0
$\begingroup$

This question already has an answer here:

Player A and B are playing a game where they roll a dice and the first player who rolls $6$ wins. What is the probability that player A wins if they goes first?

I said $\dfrac{1}{6}$ but my teacher said that was wrong. Not really the best at probability, sorry!

$\endgroup$

marked as duplicate by CY Aries, user354271, A. Goodier, gebruiker, Wouter Apr 15 '18 at 12:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

Think about this logically:

Yes, there is a ${1\over 6}$ chance for player A to win on the first move, but we need to consider what happens if player A does NOT win on the first move. Then, there is a ${5\over 6}$ chance they don't roll a 6 and a ${5\over 6}$ that player B does not roll a 6, for a total chance of ${({5\over 6})^2}$. Basically, this can keep happening forever, where both players don't roll a 6. This then turns into an infinite geometric series, with the first term being $u_1$ = ${1\over 6}$ and the common ratio being $r$ = ${({5\over 6})^2}$. The probability that A wins is the sum of this series. $$S_n = {u_1\over 1 - r}, \lvert r\rvert < 1$$ $$S_n = {{1\over 6} \over 1 - {({5\over 6})^2}}$$ $$S_n = {6\over 11}$$ So the probability that player a wins is not ${1\over 6}$ but ${6\over 11}$. Hopefully this makes sense!

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.