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Coin equivalence: THT and TTH?
This is a follow-up to the question:
http://answers.yahoo.com/question/index;_ylt=Ah6X0...
which is, in turn, a follow-up to
http://answers.yahoo.com/question/index;_ylt=Anr2n...
Suppose we flip a coin until either the sequence THT or TTH shows up. If we want to have equal probabilities for either sequence to end our coin flipping, how should we weight the coin? That is, if the probability in any toss of heads is p, what value of p creates equal probabilities (of 1/2) of ending with THT or with TTH?
[I have two more follow-ups in mind; let me know if you like this enough to see some more...]
You're approach is slightly shorter than mine, but we have the same conclusion. (If you don't start with an assumed T, you also got the "possibility" of p=0, which is likewise impossible.)
I tried briefly to come up with a second question here that fixed this (i.e., made such a p possible) by adding a positive probability of the coin landing on its side, but that didn't work out. I'll give this question another day or so for any interesting insights, then go ahead and post the more difficult next question...
Not that it really matters, but it irks me to see that I put "you're" at the beginning of that edit. Obviously that should be "your".
2 réponses
- Dr DLv 7il y a 1 décennieRéponse favorite
Have you actually worked this out? Because I'm getting p(H) = 1.
As with the previous question, we only start when we get our first T.
If we get TT, we are guaranteed to end with TTH rather than THT.
If we get TH, then there is q prob of ending with THT, and p prob of getting another H and starting all over again.
So far we have
P(TTH) = q
P(THT) = pq
P(starting over again) = p^2
Once we start over, these same probabilities keep repeating themselves.
So we require that q = pq
or q - pq = 0
q^2 = 0
p = 1
Of course physically with p = 1, we have zero chance of ending with either THT or TTH. That would be equal. I've even done a monte carlo on this. The ratio THT / TTH is only approaching 1 as p approaches 1.
- il y a 5 ans
Yes, they are different because of the way that the TTH pattern is constructed to overlap previous unsuccessful patterns. Consider only the TTH case where TT has occurred. There is a 1/2 probability that a third toss will yield H and complete the pattern. But notice that a 3rd toss of T doesn't "reset" the pattern matching, since TT remains the previous two matches. Thus, there is a 1/2 probability that a 4th will complete the pattern. And if doesn't, a 1/2 probability that a 5th will, ad nauseum. Now consider THT where TH has occurred. There is a 1/2 probability of flipping a T and completing the pattern. And there is a 1/2 probability of flipping H which *resets* the pattern. Now HH are the previous two characters. There is 0 probability that a 4th flip will complete, and zero probability that a fifth flip will complete. There is 1/8 probability that THT will occur on the 6th flip. This will occur anytime the last two characters are the same in one pattern and not in the other. UPDATE: I should have pointed out that the reason the TTH pattern will succeed sooner and more often is because, given a sequence of more than a few tosses, there will be more opportunities for TTH to succeed than there will be for THT, even though the probability of success (given an opportunity) is the same at 1/2.
