# Probability question related to dice

1. Oct 20, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
A die is thrown n times(n being odd). The probability that even face turns odd number of times is

2. Relevant equations

3. The attempt at a solution
The probability that even face appears is 1/2 and odd face appears is 1/2. From any of the n throws, the even face can appear 1, 3, 5 ..... or n times. Thus the required probability is
$\dfrac{1}{2^n} [^nC_1+^nC_3+^nC_5+............+^nC_n]$

But I don't know how to sum this series.

2. Oct 20, 2013

### Saitama

Multiply and divide by 2. Then investigate the expansion of $(1+x)^n$ and $(1-x)^n$.

3. Oct 21, 2013

### haruspex

What relationships can you find between that probability and the probability that even faces appear an even number of times?

4. Oct 21, 2013

### utkarshakash

They both add up to 1. That's what I can think.

5. Oct 21, 2013

### haruspex

Yes, they add up to 1, but there's another relationship. Try rephrasing one of the two conditions in terms of probability of a number of odd faces. Can you see a useful symmetry?

6. Oct 21, 2013

### tiny-tim

hi utkarshakash!

as a general rule, if you're stuck, try a simple case and see if you can see a pattern

eg what is the answer for n = 3?

7. Oct 22, 2013

### utkarshakash

I am not able to make out what is wrong in my original solution. For eg- If I take n=3 (as suggested by tiny-tim) then I have to find the probability that even face appears either 1 or 3 times.If it appears 1 times out of 3 throws I can write 3C1(1/2)^3. SUppose instead that if it appears 3 times out of 3 throws, the probability will be 3C3(1/2)^3. Adding the 2 probabilities should give the correct answer.

8. Oct 22, 2013

### tiny-tim

hi utkarshakash!

ah, but your original question was …
… and that's what we've all been talking about!
and that answer is … ?

9. Oct 22, 2013

### utkarshakash

$(1+x)^n-(1-x)^n=2(C_1x+C_3x^2......) \\ 2^{n-1}=C_1+C_3+C_7......$
So the answer should be $2^{n-1}/2^n$ which unfortunately is incorrect.

The correct answer is $\dfrac{1.3.5.7.........(2n-1)}{2^n n!}$

10. Oct 22, 2013

### haruspex

If that's the answer then I don't think I have understood the question.
The reasoning I was trying to lead you towards is:
- prob of odd number of even throws equals prob of odd number of odd throws, by symmetry
- prob of odd number of odd throws equals prob of even number of even throws since they are the same event
- prob of odd number of even throws and prob of even number of even throws add to 1
Putting those three together we get 1/2.
So I strongly suspect the problem has been misinterpreted somewhere along the line. Is it the exact wording? A translation? Should it be 2n throws?

11. Oct 22, 2013

### utkarshakash

Oh! I'm really sorry as I've made a mistake in posting the correct answer. Actually the correct answer for this question is 1/2 and the answer posted here is the answer of some other question. I apologize for any inconvenience that may have occurred to you.

12. Oct 22, 2013

### Ray Vickson

Since people have already told you the answer, I might as well explain one of the other approaches that was suggested to you, but which you seemed to ignore. Let $S_E = \sum{k} C(n,2k)$ and $S_O = \sum_{k} C(n,2k+1)$ be the even and odd sums. We have
$$0 = (1-1)^n = C(n,0)-C(n,1)+C(n,2)- \cdots = S_E - S_O$$
and
$$2^n = (1+1)^n = C(n,0) + C(n,1) + C(n,2) + \cdots = S_E + S_O.$$
Therefore, $S_E = S_O = 2^{n-1}$.

13. Oct 22, 2013

### utkarshakash

I did not ignore it. I followed this approach to reach at the answer. You should read my previous post.

14. Oct 22, 2013

### tiny-tim

another way to show that the odd sum equals the even sum … which is clearer if you use 2n+1 instead of n … is by using the symmetry of nCi:

2n+1C2i+1 = ∑ 2n+1C2(n-i)