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If a team has probability p of winning each game, and if the team wins/loses a match when it is 2 games ahead/behind the other team, is there an analytical formula or approximation for the probability that the team wins the match, as a function of p? One can of course resort to Monte Carlo. Here is what I tried:

p = prob of A winning each game
p1 = prob of A winning when match tied
p2 = prob of A winning when ahead by 1 game
p3 = prob of A winning when behind by 1 game

p1 = pp2 + (1-p)p3 = pp2 + p3 - pp3
p2 = p + (1-p)p1    = p + p1 - pp1
p3 = pp1 + 1 - p    = pp1 + 1 - p


plug last into first:


p1 = pp2 + (1-p)  (p*p1 + 1 - p)


plug 2nd into above


p1 = p(p + p1 - pp1) + (1-p) * (pp1 + 1 - p)
   = p^2 + pp1 - p^2p1 + pp1 + 1 - p - p^2*p1 - p + p^2


p1 = (2p^2 - 2p + 1)/(2*p^2 - p + 1)

But for p=0, this gives p1=1, which is wrong.

Fortranner
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2 Answers2

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The easy way to do the calculation is to note that you win two in a row, or you lose two in a row, or you start again.

So the probability is $p_1= \dfrac{p^2}{p^2+(1-p)^2}= \dfrac{p^2}{2p^2-2p+1}$.

Let's try it your way:

  • $p_1=p\times p_2+(1-p)\times p_3$
  • $p_2=p\times 1+(1-p)\times p_1$
  • $p_3=p\times p_1+(1-p)\times 0$ though you seem to have missed the $\times 0$

so substituting the second and third into the first and tidying up gives

$$p_1 = p^2 \times 1 +p(1-p)\times p_1+ (1-p)p \times p_1 + (1-p)^2\times 0$$ $$p_1 (1-2p(1-p))= p^2 $$ $$p_1 = \dfrac{p^2}{1-2p(1-p)} = \dfrac{p^2}{1-2p+2p^2}$$ which is the same result.

Henry
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3

Since Wimbledon is on, we could well model the game as being at deuce, and compute the probability of your winning the game as an exact parallel !

Then taking your winning probability for a point as $p$, losing as $q (=1-p)$, from deuce $d$ you either win two consecutive points or win/lose one point each to return to deuce

Thus $d = p^2 +2pq*d$

which yields $d = \Large\frac{p^2}{1-2pq}$

true blue anil
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