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Hi I need help with problem related to probability.

There is a game of basketball between you and one other person. (only 1 game). Both agree that the stake will be $\$10$ (winner gains $\$10$ more, loser loses $\$10$). At a random point in the game, the game is interrupted, and you must make either one of the following choice:

1) Continue playing with a new stake of $\$20$

2) Stop playing and lose $\$10$

What winning probability that you must/should have to choose option 1? And explain why?

I have tried to list all cases that can happen, which are $3$: you keep playing and you win $\$20$, you keep playing and you lose $\$20$, you quit and you lose $\$10$. From there I don't know how to use the data to calculate the probability? Can someone give me some guidance?

callculus42
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Hieu Chu
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1 Answers1

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Let $p$ be the probability of winning, and $X$ your gain if you keep playing.

$X=20$ with probability $p$

and

$X=-20$ with probability $1-p$

Hence your expected gain : $E(X)=20p-20(1-p)=40p-20$

If you quit, your gain is $-10$

It is worth staying in the game iff $40p-20>-10$, that is, iff $p>1/4$

If you believe you have at least one chance out of four of winning, you better keep playing.

This strategic subtelty applies to several games, including backgammon.

Evargalo
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