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Huygens problem

One of my friends is taking a probability class, and he gave me this problem to solve:

Entities A and B play game such that A takes the first turn and can win with probability 1/4. B goes second and can win with probability 1/3. What is the probability A will win before B?

I’m not sure I understand it…

Clarification Now I understand it; consider that A and B are playing a game of darts. Then P(A) and P(B), the probability that A,B will make their next throws, do not necessarily sum to unity. And if A,B alternate throws, then P(A), P(B) alternate between being zero and nonzero.

So here’s my solution:

P(A \mbox{ wins before } B) = \sum_{n=1}^\infty P(A \mbox{ wins on the $n$-th throw }) P(B \mbox{ lost all the previous turns })

since P(A \mbox{ wins on exactly the $n$-th throw }) = 0 for even n, this reduces to

\sum_{k=0}^\infty P(A \mbox{ wins on the $2k+1$-th turn }) P(B \mbox{lost all previous turns }) .

The first term is calculated as

[Unparseable or potentially dangerous latex formula. Error 5 : 702x22]
= \frac{1}{4} \prod_{l=1}^{2k} P( A \mbox{ A loses $l$-th turn } )
= \frac{1}{4} \prod_{l=1}^k P(A \mbox{ loses $2l-1$-th turn} )
= \frac{1}{4}\left(\frac{3}{4}\right)^k

The second term is calculated as

P( B \mbox{ didn’t win any of the previous $2k$ turns } ) = \prod_{i=1}^{2k} P( B \mbox{ didn’t win the $i$-th turn })
= \prod_{i=1}^k \left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^k

Therefore, the desired probability is

\sum_{k=0}^\infty \frac{1}{4}\left(\frac{3}{4}\right)^k\left(\frac{2}{3}\right)^k = \frac{1}{4}\sum_{k=0}^\infty \left(\frac{1}{2}\right)^k = \frac{1}{4} \cdot 2 = \frac{1}{2} .

This entry was posted on Sunday, September 19th, 2004 at 12:57 pm and is filed under Mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “Huygens problem”

  1. Z Says:
    September 19th, 2004 at 2:34 pm

    Another way:
    Let B_1 = A wins on first try, B_2 = A misses and B wins, B_3 = they both miss
    Let A = {A wins before B}. Then P(A) = sum over i=1 to 3 P(A intersection B_i).
    P(A int B_1) = 1/4, P(A int B) = 0, P(A int B_3) = 1/2 P(A)
    The subtle point here is that the probability(A wins before B AND they both miss) is 1/2 P(A)! That is, 1/2 is the probability they both miss in the first round and P(A) is the probability that A wins given B_3.

    So we get P(A) = 1/4 + 1/2 P(A). Thus P(A) = 1/2.

  2. Z Says:
    September 19th, 2004 at 2:37 pm

    I forgot to say that the B’s are a partition of the sample space.

  3. James H. Says:
    December 10th, 2004 at 3:42 am

    this problem would be more interestingly solved using Christian Huygens methods set out in his book “Libellu De Ratiociniis in Ludo Aleae”. Based on his work let x equal A’s expectation to win when it is A’s turn and y be A’s expectation when it is B’s turn. Also let a=the prize Then according to Huygens, When it is A’s turn, A has 1 chance of a and 4 chances of y. so x equals the some of the product of these divided by the total number of chances.. x=(1a+4y)/5
    now by similar argument when it is B’s turn, a has 1 chance of 0 (i.e. losing) and 3 chances for x thus y=(1*0+3x)/4 pluging this back into x=(1a+4y)/5, it is found that x=1/2a. Thus it is a fair play and both should be at even odds.

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