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I'm playing a game where I'm tossing a fair coin. When it's heads I gain a dollar. When it's tails I lose a dollar. I play until I have +24 or -36. What's the probability of each condition?

I'm comparing answers to chatgpt.

kashix9
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    MathOverflow is not for you to harvest answers, especially if you do not intend to engage with the content but merely compare them to a chatbot. –  Mar 31 '23 at 20:00
  • I've found my answer by running excel simulations. It appears the answer is simply 36/60 and 24/60. Chatgpt could not provide a satisfactory answer. – kashix9 Mar 31 '23 at 22:15
  • Wikipedia or any good textbook would give you the answer. You have a fair coin and thus a martingale, so with any stopping rule the expected position will always be $0$. So you need the probabilities to enable $24 \times \frac{36}{24+36}-36\times \frac{24}{24+36}=0$ – Henry Apr 01 '23 at 14:24

1 Answers1

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You don't need to simulate to get the answers.

  • The ambit of the game is $24-(-36) =60$

  • Since the odds are even, the up/down relationship will be linear, and the two ends represent the extremities of a straight line

  • It is evident that the nearer you are to your goal, the higher your probability of achieving it.

  • Thus the probability of success is equal to the fraction of the ambit from ruin, and vice-versa.

  • You can solve any gambler's ruin problem with even odds similarly.

true blue anil
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