Suppose we flip a fair coin three times. Then HTH (first heads, then tails, then heads) and TTH (first two tails, then heads) are equally likely to occur.
Now, flip a fair coin repeatedly until we flip either HTH or TTH in a row. If TTH shows up before HTH, then I win. If HTH shows up before TTH, you win.
What is the probability that you win? If we were gambling and I paid out 1:1, would you play?
otro
Favorite Answer
First, consider the general case when none of us have won yet after several throws. From this point, let P(F1,F2) be the probability of you finally win given the two previous throws are F1 and F2.
We have the following equations:
P(H,H) = 1/2·P(H,H) + 1/2·P(H,T)
P(H,T) = 1/2·0 + 1/2·P(T,T)
P(T,H) = 1/2·P(H,H) + 1/2·P(H,T)
P(T,T) = 1/2·1 + 1/2·P(T,T)
Solving:
P(H,H) = P(H,T) = P(T,H) = 1/2
P(T,T) = 1
Now, in the first two throws HH, HT, TH and TT are equally probable, so your chances of winning the game are: 1/4·1/2 + 1/4·1/2 + 1/4·1/2 + 1/4·1 = 5/8
And mine are 3/8. I wouldn't play.
Anonymous
TTH would have a higher chance of winning. So NO I would not play you.
We both have 1/8 chance of winning within the first 3 flips.
Let say after 3 flips and we are tied then the 3 flips must be one of the following:
TTT...............HHH
THT...............THH
HHT
HTT
Same number of XTT vs XHT. But 4 out of 6 ends in T, and 2 out of 6 end in H. This would put TTH in a better position.
Northstar
No matter what coin sequence the first person picks, the second person can pick one that beats it.
John Horton Conway wrote something about this a long time ago but I can't find it.