The details of the following (rather elementary) brainteaser are completely faithful to the actual scenario being described. In other words, it's a rare instance where a practical situation can be modelled exactly. The Wimbledon mens singles tournament has 128 players. The first round pairings are completely random, subject to the constraint that none of the top 32 players can be paired against each other. Two competitors, Olivier Rochus, and his brother Christophe are competing, and neither are in the elite group of 32 players. What is the probability that these brothers play in the first round (as actually occurred)? [No matter how low the probability is, I'm sure that this pairing wasn't "rigged". On similar grounds, I don't think the gaming room was rigged either.]Paul Epstein

I got the probability = 0.00350877193064151Quite low (0.3%) ... there was a 1 in 300 chance.In my usual self, I would be willing to bet upto $10 on such a thing (expecting $3000 in case of a win).And probably, I won't accept more than $1 for betting the other side (i.e. if I lose, I would need to pay atmost $300).Now, what does it tell about my risk-taking ability or loss-averse nature ?How do you think I should condition my risk-taking preferences .... to win in real-world markets?Neither of those big nos. (300 or 3000) would make me broke .... but does the statement of my preferences, directly extrapolatable to much higher number (which could make a difference of me getting financiallybroke or not)?Lastly, how much you be willing to bet on the either side, if offered the bet in real-life situation?

Total no. of ways = 32! x C(96,32) x 64! / (32! x 2^32)Total with brothers paired up = 32! x C(94,32) x 62! / (31! x 2^31)Ratio gave me the probability. Hope I did the math right.

Actually, I believe that the probability is 2/285. I'll give others the chance to work on it before explaining how I derived this figure. Here's an intuitive argument why 2/285 seems more reasonable than 0.3%.Suppose the pairings were random without restrictions. Then the answer is clearly 1/127. Since a non-elite player is more likely to play a fixed elite player than a fixed non-elite player, the answer to my prob is obviously less than 1/127. But how much less? Your claim is that the elite separation divides the probability by more than 2. Intuitively that seems wrong to me. Since most non-elite players do not play elite players, the elite separation seems to me to have a smaller effect.I could be wrong, though. It may also be the case that I worded the original problem differently to how I intended. Perhaps you could rephrase the problem in your own words to check that there has been no miscommunication about what the problem asks. Different answers in this sort of problem are often due to people working on different questions!Paul Epstein

(edited)

It's 2/285 like what's his name said. Each of the 32 top players has a 1/96 chance of being paired with a given non-top-32 player, since they're all equally likely. So conversely, each non-top-32 player has a 1/96 chance of being paired with a given top 32 player. Adding over all 32, there's a 1/3 chance a given non-top-32 player will be paired with any one in the top 32. So there's a 2/3 chance he'll get paired with a non-top 32. In this case each of the 95 nonelite players is equally likely, so for Brother 1 the chance he'll get paired with Brother 2 is 2/3*1/95 = 2/285.

yup, I agree.Bhutes reasoning, by strong force, was good as well,but it looks like he got some maths wrong.A.

yeah, I made an error in excel ... I should have used brackets to get 31 (2^32 should have become 2^(32-1), while I did 2^32-1)Hence off by a factor of 2.Maybe I should have done intuition based checking (like paul showed) & zonk's way is far too simpler.