# Determine the mathematical expectation of the number of games in such a match [closed]

The probability of victory of the younger brother over the elder is 4/7 in each party (there are no draws), and the results of all parties do not depend on each other. They play a match until one of them wins two games in a row. Determine the mathematical expectation of the number of games in such a match.

• I think party is a false friend, you really mean game. – Arnaud Mortier Dec 5 '19 at 10:39

Let $$y$$ be the number of games left in a match once the younger has just won a game and let $$e$$ be the number of games left in a match if the elder has just won a game.

Consider the situation when the younger has just won a game. With probability $$\frac{4}{7}$$ the match ends next game and with probability $$\frac{3}{7}$$ the next game is won by the elder brother. So

$$E(y)=\frac{4}{7}\times 1+\frac{3}{7}\times (1+E(e))=1+\frac{3}{7}E(e)$$.

Similarly,

$$E(e)=1+\frac{4}{7}E(y)$$.

Solving, we get $$y=\frac{70}{37},e=\frac{77}{37}$$.

Then the expected number of games is $$\frac{4}{7}\times (y+1) +\frac{3}{7}\times (e+1)= \frac{110}{37}.$$

• From my understanding, if the first match has the younger brother win, and the second match has the older brother win, then we are not in the original situation; we are in a situation where the older brother only needs to win once to win the whole game. – Mees de Vries Dec 5 '19 at 10:24
• Yes - good point. Thanks. – S. Dolan Dec 5 '19 at 10:34
• Let E be the expected number of matches so that anyone wins the two games in a row. In the first two matches, the probability that anyone wins is $\frac{25}{49}$. The probability that one of wins one match is $1-\frac{25}{49}$. Thus you have to solve the equation $E = \frac{25}{49}(2) + \frac{24}{49}(E+2)$ giving you a $E = \frac{98}{25}$ – Satish Ramanathan Dec 5 '19 at 12:39
• @Satish Ramanathan. Your answer is wrong because it assumes that after two games we are back in the initial position. (See the comment by @ Mees de Vries.) – S. Dolan Dec 5 '19 at 13:14
• You are right. in which case it is $E = \frac{25}{49}(2) + \frac{12}{49}(E+2)+ \frac{12\times 3}{49}$ gives you $E = \frac{110}{37}$ – Satish Ramanathan Dec 5 '19 at 17:40