A and B play a game of Go, and the probability of A winning is , and the probability of B winning is

If the person who wins the first game also wins the second game, the game is over. At this time, the probability of A winning is , the probability of B winning is , and if A and B each win 1 game, the game will start again, and the probability is 2

The probability of winning is x, x= 2+2 xx= 2 (1-2).

Therefore, the probability of winning is 2 (1-2).

The answer is 2 (2 2-2 +1).

The positive solution point connection is just a different expression.

The shooting game is regarded as a Go game, and the probability ab can be regarded as your probability.

This is what I used to answer in the same type of question as yours.

Don't forget, happy learning.

It is easy to analyze from the question to know that there are only two cases in which A wins: 1A wins B (repeat) after A wins 2 games in a row 2B wins A (repeat) after A wins two games in a row.

Therefore, the probability of winning the first game.

pA = 2 (A wins two games in a row) + 2 2 (A wins B wins two games in a row and B wins A wins two games in a row) + 2 2 2 2 (A and B cross in the first four games and each has two games left, and A wins two games in a row) + ....2 n n 2 (the first 2n games A and B cross each win n games, and then A wins two games in a row) +....

Therefore, p A = 2+2 2+2 2 2 2+......2α^nβ^nα^2+……

2+2α^2（αβ2β^2+……nβ^n+……The inner is the sum of the proportional sequence, and the limit is taken at the end.

2+lim[ 1- n n) (1- n approaches + )

Set up two people to play an n-set match, according to the question:

nα-nβ=2

This results in =(2+n) 2n

1-α=(n-2)/2n

The greater the n, the closer the win rate of the two will be, and close.

Hello, the probability of A winning is.

Summary. A win of 2 or 3 games can be won.

A win of 2 or 3 games can be won.

A and B play chess games, and it is known that the probability of A winning each round of the game is, and the probability of A winning the three rounds of the game is obtained.

Don't understand. The probability of winning two games is to choose two out of three games to win.

So it's c up 2 down 3

This is indicated, and two of the three rounds are chosen.

It is expressed as the probability of winning two out of three games.

This is followed by the probability of winning all three games.

A win rate p=. B win rate = Q=

p³+3qp³+6q²p³

In the best-of-five games, A may win.

A win rate p=, win rate = q=

Best-of-three p(A)=pp+pqp+qpp=best-of-five p(a)=ppp+pqpp+qppp+ppqp+pqp+pqp+pqpp+qqpp+qqppp+qpqpp+qpqpp+qppqpp+qppqp

p³+3qp³+6q²p³

Therefore, in the best of five games, A may win.

A format with more rounds is in favor of the masters.

If the first three games are all won by A, the game will end directly, and the result will be 3:0, so the fourth game must be won by A, and A will win two games in the first three games, lose one game, and the first three games will be in any order. The result is 3*2 3*2 3*1 3*2 3=8 27, and the first 3 means that there are three possibilities for the order of the first three games.

This is the knowledge of permutations and combinations, the probability of A winning is 2 3, then B wins the probability of 1 3, A 3:1 wins, there may be the following situations: B A, A B A A, A A B A, A A B A, so that there is 1 3 * 2 3 * 2 3 * 2 3 * 2 3 + 2 3 * 1 3 * 2 3 * 2 3 + 23 * 2 3 * 1 3 * 2 3 = 8 27

First of all, the score of 3:1 must lose one game in the first three games, so there is 1 3*c31 *2 3*2 3 (the two that win).

Then the 4th win *2 3

So the probability is c31*1 3*2 3*2 3*2 3=8 27

There are only three possibilities, namely A, A, B, A, A, and B, A, and A, each of which is 2 3 * 2 3 * 2 3 * 1 3 = 8 81, a total of 8 27

A 3:1 win means that you win two of the first three games and win the fourth game. c（3,2）x 2/3 x 2/3 x ( 1- 2/3) x 2/3 = 8/27

So choose A

A wins through 4 games, then A wins 3 wins and 1 loss in these four games, and the last game is a win, that is, the first three games are 2 wins and 1 loss, and the fourth game wins.

Such probability = [c(3,1)*

c(3,1)* means that one of the first three games is won by B, and the remaining 3 games are won by A, so multiply by 3 to finish! ~

This is a typical binomial distribution exercise.

The probability of A winning is that A wins two games before B wins two more games, assuming that A wins two games in a row.

A wins, 2 games, B wins, 1 game, here must be the next 2 games, A and B win each game, because it has two situations, one is that A wins first, B wins one game, and the other is that B wins first, A wins one game, so 2*

So the probability of A winning is =

(1) There are two situations for the end of the game after two rematches, that is, A wins two games in a row or B wins two games in a row. The probabilities are and the sum of is.

2) A win is to win two or three of the remaining three games, the probability is:

60%.

Because according to the known assumptions, regardless of psychological factors, each game is an independent event, and its probability is not affected, i.e. regardless of the first game, the probability of winning each game is 60%.

Please raise the bounty, this thing is not difficult to calculate, but it is quite troublesome to knock it out.

The first three sets are: SSF, SFS, FSS

In order to win, the second four sets must be won in two sets: SFS, SFFS, SS, FSS, FS, FS, FSS, FFSS, so there are seven wins in the four sets.

In the first three sets of 2-1 to win the first four games, there are a total of 3*7 = 21 situations.

(1) (2) The reason is not explained, it is completely oral, and the result may not be correct.

Simplify A to win 2 3 for A, B to win 1 3 for B, A+B=1 for Aa(A+B)+ABA+BAA=20 27 A to win and play three games ABA+BAA=8 27

A2 5

Two cases of dust skin.

1.A won the first game of the quarrel, probability = 2 5

2.A won in the second game, that is, A lost in the first game, the probability = 3 Brother Li did 5*2 5=6 25