Homework Help: 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 am thinking whether the question asks about:

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

21. May 6, 2015

haruspex

Yes.

22. May 6, 2015

ichabodgrant

so the probability of getting HH is p2?
Then the required probability is 1- p2?

23. May 6, 2015

haruspex

Yes, but can't you assume it's a fair coin?

24. May 6, 2015

ichabodgrant

then 1 - 0.52 = 3/4

25. May 6, 2015

ichabodgrant

thank you for your kindness