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John and Bill wanted to play only one game Ping-pong.The first player to reach 21 points wins the game.

John is 20% better than Bill — meaning he scores a 20% higher chance of scoring a point.

Bill said to John "well, you`ll win if we score as usual,cause you play better than me".

"how do you want us to score then?" asked John.

Bill:"any point I`ll score will be counted as(1.5) points to me,and for you a point will be as it is only one point,do you agree?"

" OK" said john,"but I`ll start the first serve".

Each player serves three points in a row and then switch server.

Now, who won that game?

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    $\begingroup$ What does "John is 20% better than Bill." mean? We need to know 1) the probability that John will win a point if John serves 2) the probability that John will win a point if Bill serves. $\endgroup$ – Rosie F Oct 1 '16 at 7:57
  • $\begingroup$ @RosieF The 20% means that John has 20% more chance to score points than Bill and serving will determine the sequence only $\endgroup$ – user26522 Oct 1 '16 at 8:31
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    $\begingroup$ @RosieF; P(J wins)=1.2*P(B wins). And P(J)+P(B)=1 $\endgroup$ – JMP Oct 1 '16 at 8:39
  • $\begingroup$ In ping pong can you only score points when serving? $\endgroup$ – gtwebb Oct 1 '16 at 14:40
  • $\begingroup$ According to the ITTF, either player can score regardless of who served. ittf.com/ittf_handbook/ittf_hb.html So it's irrelevant who has first serve unless a player is more likely to win a point when they serve. (Which is probably true in real life, but isn't specified in this question.) $\endgroup$ – Joe Oct 1 '16 at 21:52
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If "20% better", as Anjan said in comments, means P(John winning a point) = 1.2 * P(Bill winning a point) and P(J)+P(B) = 1, that means

John will win a point ~54.5% of the time, and Bill will win it ~45.5%.

Since Bill gets 1.5 points every time he wins an exchange, he needs to win 14 exchanges before John wins 21.
35 exchanges (14+21) is enough for them both to reach 21 points, so 34 exchanges are required to guarantee that one of them reaches 21 points.

Therefore, Bill must win 14 out of 34 exchanges, at a 45.5% success rate.
John must win 21 out of 34 at a 54.5% success rate.

Using the

Binomial Distribution (link here), we can find out the likelihood of this:
14 or more successes in 34 trials with 0.455 probability of success -> ~75% chance
21 or more successes in 34 trials with 0.545 probability of success -> ~25% chance.

Therefore, it's about three times more likely that

Bill wins.

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If we interpret "John is 20% better" to mean that P(John scores) = P(Bill scores) + 20%, so that P(John scores) = 60% and P(Bill scores) = 40%, and we use the method in Joe's answer, the result is $$ P(\text{Joe wins match}) = \sum_{k=21}^{34}\binom{34}{k}(0.6)^k(0.4)^{34-k}\approx 49\% $$ So this match slightly favors Bill, but is still pretty fair.

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I get a different answer to Joe:

John wins $\frac6{11}$ of the time, Bill $\frac5{11}$. After $100$ balls have been played, John will have $~54.54$ points, Bill will have $~68.18$ points. Scaling Bill's score to $21$ involves a factor of $~3.25$ so I predict a win for Bill after $~31$ balls, score (17-21).

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