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There is a 50/50 chance of winning a game. I bet small on every game and wait for a certain number of losses in a row, say 5, and then bet big (more than 32 times the original bet) on the 6th game, thus increasing the probability of success. If I still continue to lose I then double up every time until I make back that 6th bet (this would be the Martingale system). I can't see the flaw but I feel like there is one? Can anyone give a probabilistic explanation as to why this strategy would fail?

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    $\begingroup$ You always have a chance of losing again and again. Even though the expected value of the game is presumably fair, you can always run out of money (since you presumably started with a finite quantity). Once upon a time, I played a textbased mmorpg that had a gambling house in it, and I used to bet like this (and frequently came out on top), however I lost literally 37 games in a row (that is not an exaggeration) in a 50-50 game, bankrupting me several times. Fortunately, I learned my lesson in a game and not using real money. $\endgroup$
    – JMoravitz
    Jun 14, 2015 at 2:25
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    $\begingroup$ The result of those five loses in the row don't increase your chances of winning in the sixth. $\endgroup$ Jun 14, 2015 at 2:34
  • $\begingroup$ If I flip a coin 5 times and every time it lands on tails, isn't that a conditional probability problem? Wouldn't that lower the probability that it would land heads on the next flip? $\endgroup$
    – Omar
    Jun 14, 2015 at 2:37
  • $\begingroup$ No, that is incorrect, @omar. Look up conditional probabilities. Coin tosses are independent events. $\endgroup$ Jun 14, 2015 at 2:39
  • $\begingroup$ Ok. Got it. Thanks! $\endgroup$
    – Omar
    Jun 14, 2015 at 2:51

2 Answers 2

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There is no problem in what you describe. But your strategy assumes that you have an unlimited "bank" to bet from.

If you have a finite amount of cash in bank, then the probability of eventual bankrupt is equal to 1.

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  • $\begingroup$ Ok. So I understand that I would run out of cash if I continually doubled each losing bet. But what about the first part? Doesn't the fact that I've lost 5 in a row increase my odds of winning? Wouldn't betting large effectively increase my odds to greater than 50 percent? $\endgroup$
    – Omar
    Jun 14, 2015 at 2:34
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    $\begingroup$ Im not sure what you mean. If i understand you right then my answer is: Probability doesn't work that way. Yes, the probability that you are losing 5+1=6 times is very low. But losing the 6th time conditioned on you already lost 5 times , are still 50/50 im afraid. $\endgroup$
    – Conformal
    Jun 14, 2015 at 2:39
  • $\begingroup$ Got it. Thanks! $\endgroup$
    – Omar
    Jun 14, 2015 at 2:52
  • $\begingroup$ Also, the strategy assumes that the house will accept bets of unlimited size. (Even if you have unlimited funds, it doesn't mean the house does.) $\endgroup$
    – cjm
    Jun 14, 2015 at 5:46
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This is a perfectly sustainable strategy, if you can take out infinite loans. This is totally possible in a pure mathematical environment, and you can make a scenario in which you win infinitely; however, if you cannot take out loans, then the odds of your losing all of your money on $1$ run are $$\dfrac{1}{\log_2\left(\dfrac{\text{total money}}{\text{base bet}}\right)}$$ and your odds of doubling your money on $1$ run are once again $$\dfrac{1}{\log_2\left(\dfrac{\text{total money}}{\text{base bet}}\right)}$$ so you are just as likely to lose everything as to double your money on a run, so your expected payout per run is $\$0$. Please do not try this strategy in real life, because it will appear that you are steadily profitable until that awkward moment when you are not, and suddenly you have lost everything.

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    $\begingroup$ I love that "Please do not " part , as if it was drinking soda in the bus :-) $\endgroup$
    – Conformal
    Jun 14, 2015 at 2:37

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