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Topic: Nice Dice (Read 779 times) |
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Noke Lieu
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Once again, my lack of probability skills leaves me wanting...
I created a nice question that has left me scratching my head. Trying to *see* the answer resulted in gibberish. 
What's the minimum number of dice that one needs to roll to be >50% certain that there's going ot be at least one of each 1 ,2, 3 , 4, 5, 6?
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ThudnBlunder
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Re: Nice Dice
« Reply #1 on: Feb 17th, 2010, 1:01am » |
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The expected number of throws = 6#SIGMA((1/r) for r = 1 to 6
This comes to 14.7 throws.
Sometimes we will need more, sometimes less.
On average we will need more half the time and less half the time.
So I guess 15 is the answer to your question.
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towr
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Re: Nice Dice
« Reply #2 on: Feb 17th, 2010, 1:04am » |
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The probability of having at least one of each after n throws is 1 + 0n - 5*2(2 - n) + 5*3(1 - n) + 5*2n*3(1 - n) - 6(1 - n) - 5n*6(1 - n)
(Kindly relying on quickmath.com's matrix exponentiation. The 0n is solely there to get the right answer for 0 throws, so its a bit superfluous.)
[edit]13 throws is enough, this gives you a probability of 485485/944784 ~= 0.5139[/edit]
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| « Last Edit: Feb 17th, 2010, 1:14am by towr » |
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Grimbal
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Re: Nice Dice
« Reply #3 on: Feb 17th, 2010, 1:26am » |
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I got the same as towr.
f(n,k) = prob of having k different values in n throws.
f(0,0) = 1, f(0,k) = 0, f(n,0) = 0
f(n,k) = f(n-1,k-1)*(6-(k-1))/6 + f(n-1,k)*k/6
I.e. on the last throw, either you get a new value (left term) or you don't (right term).
(Kindly relying on Excel's formula evaluation)
I get f(13,6) = 0.51386
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| « Last Edit: Feb 17th, 2010, 1:29am by Grimbal » |
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towr
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Re: Nice Dice
« Reply #4 on: Feb 17th, 2010, 1:51am » |
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If you rewrite my formula as
1*(0/6)n - 6*(1/6)n + 15*(2/6)n - 20* (3/6)n + 15*(4/6)n - 6*(5/6)n + 1* (6/6)n
it's interesting to note that the coefficients form a row from pascal's triangle, 1 6 15 20 15 6 1, except with alternating sign.
Which suggest it might easily be generalized to k-sided dice, and also that there is probably an easier way to find the formula.
[edit]
For k-sided dice, p = S(n,k)*k!/kn, where S(n,k) is a Stirling number of the second kind
[/edit]
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| « Last Edit: Feb 17th, 2010, 3:28am by towr » |
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ThudnBlunder
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Re: Nice Dice
« Reply #5 on: Feb 18th, 2010, 12:12pm » |
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Ah, so a Monte Carlo simulation would give a skewed distribution about the mean of 14.7?
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ThudnBlunder
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Re: Nice Dice
« Reply #7 on: Feb 19th, 2010, 4:17am » |
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on Feb 17th, 2010, 1:01am, ThudanBlunder wrote:
On average we will need more half the time and less half the time.
So I guess 15 is the answer to your question.[/hide] |
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I obviously didn't spend too much time on that answer. 
It is clear from towr's graph why the distribution must be skewed to the left.
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