have advantage, there must be risks. Martingale works. There is a horror story, then you must recoup your losses. The zero, very deadly. Also, when you play the casino, expect there to be a straight that WILL wipe you out. Out of the bracket, a quick buck. — Preceding unsigned comment added by 23.117.16.45 (talk) 13:20, 8 December 2015 (UTC)

External links modified
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 * Added archive http://web.archive.org/web/20151013024650/https://books.google.com/books?id=0bAYl6d7hvkC&pg=PA298 to https://books.google.com/books?id=0bAYl6d7hvkC&pg=PA298

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The neglected scope
The strategy does work... in a perfect casino only.

Why it failed to work in real casinos?

For the probabilities of big games are always deliberately manipulated using out-of-the-system methods.

And the game tends to eventually end up in a shooting game between the player (and/or whose representatives) and the casino's personnel, which would very likely end up very violently...

— 154.48.253.21 (talk) 11:32, 11 February 2020 (UTC)
 * Sorry, but this is absurd on multiple levels. Most importantly, martingale does not work as explained by the article and this has nothing to do with any casino manipulation. Further, the accusation of violence is ludicrous. Please use reliable sources. O3000 (talk) 15:01, 15 February 2020 (UTC)

The main reason it would never work is certainly because of the big game manipulation.

Accusation?.. No, no, no. That's indeed a joke.

Imagine what would happen, when neither side can afford losing the game.

— Wikipedian Right (talk) 13:06, 16 February 2020 (UTC)
 * No, game manipulation has absolutely nothing to do with why Martingale doesn't work. O3000 (talk) 13:12, 16 February 2020 (UTC)

I think we might have defined "work" differently.

For whoever won the prize (to the specific extent), even in a negative expectation game, it does count as "work".

No gambling method would work in the long term, for the gambler's ruin is inevitable.

— Wikipedian Right (talk) 14:53, 16 February 2020 (UTC)
 * No gambling method would work in the long term, for the gambler's ruin is inevitable. Not true, because the gambling method could preclude the tenets of gambler's ruin. An adequately bankrolled, fractional Kelly better in a positive expectation game is not subject to ruin. And please stop using tq2 and read WP:INDENT. Your threads violate our procedures. O3000 (talk) 15:12, 16 February 2020 (UTC)
 * Wikipedia is not a place to publish these rants. Please publish in a paper (Reliable Source) then it might be cited here. Limit-theorem (talk) 15:00, 16 February 2020 (UTC)
 * Agree. O3000 (talk) 15:12, 16 February 2020 (UTC)

Seemingly plausible...

But is the 0 probability of bankruptcy ensured this way?

Unless you can convince the general public into believing that the text walls are more readable... I fear I wouldn't change my approach.

Regardless, the whole talk page thing is already a mess. Wouldn't really matter if it just "worsen" a bit.

But isn't the advancement of mankind sort of contributed by "rants" like these?

Also check: On the General Dogmatism on Wikipedia

— Wikipedian Right (talk) 17:28, 16 February 2020 (UTC)
 * Your edits are unreadable and you are clearly WP:NOTHERE. Don't expect further responses and don't ping me. O3000 (talk) 17:41, 16 February 2020 (UTC)

title: Talk:Martingale (betting system)


text:
Sulaiman Abu Ghaith

Sulaiman Jassem Sulaiman Ali Abu Ghaith (سليمان جاسم سليمان علي أبو غيث; born 14 December 1965) is a Kuwaiti regarded as one of al-Qaeda's spokesmen. He is married to one of Osama bin Laden's daughters. In 2013, Gaith was arrested in Jordan and extradited to the United States. In 2014, he was convicted in a U.S. federal court in New York for "conspiring to kill Americans and providing material support to terrorists" and sentenced to life imprisonment He is serving his sentence at the federal ADX Florence prison in Colorado.

