Probability question related to dice

  • #1
utkarshakash
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Homework Statement


A die is thrown n times(n being odd). The probability that even face turns odd number of times is

Homework Equations



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
[itex]\dfrac{1}{2^n} [^nC_1+^nC_3+^nC_5+............+^nC_n][/itex]

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

Answers and Replies

  • #2
3,812
92

Homework Statement


A die is thrown n times(n being odd). The probability that even face turns odd number of times is

Homework Equations



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
[itex]\dfrac{1}{2^n} [^nC_1+^nC_3+^nC_5+............+^nC_n][/itex]

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

Multiply and divide by 2. Then investigate the expansion of ##(1+x)^n## and ##(1-x)^n##.
 
  • #3
haruspex
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Homework Statement


A die is thrown n times(n being odd). The probability that even face turns odd number of times is
What relationships can you find between that probability and the probability that even faces appear an even number of times?
 
  • #4
utkarshakash
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What relationships can you find between that probability and the probability that even faces appear an even number of times?

They both add up to 1. That's what I can think.
 
  • #5
haruspex
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They both add up to 1. That's what I can think.
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
tiny-tim
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hi utkarshakash! :smile:

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? :wink:
 
  • #7
utkarshakash
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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?

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
tiny-tim
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hi utkarshakash! :smile:

ah, but your original question was …
But I don't know how to sum this series.

… and that's what we've all been talking about!
… 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.

and that answer is … ? :wink:
 
  • #9
utkarshakash
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hi utkarshakash! :smile:

ah, but your original question was …


… and that's what we've all been talking about!


and that answer is … ? :wink:

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

The correct answer is [itex]\dfrac{1.3.5.7.........(2n-1)}{2^n n!}[/itex]
 
  • #10
haruspex
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[itex](1+x)^n-(1-x)^n=2(C_1x+C_3x^2......) \\
2^{n-1}=C_1+C_3+C_7......[/itex]
So the answer should be [itex]2^{n-1}/2^n[/itex] which unfortunately is incorrect.

The correct answer is [itex]\dfrac{1.3.5.7.........(2n-1)}{2^n n!}[/itex]
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
utkarshakash
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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?

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
Ray Vickson
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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.

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
[tex]0 = (1-1)^n = C(n,0)-C(n,1)+C(n,2)- \cdots = S_E - S_O [/tex]
and
[tex]2^n = (1+1)^n = C(n,0) + C(n,1) + C(n,2) + \cdots = S_E + S_O.[/tex]
Therefore, ##S_E = S_O = 2^{n-1}##.
 
  • #13
utkarshakash
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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
[tex]0 = (1-1)^n = C(n,0)-C(n,1)+C(n,2)- \cdots = S_E - S_O [/tex]
and
[tex]2^n = (1+1)^n = C(n,0) + C(n,1) + C(n,2) + \cdots = S_E + S_O.[/tex]
Therefore, ##S_E = S_O = 2^{n-1}##.

I did not ignore it. I followed this approach to reach at the answer. You should read my previous post.
 
  • #14
tiny-tim
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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) :wink:
 

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