# Dice rolling and conditional probability

We are rolling a six-sided standard dice. We roll until we see an odd number appears. If this happens in one of the rollings, if this number is 5 we win and otherwise we lose - obviously the number is 1 or 3 in this case. What is the probability of the dice being at least 10 time rolled with the condition that we have won the game?

ATTEMPTED ANSWER: $$\sum_{i=9}^{∞} (\frac{1}{2})^i\frac{1}{6}$$

The idea behind my answer is this: Since we are sure that we have won, so the the last number which appears is 5. For the appearance of 5 we have the the probability of $$\frac{1}{6}$$ and for the rollings before that, surely even numbers have appeared. That means for each of them we have the probability of $$\frac{1}{2}$$.

Any idea of why my answer works or not is highly apprectiated. Is there any way to write the soloution in terms of conditional probability, for example $$P(A|B)$$?

Since you already know that you win the game, you don't need to count for the possibilities that you roll odd numbers before you win. The probability you had $$i$$ rolls, given you win, is a geometric random variable with probability of success = $$\frac14$$ (success being a roll of $$5$$ and non-success being a roll of $$2,4,$$ or $$6).$$ Letting $$X$$ be the number of rolls, given you win:
$$P(X=i) = \left(\frac34\right)^{i-1}\left(\frac14\right) \therefore P(X\ge10) = \sum\limits_{i=10}^\infty \left(\frac34\right)^{i-1}\left(\frac14\right) = \left(\frac34\right)^9$$
• @FarhadRouhbakhsh If $X$ = # of flips of a fair coin until Heads appears, $P(X=i) = (\frac12)^{i-1}(\frac12)$ . You know your last flip is a Heads and every other flip is a Tails; even so, the specific event of $i$ flips calculates the probability that each flip before the $i^{th}$ was not Heads. Commented May 3, 2020 at 22:15