Activities during the 1991 Gulf War
Abu Ghaith, who grew up with the Muslim Brotherhood, first gained attention during the 1990–1991 Iraqi invasion and occupation of Kuwait. His sermons denouncing the occupation and Iraqi President Saddam Hussein gained him some degree of popularity amongst the Kuwaiti people. In 1992, he went to Bosnia and Herzegovina for nearly a month to do some "relief services" there. He later joined Muslim guerillas in the Bosnian War in summer 1994. The Kuwaiti government subsequently removed him from the mosque and banned him from giving sermons, as he used to criticize the Kuwaiti government and other Arab governments stridently, then he became a high school teacher of religion.

Arrival in Afghanistan in June 2000
In June 2000, he left Kuwait for Afghanistan, where he met Osama bin Laden and joined his al-Qaeda organization. His affinity for public speaking and comparative youth put him at the head of al-Qaeda's attempt to widen its appeal from ultra-conservative and mostly elderly clerics to the general population and especially the youth of majority-Muslim countries; in this capacity, he quickly became the organization's spokesman.

Al Wafa
According to documents in the unclassified dossier from Adel al Zamel's Combatant Status Review Tribunal Sulaiman Abu Ghaith was also a founder of Al Wafa al Igatha al Islamia, a charity the USA asserts provided a plausible front for al Qaeda's fund-raising efforts. One of the allegations against Al Zamil, who was also accused of being a founder of al Wafa, was that he helped Abu Ghaith's family leave Afghanistan around the time of the attacks of 9–11.

Al Qaeda video after 9/11
He rose to worldwide attention following the 11 September 2001, attacks. On 10 October 2001, he appeared on two widely circulated videos (first broadcast on al Jazeera television) to defend the attacks and threaten reprisals for the subsequent US invasion of Afghanistan, saying, "Americans should know, the storm of the planes will not stop... Thousands of the Islamic nation's youths are eager to die just as the Americans are eager to live." These statements caused the Kuwaiti government to strip him of his citizenship.

In 2002, while living in Iran, he posted a statement that al Qaeda has "the right to kill four million Americans, including one million children, displace double that figure, and injure and cripple hundreds and thousands."

Alleged connection to the Faylaka Island attackers
According to The Long War Journal American officials assert that Sulaiman Abu Ghaith attended al Qaeda's airport training camp with Anas al Kandari and Faiz al Kandari. Anas al Kandari was a young Kuwaiti who fired upon a squad of marines, killing one, in the Faylaka Island attack in 2002. Faiz al Kandari is another Kuwaiti, who was held in extrajudicial detention in Guantanamo from 2002 to 2008. In 2008, charges were prepared against him to be referred to a Guantanamo military commission. According to The Long War Journal in his book The Martyr's Oath, Stewart Bell asserted Sulaiman Abu Ghaith recruited Anas al Kandari and the other shooter to launch the Faylaka Island attacks.

Presence in Iran 2002–2013
As he moved around to escape capture by the United States in the following months, his whereabouts are unclear. According to the Long War Journal, by 2002, Sulaiman was living in Iran.

In July 2003, a Kuwaiti minister announced that the Iranian government was holding Abu Ghaith and that Kuwait had refused an offer from Iran to extradite him to Kuwait.

In September 2010, the Long War Journal falsely reported that Abu Ghaith had been freed by Iran and had left the country for Afghanistan.

In March 2013, it was reported that Abu Ghaith had spent most of the last ten years in Iran, under house arrest.

Presence in Turkey 2013
In late January 2013, Abu Ghaith entered Turkey from Iran, staying in a hotel in Ankara. For a brief period, he was detained at the request of the United States but was released since he had committed no crime in Turkey. Turkish authorities held him as "a guest" since he did not have a passport at the time. Rather than extradite him to the United States, the Turkish authorities decided to deport him to his home country, Kuwait.

Arrest, extradition to the United States, and trial
On a stopover in Amman, Jordan, Abu Ghaith was arrested by Jordanian officials and turned over to US authorities on 7 March 2013. He was subsequently extradited to the United States and housed in a federal prison in New York.

Abu Ghaith was indicted on charges of conspiring to kill Americans and tried in the Federal District Court in Manhattan (U.S. v. Abu Ghayth, U.S. District Court, Southern District of New York, No. 98-cr-01023). He pleaded not guilty on 8 March 2013.

