
Problem 3: Conditional Probability a) Assume that we roll two fair six-sided dice. What is \( P \) (sum is \( 5 \mid \) first die is 3\( ) \) ? What is \( P( \) sum is \( 5 \mid \) first die is 1\( ) \) ? b) Assume that we roll two fair four-sided dice. What is \( P \) (sum is at least 3\( ) \) ? What is \( P( \) First die is 1\( ) \) ? What is \( P( \) sum is at least \( 3 \mid \) first die is 1\( ) \) ? c) Suppose two players each roll a die, and the one with the highest roll wins. Each roll is considered a "round" and further suppose that ties magically don't happen (or those rounds are simply ignored) so there is always a winner. The best out of 7 rounds wins the match (in other words the first to win 4 rounds wins the match). Let \( W \) be the event that you win the whole match. Let \( S=(i, j) \) be the current score where you have \( i \) wins and the opponent has \( j \) wins. Compute the probability that you win the match, given the current score, i.e., \( P(W \mid S=(i, j)) \) for each of the 16 possible values of \( S=(i, j) \). To help you out a bit with part (c) above I am providing the following hints: 1) If the score is equal-that is \( S=(k, k) \)-then there is equal chance of either player winning the match, \( P(W \mid S=(k, k))=\frac{1}{2} \) for \( k=0,1,2,3 \). 2) Let \( R_{i} \) be a random variable where \( R_{i}=1 \) if you win round \( i \) and \( R_{i}=0 \) if you lose that round. Note that \( P\left(R_{i}=0\right)=P\left(R_{i}=1\right)=\frac{1}{2} \). 3) Recall that by the law of total probability \( P(W \mid S=(i, j))=P\left(W, R_{i+j+1}=1 \mid S=\right. \) \( (i, j))+P\left(W, R_{i+j+1}=0 \mid S=(i, j)\right) . \) 4) By the probability chain rule \( P\left(W, R_{i+j+1} \mid S=(i, j)\right)=P\left(W \mid R_{i+j+1}, S=(i, j)\right) P\left(R_{i+j+1} \mid\right. \) \( S=(i, j)) \) 5) Find a way to write down \( P(W \mid S=(i, j)) \) as a function of \( P(W \mid S=(i+1, j)) \) and \( P(W \mid S=(i, j+1)) \). This will help you compute the answers in a recursive manner.