Finding the probabiliy that event THTH occurs before HTHH

  • #1
I was refreshing myself of the basics of probability on wolfram for coin toss.
mathworld.wolfram.com/CoinTossing.html

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

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi SeekerofMath! Welcome to PF! :smile:
I was refreshing myself of the basics of probability on wolfram for coin toss.
mathworld.wolfram.com/CoinTossing.html

At 1) , how did they get 9/14 for the probability that THTH occurs before HTHH ?
By working it out! :rolleyes:

Have a go … show us what you get. :smile:
 
  • #3
I'd like some direction...
 
  • #4
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I'd like some direction...
if there were only 4 tossing THTH and HTHH would be equally likely but if you consider infinite tossings then things changes:

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
haruspex
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No idea if there's a better way, but here's a method which I believe would give the right answer.
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
haruspex
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No idea if there's a better way,
Indeed there is...:blushing:

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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