Probability that I will win the game?

Question

A game is played similar to tennis as follows:

To win the game, a player must score at least 4 points with a margin of 2 or more points over his opponent. e.g. he can win 4-0, 4-1, 4-2, but not 4-3. If it's tied at 3-3, 4-4 or higher, then they keep playing indefinitely until one person gains a 2 point lead.

If the probability of me winning a point is p, calculate the probability that I will win the game.

Bad Santa: For the case where p = 1/2, your answer evaluates to 1/128. Shouldn't it be closer to 1/2?
Also winning outright doesn't mean winning 4 of the first 6 points, since 4-0 is considered winning outright.

But interestingly it matches perfectly with Scarlet's answer, except for the 1/64, as Andy pointed out.

Answer 1

I'll write q for 1-p to save space and confusion (hopefully).

Probability of winning without ever getting to 3-3:
p^4 + 4p^4q + 10p^4q^2
(note for 4-1, must win 3 of the first 4 games and the last one; for 4-2, must win 3 of the first 5 games and the last one, so we get 5C3 p^3 q^2 p = 10p^4q^2)

Probability of getting to 3-3 is 20p^3q^3.
Once at 3-3 or a tied state above that, there is a probability of 2pq of going to the next tied state, p^2 of winning without going to the next tied state, q^2 of losing without going to the next tied state. So the probability of ultimately winning, having reached 3-3, is
p^2 + (2pq) p^2 + (2pq)^2 p^2 + (2pq)^3 p^2 + ...
= p^2 / (1 - 2pq)

So the ultimate probability of winning is
p^4 + 4p^4q + 10p^4q^2 + 20p^3q^3 (p^2 / (1 - 2pq))
= p^4 + 4p^4q + 10p^4q^2 + 20p^5q^3 / (1 - 2pq)

Some reasonableness checks: p=1, q=0 gives 1; p=0, q=1 gives 0; p = q = 1/2 gives 1/2 after a little arithmetic. The graph looks about as expected also.

Answer 2

I created a large transition matrix and wrote a computer program to calculate a large power of it. This doesn't give a formula for the answer, but allows me to cross check results. Bad Santa's formula is clearly wrong, because if p=1 we get 1/64. Clearly, it should be 1. However, a closer inspection reveals that if we simply remove the division by 64, the results correspond exactly to my computer simulation.

I agree with Dr D --- I don't know why there should be a p^6 in the first part of the answer, because once we get 4 points it is a win. But the formula seems to clearly work. I'm not sure why, but I'll post more if I find an explanation.

EDIT: Nice work, Scarlet Manuka. A much clearer explanation, and your formula corresponds exactly to my computer simulation.

Answer 3

First do your binominal expansion for six points (1,6,15,20,15,6,1)

Your probablity of winning outright (winning 4 of the first six points) is = 1/64 *( p^6 + 6*p^5*(1-p) +15*p^4(1-p)^2 )

Your probablity of being tied at 3-3 is = 1/64*(20*p^3(1-p)^3

To figure the odds if tied at 3-3, do a binomial exansion for 2 (1,2,1)

Your odds of winning the next two points is W3 =1/4 p^2
Your opponents odds of winning the next two points is L3=1/4 (p-1)^2

Your overall odds of winning if the game is tied at 3-3 is
=W3/(W3+L3) = p^2/(p^2 + (1-p)^2)

Therefore your overall odds of winning are:

= 1/64 *[ p^6 + 6*p^5*(1-p) +15*p^4(1-p)^2 ) +*20*p^3(1-p)^3 * (p^2/(p^2 + (1-p)^2))]

I'm sure you always wanted to know this. ;-)

>>>>>>>>>>>>>>
Ahhg.......

The 1/64 is spurious. My bad math. Should not be there. Should not be there. (The 1/64th was to normalize the binomial expansion, which I did not need to do since the "p"s do that for me). The correct answer is:

= p^6 + 6*p^5*(1-p) +15*p^4(1-p)^2 ) +*20*p^3(1-p)^3 * (p^2/(p^2 + (1-p)^2)).

This can be solved to:

=p^6 - 6*p^6+6*p^5 +15*p^6-30p^5+15p^4 +20*p^3(1-p)^3 * (p^2/(p^2 + (1-p)^2)).

or

=10p^6-24p^5+15p^4 + 20*p^3(1-p)^3 * (p^2/(p^2 + (1-p)^2)

The reason I used played six games was it was easier to do the math. It fits right into the binominal equation which makes it easy to solve without thinking. So long as you win (or would win) at least 4 games, you win. This means that, say, you won 4-1, then if you played two more games, you would be ahead either 6-1, 5-2 or 4-3, but in all cases, you won. That way I can fit it easily into the binomial expansion to calculate the odds.

To me, it is easier to calculate the odds of winning at least 4 games out of six or alternatively going 3-3 in one simple straightforward calculation rather than calculating the separate odds of winning in 4-0, 4-1 and 4-2 or getting tied at 3-3. But everyone has his or her way of tackling a problem.

What the binomial expansions says is that if p=.5, then the odds of getting to 3-3 are 20/64ths, that means that if Dr. D was going to win at least 4 out of 6, he would win no matter what, and if he was not going to win at least 3 out of six, then Pete Sampras wins.

Scarlett M's solution looks good and is probably the same as mine. The p^6 ends up being part of the equation because you can play 6 games. Scarlett's solution would end up with p^6s if you substitute in (1-p) for qs and then did some solving.

Scarlett's solution is:

= p^4 + 4p^4q + 10p^4q^2 + 20p^5q^3 / (1 - 2pq)
= p^4 + 4p^4(1-p) + 10p^4(1-p)^2 + 20p^5q^3 / (1 - 2pq)
= p^4 + 4p^4-4p^5 + 10p^6-20p^5+10p^4 + 20p^5q^3 / (1 - 2pq)
=10p^6-24p^5+15p^4 + 20p^5q^3 / (1 - 2pq)

As you will note, "10p^6-24p^5+15p^4" factors are identical for both Scarlett's and my equations, and I assume the last term would also prove identical.

Answer 4

p/2