Re: probability puzzle
- From: Bill <b92057@xxxxxxxxx>
- Date: Wed, 19 Sep 2007 14:43:15 -0700
On Sep 19, 1:40 pm, Bill <b92...@xxxxxxxxx> wrote:
On Sep 15, 3:10 pm, br...@xxxxxxx (Brian Tung) wrote:The probability for each term is;
This came out of a side discussion on cryptography.
A machine starts flipping a fair coin. It halts after 2n flips if and
only if the last n flips were all heads. Otherwise, it continues
flipping. For example, each of these would be complete flipping
sequences:
HH
TH
HTHH
HTTHHH
TTHTHHTHHHHHHH
What is the probability that the machine eventually halts?
P(H) approaches 1 as a limit but never actually reaches the limit.
Let P(n) is probability of halting in 2n flips/if not halted in
2(n-1) flips.
Easily P(n) = 1/(2^n). Sum{P(n); n = 1,2,3, ... < 1
Bill J
Whoops|
PP(n) = P(n)*(the probability that the sequence is not already
terminated}; ie,
PP(1) = P(1)*(1 - P(0)) = .5
PP(2) = P(2)*(1 - P(1)) = .25 * (1 - .5) = .125PP(3) = P(3)*(1 - .625) I think?
Bill J
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Brian Tung <br...@xxxxxxx>
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