statistic

The first two points must be split. The probability of that happening is

1 - p^2 - (1-p)^2

which is just 100% minus the probability that A wins twice minus the probability that B wins twice.

The next two points must go to the same person. The probability of that is

p^2 + (1-p)^2

The overall chance of a four point match win is the product of those two probabilities.

The probability of A winning in four is the first part times the probability of A taking the last two points

(1 - p^2 - (1-p)^2 ) p^2

If in part (b), you want the probability of A winning a match of any length (call it P), then it can be solved recursively with:

P = p^2 + 2p(1-p)P

which works out to:

P = p^2 / (2p^2 - 2p + 1)

p^2 is the probability that A wins the first two games and ends the match.
2p(1-p) is the probability that they split the first two and, essentially, start over (which is why it is multiplied by P).

where is the P that you multiplied from. is it the probability that A win two point when the game start over. i don't get it why you go from P=p^2 +2p(1-p)P which is multiplication and addition to P=p^2/(2p^2 -2p+1) which is division

If you want the chance that A wins regardless of the number of matches, you get a geometric series. The change that they win in two is p^2. The chance that they win in four is (p^2)(2p(p-1)). The chance they win in six is (p^2)(2p(p-1))^2 and so on. Use the formula for an infinite series to get the total probability as

P = p^2 / (1- 2p(p-1) )

engineer explained the geometric series associated with the problem, although his (p-1) should be (1-p).

p is the probability that A wins any point.
P is the probability that A wins any match (unknown length).

The first equation is a recursive implementation of what engineer explained. The second equation is derived by solving the first equation for P.

get it, thanks a lot