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This discussion popped up on a different forum and I'd like to hear some opinions on this.

Suppose player A and B are playing each other in a best of 7 match. Player A has probability p of winning against B. We want to calculate the probability of a player

The probability of this may be calculated as such: We calculate the probability of A winning the first game and the entire match. Then A wins the first and last game, and B wins either of k out of 2+k matches in between, for k = 0,1,2,3. Then the probability becomes

##\sum^3_{k=0} p^4(1-p)^k {2+k \choose k}##

Conversely, the same happening for B has probability:

##\sum^3_{k=0} p^k(1-p)^4 {2+k \choose k}##

Summing these yields

##2p^6 - 6p^5 + 5p^4 - p + 1##.

The observed probability of this happening is 0.73. On to the questions:

Solving ##2p^6 - 6p^5 + 5p^4 - p + 1 = 0.73## yields ##p \approx 0.3## or ##0.7##.

On to the questions:1) Can one interpret 0.7 as the expected probability of the stronger player winning the first game and the entire match? I'm not so sure.

2) Given p as a random variable, how can one calculate the probability of the stronger player winning the entire match and the first match?

3) Given p as a random variable, how can one calculate the probability of a player winning the entire match GIVEN he wins the first match?

4) What is the difference between 2) and 3), i.e. how can they be interpreted?

5) What can the entire data (list of observed results) say about the distribution of p, or any of the above?

I am sure there is a lot of confusion here. I'd appreciate someone setting this straight.

Suppose player A and B are playing each other in a best of 7 match. Player A has probability p of winning against B. We want to calculate the probability of a player

**winning the first game and the entire match**.The probability of this may be calculated as such: We calculate the probability of A winning the first game and the entire match. Then A wins the first and last game, and B wins either of k out of 2+k matches in between, for k = 0,1,2,3. Then the probability becomes

##\sum^3_{k=0} p^4(1-p)^k {2+k \choose k}##

Conversely, the same happening for B has probability:

##\sum^3_{k=0} p^k(1-p)^4 {2+k \choose k}##

Summing these yields

##2p^6 - 6p^5 + 5p^4 - p + 1##.

The observed probability of this happening is 0.73. On to the questions:

Solving ##2p^6 - 6p^5 + 5p^4 - p + 1 = 0.73## yields ##p \approx 0.3## or ##0.7##.

On to the questions:1) Can one interpret 0.7 as the expected probability of the stronger player winning the first game and the entire match? I'm not so sure.

2) Given p as a random variable, how can one calculate the probability of the stronger player winning the entire match and the first match?

3) Given p as a random variable, how can one calculate the probability of a player winning the entire match GIVEN he wins the first match?

4) What is the difference between 2) and 3), i.e. how can they be interpreted?

5) What can the entire data (list of observed results) say about the distribution of p, or any of the above?

I am sure there is a lot of confusion here. I'd appreciate someone setting this straight.

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