Help me convince friend that Martingale Betting Strategy Doesn't work

[Nottom]

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I just started covering martingales in a class I'm teaching and this betting strategy was an example. Also, coincidentally, I was just discussing this with my wife at dinner last night.

I had a friend who was also convinced of this. She has a masters degree in mathematics. I was also unable to convince her it was a bad idea. Strange how stubborn some folks can be.


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I think part of the reason your fried could possibly be confused (if she is used to think abou tmath on a higher level) is because the Matingale system does work given an infinate bankroll. In reality, the Casinos have betting limits and most don't have the bankroll to cover a string of 20 loses even if the casino would let them.

[Bytestream]

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If you don't understand that, then you don't understand how to treat infinities in mathematical models.

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This is not an infinite trial. Each betting cycle has a finite end, that is following a win.

[reubenf]

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This is not an infinite trial. Each betting cycle has a finite end, that is following a win.

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A win is not guaranteed with an infinite bankroll and no table limits. An infinite sequence of losing bets is conceivable. It happens with probability 0. This is not the same as saying it does not happen.

[Bytestream]

Perhaps stating the case differently, you might understand.

Assumptions: There is no table maximum bet limit. There is no constraint on the size of your bankroll. The plan is to start with a $1 bet, and to double the previous bet following a loss. The only time the game will end is after you win a bet, then you immediately quit.

Then, there is a 100% chance (probablitly 1), that when you quit, you are up $1.

I'm sure you don't need the use the first principle of mathematical induction to realize that this will always be the case.

[reubenf]

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Then, there is a 100% chance (probablitly 1), that when you quit, you are up $1.

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This is true. But probability 1 does not mean "guaranteed" or "always happens." It just means if you perform the act sufficiently many times, the percentage of the time it happens will get arbitrarily close to 100%.

This is exactly identical to the chance that a random real number between 0 and 1 is not equal to 0.5. The probability is 1 that the number is not equal to 0.5. But it just might be equal to 0.5 anyway. The probability is 1 that is is not equal to any specific number. But it is equal to -some- specific number.

[Kenrick]

I think we just had this thread. It even had "roulette" in the subject title.

[CMonkey]

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Then, there is a 100% chance (probablitly 1), that when you quit, you are up $1.


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No. There is a 100% chance that IF YOU ARE ABLE TO QUIT, you are up $1. The chance that you will win once and be able to quit is never equal to 100% no matter how many consecutive trials you consider.

If you want to take the limit as your chance of winning at least once goes to 100% over an arbitrarily large number of trials, you must also model the infinite nature of the bet size increases properly. You cannot arbitrarily take the limit of you winning once (which equals 100%) and ignore how the bet sizes are increasing over the same infinity by just saying "Well, no matter what, your bet is always one larger than all your previous losses." EV is (Outcome)*(Probability of that outcome) so you have to concretely model the wagers and the probabilities first, and then take the limit of the quantity. You cannot simply consider the limit of the each individual quantity (wager/outcome, probability) in isolation. If you take the limit of the quantities first and then combine them, all you get is an undefined solution ( Infinity*1 - Infinity*0 ).

[pzhon]

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If you don't understand that, then you don't understand how to treat infinities in mathematical models.

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This is not an infinite trial. Each betting cycle has a finite end, that is following a win.

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In general, when something is infinite in a mathematical model, the limit of the behavior in finite cases is what should be analyzed. Only when there is continuity at infinity can you analyze the idealized situation directly and hope to extract something meaningful. Here, there is a discontinuity at infinity. In the finite cases, playing the martingale ends with a significant chance that the casino has all of your money.

You say the martingale works. You mean,
<ul type="square">If you have an infinite bankroll, and the casino has an infinite bankroll, you can be sure to win $1.[/list]This is a sophomoric triviality, and irrelevant to analyzing the real world. (If you have an infinite bankroll, you can also rearrange your chips to have an infinite bankroll plus an additional $1 without betting anything.) That this is irrelevant for analyzing what happens when real people believe the martingale works should be obvious to you because of the limiting behavior as bankrolls for both sides go to infinity, which is for more and more severe failures.

I say the martingale fails. I mean,

<ul type="square">If you play the martingale, you will lose money on average. You have much less than a 50% chance to double up by playing a martingale. Casinos have nothing to fear from people playing a martingale. The martingale does allow you to obtain an advantage.[/list]Now, which of these is relevant to the original poster whose friend believes he can beat the casinos in the real world by playing a martingale?

[pzhon]

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If you play the martingale, you will lose money on average. You have much less than a 50% chance to double up by playing a martingale. Casinos have nothing to fear from people playing a martingale. The martingale does allow you to obtain an advantage.


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Of course, that should have been, "The martingale does NOT allow you to obtain an advantage."

[thylacine]

Your friend is absolutely correct. This strategy is absolutely infallible. (Disclaimer: In theory and assuming that certain, possibly unrealistic, conditions hold.)