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Find the probability that a total of $4$games are played

First ditermine which order of wins cause precisely $4$games.

A wins the first game:

B wins the second game( if A wins again, Awon and the game is over)

The next two games can be won by either A or B, so that one of them has the lead of two.

Equivalently for case where B wins the first game.

To Specify, the orders can be:ABAA,ABBB,BAAA,BABB

outcomes of different games are independent.

$P("4\text{games precisely")}=P("ABAA")+P("ABBB")+P("BAAA")+P("BABB")$

$=p^3(1鈭�p)+p(1鈭�p{)}^3+p^3(1鈭�p)+p(1鈭�p{)}^3$

$=2\left[p^3\right(1鈭�p)+p(1鈭�p{)}^3]$

The probability that a total of $4$ games are played is

Find the probability that A is the winner of the series .

$\text{The first two games can end in one of three outcomes -}A\text{wins (2:0),}B\text{wins or it is even (1:1).}$

If the game is continued, the result is even, and since the total of wins is not important, only the difference, this is equivalent to $0:0$.

Each couple of games, an odd and an even one, respectively, from the start of playing, can end independently in the same three ways as the first pair. Therefore, the probability that A wins is the probability that he/she wins after i=$0$,$1$,... pairs of games that end even.

A pair of games end with A winning both with probability P²

A pair of games end with B winning both with probability ($1-p$)²

A pair of games end with A:B= $1:1$with probability $2P(1-P)$

A wins in total in an infinite number of mutually exclusive sequences - that the first k=$0,1,2,$, ... pairs end so that A and B are even, and in the next pair A wins both.

$P\left(A\text{wins"}\right)=\underset{k=0,1,2,鈥�}{鈭�}P(k\text{pairs end without lead,}k+1.\text{is won by}A)$

$=\underset{k=0,1,2,鈥�}{鈭�}\left[P\right(\text{apairofgameshasnowinner}){]}^k鈰�P("AA")$

$=\underset{k=0,1,2,鈥�}{鈭�}\left[2p\right(1鈭�p){]}^k鈰�p^2$

This is a geometric sequence, and the following formula is known

$\underset{k=0,1,2,鈥�}{鈭�}{q}^k=\frac{1}{1鈭�q}$

Thus,

$P\left(A\text{wins"}\right)=p^2鈰�\frac{1}{1鈭�2p(1鈭�p)}$

$=\frac{p^2}{p^2+(1鈭�p{)}^2}$

The probability that A is the winner of the series is $=\frac{p^2}{p^2+(1鈭�p{)}^2}$$=\frac{p^2}{p^2+(1鈭�p{)}^2}$