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mathworld.wolfram.com/CoinTossing.html

At 1) , how did they get 9/14 for the probability that THTH occurs before HTHH ?

- Thread starter SeekerofMath
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- #1

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mathworld.wolfram.com/CoinTossing.html

At 1) , how did they get 9/14 for the probability that THTH occurs before HTHH ?

- #2

tiny-tim

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Hi SeekerofMath! Welcome to PF!

By working it out!

mathworld.wolfram.com/CoinTossing.html

At 1) , how did they get 9/14 for the probability that THTH occurs before HTHH ?

- #3

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I'd like some direction...

- #4

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if there were only 4 tossing THTH and HTHH would be equally likely but if you consider infinite tossings then things changes:I'd like some direction...

Now through this infinite serie you will find situations like:

THTHTH (two appearances of THTH in 6 tosses)

Which is not possible for HTHH in just 6 tosses and the best you can do is

HTHHTHH (two appereances of HTHH in 7 tosses)

This asymmetry makes THTH more likely to appear before HTHH, anyway, the actual calculations should be in the reference given by the article you posted, that is (Gardner 1988, p. 64)

Good Luck :)

- #5

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Consider 10 states, X0 to X7, Y, Z.

X0 to X7 represent the outcomes of the last three tosses, TTT, TTH, .. , HHH.

Y, Z represent THTH, HTHH having occurred, first, at some point, respectively.

We take as initial states X0 to X7 equally likely (i.e. result of first 3 tosses).

Thereafter, each toss gives a transition to a new state. If we ever enter Y or Z we stay there. This gives a 10x10 transition matrix.

Extracting the eigenvalues allows us to find the long term odds of being in state Y or Z.

- #6

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Indeed there is...No idea if there's a better way,

Let p0 be the prob of THTH before HTHH if the last 3 were TTT.

Let p1 be the prob of THTH before HTHH if the last 3 were TTH.

etc. to

Let p7 be the prob of THTH before HTHH if the last 3 were HHH.

By considering the next toss in each case, we get:

p0 = p0/2 + p1/2

p1 = p2/2 + p3/2

p2 = 1/2 + p4/2

p3 = p6/2 + p7/2

p4 = p0/2 + p1/2

p5 = p2/2

p6 = p4/2 + p5/2

p7 = p6/2 + p7/2

This simplifies quite easily to give

p0 = p1 = p4 = 5/7

p2 = 6/7

p3 = p6 = p7 = 4/7

p5 = 3/7

The overall probability of THTH before HTHH is the average of these, 9/14.

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