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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.

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  • $\begingroup$ I think party is a false friend, you really mean game. $\endgroup$ – Arnaud Mortier Dec 5 '19 at 10:39
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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 $$ \frac{4}{7}\times (y+1) +\frac{3}{7}\times (e+1)= \frac{110}{37}. $$

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    $\begingroup$ 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. $\endgroup$ – Mees de Vries Dec 5 '19 at 10:24
  • $\begingroup$ Yes - good point. Thanks. $\endgroup$ – S. Dolan Dec 5 '19 at 10:34
  • $\begingroup$ 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 $E = \frac{98}{25}$ $\endgroup$ – Satish Ramanathan Dec 5 '19 at 12:39
  • $\begingroup$ @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.) $\endgroup$ – S. Dolan Dec 5 '19 at 13:14
  • $\begingroup$ 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 $E = \frac{110}{37}$ $\endgroup$ – Satish Ramanathan Dec 5 '19 at 17:40

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