# Probability Coin Toss

1. Sep 4, 2011

### ArcanaNoir

1. The problem statement, all variables and given/known data
A coin is tossed until the same result appears twice in a row. Find the probability that this event occurs on the nth toss.

(the answer is $\frac{1}{2^{n-1}}$)

2. Relevant equations

3. The attempt at a solution

I made a tree and first only considered the results of the 2, 3, 4..ect toss that did not have the same result twice in a row. I kept getting 1/2. When I included ALL the results on the 3, 4, 5 etc toss, I did not get something equivalent to the answer.

2. Sep 4, 2011

### Dickfore

In the first n - 1 tosses, what must the distribution of outcomes be?

3. Sep 4, 2011

4. Sep 4, 2011

### Dickfore

That is true, but not sufficient. Can you have:

HHHTTT

for example?

EDIT:
Actually, if n - 1 is odd, it isn't even true.

5. Sep 4, 2011

Of course

6. Sep 4, 2011

### Dickfore

Of course what?

7. Sep 4, 2011

### ArcanaNoir

Of course you can have HHHTTT

8. Sep 4, 2011

### ArcanaNoir

but if you toss unti lyou get the same twice in a row, maybe you cant have hhhttt

9. Sep 4, 2011

### Dickfore

So, can you or can't you? Think before you type.

10. Sep 4, 2011

### ArcanaNoir

Okay how about I don't know? When I tried both ways I still got the wrong numbers.

11. Sep 4, 2011

### Dickfore

Lol. So, let's say that in the 7th toss you got the same outcome as in the 6th toss, but never before that. What are the possible ways of tossing the coin (in all previous trials)?

12. Sep 4, 2011

### ArcanaNoir

HTHTHTT or THTHTHH

13. Sep 4, 2011

### Dickfore

True. What is the probability that any of these events will occur (use conditional probability)?

14. Sep 4, 2011

### ArcanaNoir

(2/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2) ?

15. Sep 4, 2011

### Dickfore

Why do you have 2/2 in the first factor?

16. Sep 4, 2011

### ArcanaNoir

Cause we could start either heads or tails.

17. Sep 4, 2011

### Dickfore

Oh, so you went one step beyond my question and evaluated the TOTAL probability of the two union of the two favorable events. Ok, so if you count the powers of 2 in the denominator, you will get 1/26.

Can you generalize this pattern if we change 7 in our example with a general integer n? What do you get?

18. Sep 4, 2011

### ArcanaNoir

I got, it right? I easily see how (2/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2) = $\frac{1}{2^{7-1}}$

19. Sep 4, 2011

### Dickfore

Yes, so the point is that if you knew that you had a double toss in the nth trial and you know what you had, you can go back and trace your whole tossing history. But, you can have HH or TT as a double toss, so this increases your favorable outcomes by a factor of 2.

Note: It is not that easy to do it with a dice, for example.

20. Sep 4, 2011

### ArcanaNoir

thank you very much for the help!