Flipping Coins Question!(puzzle) - Page 3

Steve-O · #21 (**permalink**) 07-03-2005, 10:05 PM

TTT

HHH

THH

THT

HHT

TTH

THT

HTT

there is also

HTH

SteveY · #22 (**permalink**) 07-03-2005, 10:09 PM

I just noticed that you have THT twice. Yea im getting confused now. ;/

Steve-O · #23 (**permalink**) 07-03-2005, 10:11 PM

yeah its too late for this!!!!! Bonch is smoking something

SteveY · #24 (**permalink**) 07-03-2005, 10:17 PM

K to simplify problem, both have to hit heads first.

Assuming both hit heads already.
Player A
He hits T (50%) - he wins
He hits H (50%) - he doesnt win, but now he still has heads, so he's 50% again on the next flip.

PlayerB
He hits T (50%) - he doesnt win, and has to restart from the first heads. Then he has a 25% chance to go HH to win.
He hits H (50%) - he wins

Steve-O · #25 (**permalink**) 07-03-2005, 10:20 PM

Quote:

He hits H (50%) - he doesnt win, but now he still has heads, so he's 50% again on the next flip.

no because player B would have won

SteveY · #26 (**permalink**) 07-03-2005, 10:24 PM

They flip their own coins.

But yeah thats where I'm getting confused. If its just one coin, its a wash? but if its two coins A has adv? Weird. Maybe I am missing something

Steve-O · #27 (**permalink**) 07-03-2005, 10:26 PM

I think bonch needs to be more specific with his questions in the future. the wording is such that the rules can be interpreted in a # of different ways

tihsihtkcuf · #28 (**permalink**) 07-03-2005, 11:14 PM

To answer these questions we need to calculate, for each pair of triplets, the probability that one triplet appears before the other. Given that each triplet is equally likely, it may initially seem that each is equally likely to appear first. For an example of why this is not so, consider the triplets HHH and THH. The only way for HHH to appear before THH is if the first three tosses come up heads. Any other result will allow THH to block HHH. Therefore, the probability that HHH appears before THH is 1/8.

We may calculate the probabilities for each pair in a similar manner. Consider, for example, HTT vs. HHT. The probability HTT appears first is the mean of that probability over the four possibilities for the first two coin tosses. Let, for example, p(HT) be the probability HTT appears first following HT. Then we have:

(1) p(HH) = p(HH)/2
(2) p(HT) = p(TH)/2 + 1/2
(3) p(TH) = p(HH)/2 + p(HT)/2
(4) p(TT) = p(TH)/2 + p(TT)/2

(1) p(HH) = 0. (HTT can avoid losing only by hoping for an infinite string of heads!)
(3) p(TH) = p(HT)/2
(2) p(HT) = p(HT)/4 + 1/2 p(HT) = 2/3
(3) p(TH) = 1/3
(4) p(TT) = p(TH) p(TT) = 1/3

The mean of these four results gives us: probability of HTT appearing before HHT = 1/3.

tihsihtkcuf · #29 (**permalink**) 07-03-2005, 11:15 PM

therefore meaning HT will become before HH

helmuthpoker201 · #30 (**permalink**) 07-06-2005, 09:59 AM

There is probably a simpler way, but i know you can explain the answer with stochastic statistics somehow. Let me think about it a day or two more, i know its not 50/50 though.