I found the following question on the internet:

A player sits down at a gambling table to play a game with 1000 dollars. He bets 1 dollar at a time with probability of winning being equal to 49.3212% with a 1:1 payout. What is the probability that the player's bankroll reaches 1200 before he goes bust?

I think the question is pretty interesting. Here's the answer:

$$\frac{\left (\frac{1-p}{p} \right ) ^{1000} - 1}{\left (\frac{1-p}{p} \right ) ^{1200} - 1}= \frac{\left (\frac{0.506788}{0.493212} \right ) ^{1000} - 1}{\left (\frac{0.506788}{0.493212} \right ) ^{1200} - 1} \approx .004378132$$

Now the answer also states that if instead of betting 1 dollar each time the player decides to bet the following amounts, the probability of reaching 1200 before losing the initial 1000 dollars also changes as follows:

5: 0.336507

10: 0.564184

25: 0.731927

50: 0.785049

100: 0.809914

How are those numbers found? The initial formula doesn't include the amount wagered as a variable. So how are those different probabilities are found?


The value 1000 in the formulas is \$1000 (the original amount) divided by \$1 (the wager). If you increase the bets to \$25, you have \$1000/\$25 = 40 bets' worth of money. I trust you can take it from here?

  • $\begingroup$ I got that part. I mean like, if instead of paying 1:1 it paid 2:1 when won. (Like betting on the first 12 numbers) $\endgroup$ – SarpSTA Jan 5 '16 at 21:54
  • $\begingroup$ @SarpSTA: I see. I think that's different enough to be a new question -- I'm not sure you can use the same setup when you want a different number of rounds depending on the outcome. I can't immediately see the answer although I'm sure someone can help you. $\endgroup$ – Charles Jan 5 '16 at 22:03
  • $\begingroup$ I'll be posting a new one then. Thank you! $\endgroup$ – SarpSTA Jan 5 '16 at 22:10
  • $\begingroup$ @SarpSTA: Good luck with your new question. $\endgroup$ – Charles Jan 5 '16 at 22:16
  • 1
    $\begingroup$ @SarpSTA: with no house edge and probability $1$ of ending at $0$ or $1200$, expectations can be used to show the probability of ending at $1200$ must be $\frac{1000}{1200}$ no matter what size bets you make (you can even mix bet sizes) $\endgroup$ – Henry Jan 6 '16 at 1:23

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