Johnny and Mike meet at the races, and Johnny makes Mike the following proposition:
“Mike, you pick one horse, and I’ll pick three horses. If two or three of my horses finish ahead of your one horse, I win. If one or none of my three horses finish ahead of your one horse, you win.”
Assume that (1) there are ten horses total in the race, and (2) all horses are equally skilled.
OK, it doesn’t matter what the other horses do, so it doesn’t matter how many others there are. It also doesn’t matter what order Johnny’s horses are relative to each other. So we have four relevant horses, Mike’s one, and Johnny’s three. Of those four horses, Mike’s can be first, second, third, or fourth, and if all the horses are equal, then it’s equally likely to be any of those. If Mike’s horse is first or second, then Mike wins. If Mike’s horse is third or fourth, then Johnny wins. So assuming everything is on the up-and-up, it’s a fair bet.
But that’s a big assumption. As a very good rule for life, when someone offers you a bet with convoluted terms, he’s got an edge that isn’t obvious to you. Don’t take his bet, or you’ll end up with cider in your ear.
Actually, as worded, the odds slightly favor Mike.
**Chronos ** is right in that if Mike and Johnny pick different horses the odds are even. However this doesn’t consider the possibility that ONE of the three horses that Johnny picks IS Mike’s horse.
If that happens Mike will win 2/3 of the time and Johnny will win 1/3 of the time.
The fewer horses in the race, the greater the chance that this overlap will occur.
True, Poccacho, I was assuming a constraint on the problem that they couldn’t pick the same horse, but that isn’t explicitly specified. So Mike can, in fact, get a positive expectation. Again, assuming no funny business.