Probability question related to dice

  • #1
utkarshakash
Gold Member
855
13

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
Saitama
4,244
93

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
Science Advisor
Homework Helper
Insights Author
Gold Member
38,453
7,965

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
Gold Member
855
13
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
Science Advisor
Homework Helper
Insights Author
Gold Member
38,453
7,965
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
Science Advisor
Homework Helper
25,838
255
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
Gold Member
855
13
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
Science Advisor
Homework Helper
25,838
255
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
Gold Member
855
13
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
Science Advisor
Homework Helper
Insights Author
Gold Member
38,453
7,965
[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
Gold Member
855
13
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
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
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
Gold Member
855
13
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
Science Advisor
Homework Helper
25,838
255
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:
 

Suggested for: Probability question related to dice

Replies
11
Views
131
Replies
2
Views
561
Replies
3
Views
633
Replies
7
Views
420
  • Last Post
Replies
22
Views
677
  • Last Post
Replies
1
Views
875
Replies
4
Views
1K
Replies
8
Views
181
Replies
11
Views
680
Top