On 8 April 2013, Abu Ghaith's attorneys were considering a request for change of venue, since New York City received the greatest loss from the attacks on 11 September 2001. Abu Ghaith's lawyers sought the right to call upon Khalid Sheikh Mohammed as a witness. On 18 March 2014, this request was denied by a New York federal judge.

On 26 March 2014, Abu Ghaith was convicted of "conspiring to kill Americans and providing material support to terrorists" and he was subsequently sentenced to life in prison by U.S. District Judge Lewis A. Kaplan.

On 28 September 2017, the 2nd U.S. Circuit Court of Appeals in Manhattan upheld his conviction.

Personal life
He married his first wife, Fatima, in Kuwait, with whom he had six daughters and a son. He then married an Egyptian woman named Amal, who was detained with him in Iran, with whom he had two daughters; then he married Osama bin Laden's daughter, Fatima, with whom he had a daughter and a son.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Sulaiman Abu Ghaith


text:
DNA methyltransferase

In biochemistry, the DNA methyltransferase (DNA MTase, DNMT) family of enzymes catalyze the transfer of a methyl group to DNA. DNA methylation serves a wide variety of biological functions. All the known DNA methyltransferases use S-adenosyl methionine (SAM) as the methyl donor.

Substrate
MTases can be divided into three different groups on the basis of the chemical reactions they catalyze:
 * m6A - those that generate N6-methyladenine
 * m4C - those that generate N4-methylcytosine
 * m5C - those that generate C5-methylcytosine

m6A and m4C methyltransferases are found primarily in prokaryotes (although recent evidence has suggested that m6A is abundant in eukaryotes ). m5C methyltransferases are found in some lower eukaryotes, in most higher plants, and in animals beginning with the echinoderms.

The m6A methyltransferases (N-6 adenine-specific DNA methylase) (A-Mtase) are enzymes that specifically methylate the amino group at the C-6 position of adenines in DNA. They are found in the three existing types of bacterial restriction-modification systems (in type I system the A-Mtase is the product of the hsdM gene, and in type III it is the product of the mod gene). These enzymes are responsible for the methylation of specific DNA sequences in order to prevent the host from digesting its own genome via its restriction enzymes. These methylases have the same sequence specificity as their corresponding restriction enzymes. These enzymes contain a conserved motif Asp/Asn-Pro-Pro-Tyr/Phe in their N-terminal section, this conserved region could be involved in substrate binding or in the catalytic activity. The structure of N6-MTase TaqI (M.TaqI) has been resolved to 2.4 A. The molecule folds into 2 domains, an N-terminal catalytic domain, which contains the catalytic and cofactor binding sites, and comprises a central 9-stranded beta-sheet, surrounded by 5 helices; and a C-terminal DNA recognition domain, which is formed by 4 small beta-sheets and 8 alpha-helices. The N- and C-terminal domains form a cleft that accommodates the DNA substrate. A classification of N-MTases has been proposed, based on conserved motif (CM) arrangements. According to this classification, N6-MTases that have a DPPY motif (CM II) occurring after the FxGxG motif (CM I) are designated D12 class N6-adenine MTases. The type I restriction and modification system is composed of three polypeptides R, M and S. The M (hsdM) and S subunits together form a methyltransferase that methylates two adenine residues in complementary strands of a bipartite DNA recognition sequence. In the presence of the R subunit, the complex can also act as an endonuclease, binding to the same target sequence but cutting the DNA some distance from this site. Whether the DNA is cut or modified depends on the methylation state of the target sequence. When the target site is unmodified, the DNA is cut. When the target site is hemimethylated, the complex acts as a maintenance methyltransferase, modifying the DNA so that both strands become methylated. hsdM contains an alpha-helical domain at the N-terminus, the HsdM N-terminal domain.

