Calculating Probability of Sixth Die Face on Nth Consecutive Throw

  • Thread starter LittleWolf
  • Start date
In summary: I think this is the right answer. In summary, the probability of getting the desired result is 5/324, or 1 in 24.
  • #1
LittleWolf
38
0
Does anyone know how to determine the probability that the sixth side of a fair die will appear on the Nth consecutive throw.
 
Physics news on Phys.org
  • #2
I suggest you sit down and try to reason this through, but it's just

[tex]P(N)=\frac{1}{6^N}[/tex]
 
  • #3
The problem I was trying to ask was " You throw a die until all six faces have shown up and then you stop." What is the probability that you stop on the Nth throw. I believe that P(N=6)=6!/(6^6) and P(N<6)=0 but I am confused about the probability of stopping for N>1.
 
  • #4
I don't expect that to have a simple answer. It's easy enough to answer:

"What is the probability of getting u 1's, v 2's, w 3's, x 4's, y 5's, and z 6's in N (= u+v+w+x+y+z) throws"

But then to get the answer to your problem, you have to add up a lot of individual cases. :frown:
 
  • #5
Counting the number of ways that all the faces can come up on n rolls is the same as counting the number of onto functions from {1, 2, ... n} to {1, 2, 3, 4, 5, 6}, since each time you roll the die you effectively map the number of the roll to one of the faces on the die.
 
Last edited:
  • #6
There is a formula for counting the number of onto functions.
 
  • #7
I think a good strategy would be counting the number of ways to get at least 1 of each of 5 numbers in N-1 throws and exactly 0 of the last. This way order doesn't matter, which simplifies the problem (I think/hope). This should be exactly six times P(N).
 
Last edited:
  • #8
Musing aloud

Let [tex]F_k(n)[/tex] represent the number of ways a die with [tex]k[/tex] equiprobible faces can land once on all but one faces at least once, with the last face never coming up, out of [tex]n+k-1[/tex] throws. Clearly the probability of the desired event is [tex]P(n)=\frac{F_6(n-6)}{6^n}[/tex] as long as n>=6.

[tex]F_6(0)=6\cdot5![/tex], since there are 5! ways for the 5 objects to be chosen once each, and 6 ways to choose which element is last. This brings us to P(6)=6!/6^6=5/324, as LittleWolf pointed out.

In fact, in general [tex]F_k(0)=\frac{k!}{k^k}[/tex], since each of the faces other than the last must be landed on exactly once.

[tex]F_2(n)=2[/tex], since either all the results are 1 or all the results are 2.

[tex]F_3(n)=3\left(\sum_{l=1}^{n+1}{n+2\choose l}\right)=3\left(\sum_{l=0}^{n+2}{n+2\choose l}-{n+2\choose0}-{n+2\choose n+2}\right)=3\left(2^{n+1}-2\right)[/tex]

Is there some good pattern, or some Bell-related OEIS sequence I'm missing here?
 
  • #9
I said there was already a formula for counting the number of onto functions. Given that LittleWolf was asked the problem, he probably already has that formula. I looked it up and this is it, using ^ for exponentiation and C(a, b) for "a choose b":

The number of onto functions from a set of size m to a set of size n is equal to the sum for i = 0 to n of ((-1)^i * C(n, i) * (n - i)^m).

So to find the answer to LittleWolf's question, you substitute N for m and 6 for n in that expression.
 
  • #10
BicycleTree said:
The number of onto functions from a set of size m to a set of size n is equal to the sum for i = 0 to n of ((-1)^i * C(n, i) * (n - i)^m).

So to find the answer to LittleWolf's question, you substitute N for m and 6 for n in that expression.

First: Thanks for the formula, I couldn't think of that one. Actually, I'm not sure if I even knew of it before... it looks vaguely familliar, but no more. If you don't mind I'll [tex]\TeX[/tex] it so I can follow it a little better...

[tex]\sum_{i=0}^{n}(-1)^i{n\choose i}(n-i)^m=\sum_{i=0}^{n-1}(-1)^i{n\choose i}(n-i)^m[/tex]

[tex]\sum_{i=0}^{6}(-1)^i{6\choose i}(6-i)^N= {6\choose 0}6^N-{6\choose 1}5^N+{6\choose 2}4^N-{6\choose 3}3^N+{6\choose 4}2^N-{6\choose 5}= 6^N-6\cdot5^N+15\cdot4^N-20\cdot3^N+15\cdot2^N-6[/tex]

OK... this is the number of ways to have all 6 faces show up by the Nth turn. The probability of it turning up on the Nth turn is then
[tex]P(N)=6^{-N}\left((6^N-6\cdot5^N+15\cdot4^N-20\cdot3^N+15\cdot2^N-6)-(6^{N-1}-6\cdot5^{N-1}+15\cdot4^{N-1}-20\cdot3^{N-1}+15\cdot2^{N-1}-6)\right)[/tex]

Right?
 
