Thread: Tennis players with 50% chance to win each point...

1. Tennis players with 50% chance to win each point...

If each player has a .5 chance of winning each point, what will be the average length of a game, (i.e. first one to 4 points, with a margin of 2)?

2. You can find the distribution of L=number of games played until a winner is declared.
But the tie breaking part is a mess. Let's say we have player A and B.

$P(L=4)=P(AAAA)+P(BBBB)=2\left({1\over 2}\right)^4$

$P(L=5)=P(BAAAA)+P(ABBBB)=2{4\choose 1}\left({1\over 2}\right)^5$

The choose is there, because A must win that last point and B the last point in those strings.

$P(L=6)=P(BBAAAA)+P(AABBBB)=2{5\choose 2}\left({1\over 2}\right)^6$

3. Thanks for the help. Sadly I'm not well trained in math and I didn't get your equations.

Here was my simplistic try at solving it:

The first 4 points give us 16 permutations, which I'll try to use to average percentages.

Okay, the probability of a game ending in four straight points is 2/16, or 12.5% of the time game ends in 4.

Now out of the 14/16 different configurations not ending in 4, how many might end in 5—any with 3 points per side, which is... 8 of them have a 50% chance of the game ending, so 4/16 end in 5 points, or about 25%.

Now we have 10/16 left, how many ends in 6 points. Only 10 can possibly end in 6 points, and on average 5 will end. So 5/16 end in 6 points, or 31.25%. We have now accounted for 68.75% of all games.

Now we are theoretically at duece with 3-3, and the game cannot end. So 31.25% of all games will go to deuce on average. After the first point we are at 8 points; from then on I'm still trying to figure it out...

points - percentage
4 - 12.5
5 - 25
6 - 31.25
?? - 31.25

Now we simply average that out.

4*12.5 +5*25+ 6*31.25+??*31.25 =
??/100=

?? points a game, on average.