Puzzle time: The Williams sisters

[jon-nyc]

Some tennis fans get excited when Venus and Serena Williams meet in a tournament. The likelihood of that happening depends on seeding and talent, so let's instead construct an idealized elimination tournament of 64 players, each as likely to win as to lose any given match, with bracketing chosen uniformly at random. Assuming that Venus and Serena are among the 64, what is the probability that they wind up playing each other?

[Klaus]

That's a lot of coin flipping...

[Klaus]

:::

1/32.

The total number of games that is played is 2^6-1 = 63.
The total number of possible games is 64 over 2, or 64 * 63/2.
The probability of a specific game to be played, including V vs S, is hence 63 / (64*63/2) = 1/32.

:::

[jon-nyc]

That was my approach and my answer. Much easier than the bottom-up recursion approach.

[jon-nyc]

:::

Another approach is to multiply expected number of games for any team (1+1/2+1/4+1/8+1/16+1/32) times 1/63 odds that the other sister is your partner. Gives you the same answer.

:::

[Horace]

:::

There are six rounds of the tournament, and the sisters have to co-exist in a given round, to meet in it. They co-exist in the first round, where they have a 1/63 chance to meet, by the logic that wherever the first sister is placed in the 64 spots, the second has a 1/63rd chance of being in opposition to her spot. Thus if the sisters co-exist in a round, they have a 1/(n-1) chance of meeting in that round, where n is the number of competitors in that round. If they do not meet, then the chance of both advancing is .5 * .5. So the chance they both advance to the next round, and meet there, is (.5^2) * (1/31). The chance they do not meet there and advance to the third round and meet there is (.5^(2 * 2)) * 1/15. Next round you get (.5^(2 * 3)) * 1/7, and so on. Sum the probabilities for the total chance they ever meet.

:::