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Topic Title: K heads in a row, before 3 tails in a row
Created On Thu Nov 15, 07 03:30 PM

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Kaibab
Junior Member

Posts: 16
Joined: Jun 2003

Thu Nov 15, 07 03:30 PM

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Flip a fair coin until either
(A) it comes up Heads K times in a row; or
(B) it comes up Tails 3 times in a row.

Prove that the probability of (Heads K times in a row)
occurring BEFORE (Tails 3 times in a row), is

p = 7 / (6 + (2^K)) for K = 0, 1, 2, ...

(a result which I personally find surprising)

---------------------------------------
1st posting here, hope this is apropos; I searched for
similar messages but failed to find any. If this is in fact
a well-beaten dead horse please accept my apologies,
and I would appreciate hyperlink pointers to other
messages which analyze the problem or similar ones.
thx

Kaibab
Junior Member

Posts: 16
Joined: Jun 2003

Fri Nov 16, 07 02:51 PM

User is offline

got a PM that suggested this was indeed an old and well-known problem, directing me (here). *

Probability of (m heads in a row) occurring before (n tails in a row) =

p = ((2^n) - 1) / ( (2^n) + (2^m) - 2 )

which in the case I was interested in (n=3), means

p = 7 / ( 6 + (2^m) )

as desired.

* beware, there is a small typo on page 2. When setting up the simultaneous equations in p and q, the formula given for q is wrong:

WRONG: q = p * (1 / (2^(t-1)))
CORRECTION: q = p * (1 - (1 / (2^(t-1))))

This is just a typo; the solution values for p and q are correct, and so is their combination into a final "H wins with probability ____" answer.

Edited: Sat Nov 17, 07 at 02:00 PM by Kaibab

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