Among the m6A methyltransferases (N-6 adenine-specific DNA methylase) there is a group of orphan MTases that do not participate in the bacterial restriction/methylation system. These enzymes have a regulatory role in gene expression and cell cycle regulation. EcoDam from E. coli and CcrM from Caulobacter crescentus are well characterized members of these family. More recently, CamA from Clostridioides difficile, was shown to play key functional roles in sporulation, biofilm formations and host-adaptation.

m4C methyltransferases (N-4 cytosine-specific DNA methylases) are enzymes that specifically methylate the amino group at the C-4 position of cytosines in DNA. Such enzymes are found as components of type II restriction-modification systems in prokaryotes. Such enzymes recognise a specific sequence in DNA and methylate a cytosine in that sequence. By this action they protect DNA from cleavage by type II restriction enzymes that recognise the same sequence

m5C methyltransferases (C-5 cytosine-specific DNA methylase) (C5 Mtase) are enzymes that specifically methylate the C-5 carbon of cytosines in DNA to produce C5-methylcytosine. In mammalian cells, cytosine-specific methyltransferases methylate certain CpG sequences, which are believed to modulate gene expression and cell differentiation. In bacteria, these enzymes are a component of restriction-modification systems and serve as valuable tools for the manipulation of DNA. The structure of HhaI methyltransferase (M.HhaI) has been resolved to 2.5 A: the molecule folds into 2 domains - a larger catalytic domain containing catalytic and cofactor binding sites, and a smaller DNA recognition domain.

Highly conserved DNA methyltransferases of the m4C, m5C, and m6A types have been reported, which appear as promising targets for the development of novel epigenetic inhibitors to fight bacterial virulence, antibiotic resistance, among other biomedical applications.

De novo vs. maintenance
De novo methyltransferases recognize something in the DNA that allows them to newly methylate cytosines. These are expressed mainly in early embryo development and they set up the pattern of methylation. De novo methyltransferases are also active when a signal-responsive cell, such as a neuron, needs to alter protein expression. As an example, when fear conditioning creates a new memory in a rat, 9.17% of the genes in the rat hippocampus neuron genome are differentially methylated.

Maintenance methyltransferases add methylation to DNA when one strand is already methylated. These work throughout the life of the organism to maintain the methylation pattern that had been established by the de novo methyltransferases.

Mammalian
At least four differently active DNA methyltransferases have been identified in mammals. They are named DNMT1, two isoforms transcribed from the DNMT3a gene: DNMT3a1 and DNMT3a2, and DNMT3b. Recently, another enzyme DNMT3c has been discovered specifically expressed in the male germline in the mouse.

Manzo et al. observed differences in genomic binding of DNMT3a1, DNMT3a2 and DNMT3b. They found 3,970 regions exclusively enriched for DNMT3a1, 3,838 exclusively enriched for DNMT3a2 and 3,432 exclusively enriched for DNMT3b.

The DNMT enzymes are not only regulated in their methylating locations on the genome by where they bind to DNA, but they are also regulated by the post-translational modifications on the histone proteins of the nucleosomes around which the genomic DNA is wrapped (see Figures). Rose and Klose reviewed the relationship between DNA methylation and histone lysine methylation. For example, they indicated that H3K4me3 appears to block DNA methylation while H3K9me3 plays a role in promoting DNA methylation.

DNMT3L is a protein closely related to DNMT3a and DNMT3b in structure and critical for DNA methylation, but appears to be inactive on its own.

DNMT1
DNMT1 is the most abundant DNA methyltransferase in mammalian cells, and considered to be the key maintenance methyltransferase in mammals. It predominantly methylates hemimethylated CpG di-nucleotides in the mammalian genome. The recognition motif for the human enzyme involves only three of the bases in the CpG dinucleotide pair: a C on one strand and CpG on the other. This relaxed substrate specificity requirement allows it to methylate unusual structures like DNA slippage intermediates at de novo rates that equal its maintenance rate. Like other DNA cytosine-5 methyltransferases the human enzyme recognizes flipped out cytosines in double stranded DNA and operates by the nucleophilic attack mechanism. In human cancer cells DNMT1 is responsible for both de novo and maintenance methylation of tumor suppressor genes. The enzyme is about 1,620 amino acids long. The first 1,100 amino acids constitute the regulatory domain of the enzyme, and the remaining residues constitute the catalytic domain. These are joined by Gly-Lys repeats. Both domains are required for the catalytic function of DNMT1.

