- #36
humanino
- 2,527
- 8
From what I gather, the original problem should have specified that each player tosses his own coin.
Looking for the second win (and by the other person) changes the problem significantly. The two players have an equal probability of winning. Moreover, the expected value of the number of tosses before a win is equal for the two players: five throws. Making it so that the coin tosses continue until a winning configuration is found for each player is a different problem.Borek said:Why do you look for the second? First configuration found, game over.
D H said:Looking for the second win (and by the other person) changes the problem significantly. The two players have an equal probability of winning. Moreover, the expected value of the number of tosses before a win is equal for the two players: five throws. Making it so that the coin tosses continue until a winning configuration is found for each player is a different problem.
I just thought initially there was something interesting going on. At this point, the problem has been explained by several methods, and I take the opportunity to thank Hurkyl for his contribution. I said several times that it was a pity I misstated the initial problem.Borek said:That's exactly what I was pointing at, humanino was just solving completely different question
What do you get ?Redbelly98 said:The problem becomes more interesting if the choices are htt vs. ttt. The odds are not even close to 50/50.