# Probability question on fair coin

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1. May 5, 2015

### ichabodgrant

1. The problem statement, all variables and given/known data
A fair coin is continually flipped. What is the probability that the pattern T,H occurs before the pattern H,H, where T and H respectively denote Tail and Head of a coin?

2. Relevant equations
Prob. = (n r) (pr)(1-p)n-r

3. The attempt at a solution

i. prob. of TH vs HT vs TT vs HH
ii. THHHHHHH...or TTHHHHHH... or HTHHHHHH... or HTTHHHHH... or HTTTTTHHH... or ....

If it is the 2nd one, how can we calculate the prob.? It looks like an infinitely long series... or should I let there be n trials? Then I use binomial distribution to find?

Last edited: May 6, 2015
2. May 5, 2015

### haruspex

Is there any reason why one of the two results is more likely than the other?

3. May 5, 2015

### ichabodgrant

I don't quite understand your question...
The prob. of getting a head or a tail is 1/2? Is this what you are asking?

4. May 5, 2015

### haruspex

I'm asking whether you can think of any reason why getting HT is more or less likely than getting TH.

5. May 6, 2015

Same?

6. May 6, 2015

Yes.

7. May 6, 2015

### ichabodgrant

So simply calculate the first case?

8. May 6, 2015

### haruspex

What will the two probabilities add up to?

9. May 6, 2015

### ichabodgrant

Which two? TH and HT?

0.5*0.5 + 0.5*0.5 = 0.5?

10. May 6, 2015

### haruspex

No.
The two events are,
• that TH occurs before HT in an arbitrarily long sequence,
• that HT occurs before TH in an arbitrarily long sequence
What must those two probabilities add up to?

11. May 6, 2015

### ichabodgrant

1?

12. May 6, 2015

### ichabodgrant

but it seems 0.5*0.5*0.5*... , gets you 0...

13. May 6, 2015

### haruspex

Yes.
What exactly is the event for which that calculation applies?

14. May 6, 2015

### ichabodgrant

Oh Sorry...

I have a typo in the question...
The correct question is "T,H before HH"

15. May 6, 2015

### ichabodgrant

Should I indeed use conditional probability?
In case I have a T at the 1st trial, then it already achieves the event that TH appears before HH?

16. May 6, 2015

### haruspex

Ok, that makes it a lot more interesting.
Assign unknowns to those two probabilities. Consider the first two tosses. There are four situations at that point, equally likely. Consider the probabilities of the two outcomes of interest when continuing from each of those four positions.
See what equations you can extract.

17. May 6, 2015

### ichabodgrant

Actually TT, HT,TH all means TH must occur before HH, right?

18. May 6, 2015

### ichabodgrant

so if there is no HH, then it is alright?
Let the probability of getting a H be p.

Then the required prob. is 1 - p2?

19. May 6, 2015

### haruspex

What do you mean "it's alright"?
There are four equally likely states after two tosses:
HH
HT
TH
TT
In two of those, the outcome is already determined, yes?
Look at the remaining two. Can you predict what the eventual outcome will be for those?

20. May 6, 2015

### ichabodgrant

If it is HT, then no matter the next one is T or H, TH already occurs before HH, right?
If it is TT, then also no matter the next one is T or H, TH already occurs before HH, right?
If it is TH, then obviously no matter the next one is T or H, TH already occurs before HH, right?
Just the case if you obtain all H and no T at the first 2 tosses will give you HH before TH..

I am thinking in this way