DNMT1 has several isoforms, the somatic DNMT1, a splice variant (DNMT1b) and an oocyte-specific isoform (DNMT1o). DNMT1o is synthesized and stored in the cytoplasm of the oocyte and translocated to the cell nucleus during early embryonic development, while the somatic DNMT1 is always found in the nucleus of somatic tissue.

DNMT1 null mutant embryonic stem cells were viable and contained a small percentage of methylated DNA and methyltransferase activity. Mouse embryos homozygous for a deletion in Dnmt1 die at 10–11 days gestation.

TRDMT1
Although this enzyme has strong sequence similarities with 5-methylcytosine methyltransferases of both prokaryotes and eukaryotes, in 2006, the enzyme was shown to methylate position 38 in aspartic acid transfer RNA and does not methylate DNA. The name for this methyltransferase has been changed from DNMT2 to TRDMT1 (tRNA aspartic acid methyltransferase 1) to better reflect its biological function. TRDMT1 is the first RNA cytosine methyltransferase to be identified in human cells.

DNMT3
DNMT3 is a family of DNA methyltransferases that could methylate hemimethylated and unmethylated CpG at the same rate. The architecture of DNMT3 enzymes is similar to that of DNMT1, with a regulatory region attached to a catalytic domain. There are at least five members of the DNMT3 family: DNMT3a1, DNMT3a2, 3b, 3c and 3L.

DNMT3a1, DNMT3a2 and DNMT3b can mediate methylation of CpG sites in gene promoters, resulting in gene repression. These DNA methyltransferases can also methylate CpG sites within the coding regions of genes, where such methylation can increase gene transcription. Work with DNMT3a1 showed it preferentially localized to CpG islands bivalently marked by H3K4me3 (a transcription promoting mark) and H3K27me3 (a transcription repressive mark), coinciding with the promoters of many transcription factors. Work with DNMT3a2, in neurons, found that the DNA methylation changes caused by DNMT3a2 predominantly occur in intergenic and intronic regions. These intergenic and intronic DNA methylations were thought to likely regulate enhancer activity, alternative splicing or the expression of non-coding RNAs.

DNMT3a1 can co-localize with heterochromatin protein (HP1) and methyl-CpG-binding protein (MeCBP), among a number of other factors. They can also interact with DNMT1, which might be a co-operative event during DNA methylation. DNMT3a prefers CpG methylation to CpA, CpT, and CpC methylation, though there appears to be some sequence preference of methylation for DNMT3a and DNMT3b. DNMT3a methylates CpG sites at a rate much slower than DNMT1, but greater than DNMT3b.

The expression of DNMT3a2 differs from DNMT3a1 and DNMT3b because DNMT3a2 expression occurs in the pattern of an immediate early gene. DNMT3a2 is induced to express in neurons, for instance, by new neuronal activity. This may be of importance in establishing long-term memory. In a rat, high levels of new DNA methylations in neurons of the hippocampus occur after a memorable event is imposed on a rat, such as contextual fear conditioning. Bayraktar and Kreutz found that DNMT inhibitors, applied in the brain, prevented long-term memories from forming.

DNMT3L contains DNA methyltransferase motifs and is required for establishing maternal genomic imprints, despite being catalytically inactive. DNMT3L is expressed during gametogenesis when genomic imprinting takes place. The loss of DNMT3L leads to bi-allelic expression of genes normally not expressed by the maternal allele. DNMT3L interacts with DNMT3a and DNMT3b and co-localized in the nucleus. Though DNMT3L appears incapable of methylation, it may participate in transcriptional repression.

DNMT inhibitors
Because of the epigenetic effects of the DNMT family, some DNMT inhibitors are under investigation for treatment of some cancers:
 * Vidaza (azacitidine) in phase III trials for myelodysplastic syndromes and AML
 * Dacogen (decitabine) in phase III trials for AML and CML. EU approved in 2012 for AML.
 * guadecitabine, an experimental drug under development by Astex Pharmaceuticals and Otsuka Pharmaceutical. It failed to meet primary endpoints in a 2018 Phase III AML trial.

title: DNA methyltransferase