  • #11
Looks right to me.

I think the reason it might be familiar to you is that (so says my book) the summation is a Stirling number of the second kind, S(m, n), multiplied by n!.
http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html
The Stirling numbers count the number of ways to divide a set of size m into n nonempty sets, so to count functions you want to have the order of the n sets important so you multiply by n!.
 
Last edited:
  • #12
BicycleTree said:
Looks right to me.

I think the reason it might be familiar to you is that (so says my book) the summation is a Stirling number of the second kind, S(m, n), multiplied by n!.

I thought that Stirling numbers (of the second kind; I've never used Stirling numbers of the first kind) should come up, but I didn't actually follow through with that.

Thanks for the formula!
 
  • #13
Modifying your notation, the probability P(N,6) is the cumulative distribution function. The probability that the last face shows up on the Nth throw is 6*P(N,5). If you consider the recursion formula for Stirling#s of the 2nd kind, you can see why P(N,6) is the CDF. Thanks to everyone for clearing up this problem.
 
  • #14
a related problem is
what is the the expected number of throws to get a six on a fair dice

this was raised at
https://www.physicsforums.com/archive/topic/t-45262_The_expected_value_of_a_Geometric_Series.html
which seems to be closed now.

yes the answer is 6 but how do you prove it.

if p=1/6 and q=1-p
x is the number of throws

then E(x) is sum(x.P(X=x)) = sum(x.p.q^(x-1))
the sums are for x=1 to infinity
i.e. E(x) is p(1 +2q +3q^2 +4q^3+...)

let S= E(x)/p = 1 +2q +3q^2 +4q^3+...

rearranging the sum as
S=
1 + q + q^2 + q^3 + ...
+ q + q^2 + q^3 + ..
+ q^2 + q^3 +...
then rearrange these lines as

S= 1 ( 1 + q + q^2 + q^3 + ...)
+q (1 + q + q^2 + q^3 + ...)
+q^2 ( 1 + q + q^2 + q^3 + ...)
...

so S= (1 + q + q^2 + q^3 + ...)*(1 + q + q^2 + q^3 + ...)

i.e S= (1 + q + q^2 + q^3 + ...)^2
this geometric series sums to 1/(1-q)

i.e. S= (1/(1-q))^2
i.e. = 1/p^2
So E(x) =pS = 1/p
in the case of six sided die, p=1/6, E(x) =6


this solutiuon found at http://math.berkeley.edu/~mchrist/Math55/Lectures/L28.pdf [Broken]

probably a good source several problems in this forum
 
Last edited by a moderator:

1. What is the probability of rolling a specific number on a six-sided die?

The probability of rolling a specific number on a six-sided die is 1/6 or approximately 16.67%. This is because there are six possible outcomes (each face of the die) and only one of them will result in the desired number.

2. How many different combinations can be rolled on a six-sided die?

There are 6 possible outcomes for each roll of a six-sided die. Therefore, there are 6^6 (or 46,656) different combinations that can be rolled on a six-sided die.

3. What is the expected value of rolling a six-sided die?

The expected value of rolling a six-sided die is 3.5. This is calculated by taking the sum of all possible outcomes (1+2+3+4+5+6) and dividing it by the number of outcomes (6).

4. Can the outcomes on each face of a die be predicted?

No, the outcomes on each face of a die cannot be predicted. The outcome of each roll is completely random and is not affected by previous rolls or external factors.

5. How does the shape of a die affect the probability of rolling a specific number?

The shape of a die, specifically the number of faces it has, is directly related to the probability of rolling a specific number. The more faces a die has, the lower the probability of rolling a specific number. For example, a ten-sided die has a lower probability of rolling a 6 compared to a six-sided die.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
20
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
7
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
20
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
2K
  • Precalculus Mathematics Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
31
Views
3K
  • Precalculus Mathematics Homework Help
Replies
12
Views
2K
Replies
1
Views
453
Back
Top