#1
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Probability of rolling a double with n dice
Can someone point out the method to compute the probability of rolling a double when you have "n" dice? i.e. If you have 2 dice, the probability is 1/6, if you have three dice, it is...? and if you have 7 dice, the probability must be 1.
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#2
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Re: Probability of rolling a double with n dice
and I mean the chance of rolling AT LEAST one double.
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#3
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Re: Probability of rolling a double with n dice
[ QUOTE ]
Can someone point out the method to compute the probability of rolling a double when you have "n" dice? i.e. If you have 2 dice, the probability is 1/6, if you have three dice, it is...? [/ QUOTE ] This is 1 minus the probability of all dice being different. For 3 dice: 1 - 6*5*4/6^3 = 44.4% For 4 dice: 1 - 6*5*4*3/6^4 For 5 dice: 1 - 6*5*4*3*2/6^5 For 6 dice: 1 - 6*5*4*3*2*1/6^6 In general for n < 7: 1 - P(6,n)/6^n Where P(6,n) is permutations of 6 things taken n at at time, or 6!/(6-n)! [ QUOTE ] and if you have 7 dice, the probability must be 1. [/ QUOTE ] Correct. |
#4
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Re: Probability of rolling a double with n dice
thank you for your timely response, Bruce.
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#5
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Re: Probability of rolling a double with n dice
Sorry to bring this up again: What about the probability of making EXACTLY ONE double when rolling n dice?
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#6
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Re: Probability of rolling a double with n dice
[ QUOTE ]
Sorry to bring this up again: What about the probability of making EXACTLY ONE double when rolling n dice? [/ QUOTE ] There are C(n,2) ways to pick the 2 dice that pair, times 6 possible pairs, times the number of ways the remaining dice can be different. Divide by 6^n total ways to throw the dice. Note that this considers 3-of-a-kind as more than 1 pair. 3 dice: 6*C(3,2)*5/6^3 = 6*(3*2/2)*5/6^3 = 41.7% 4 dice: 6*C(4,2)*5*4/6^4 = 6*(4*3/2)*5*4/6^4 = 55.6% 5 dice: 6*C(5,2)*5*4*3/6^5 = 6*(5*4/2)*5*4*3/6^5 = 46.3% 6 dice: 6*C(6,2)*5*4*3*2/6^6 = 6*(6*5/2)*5*4*3*2/6^6 = 23.1% 7 dice: 6*C(7,2)*5*4*3*2*1/6^7 = 6*(7*6/2)*5*4*3*2*1/6^7 = 5.4% General (n < 8) 6*C(n,2)*P(5,n-2)/6^n For n >=8 there must be at least 2 pairs (counting three-of-a-kind as more than 1 pair). Where C(n,2) is combinations of n things taken 2 a time = n!/(n-2)!/2! and P(5,n-2) is permutations of 5 things taken n-2 at a time = 5!/(n-2)! |
#7
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correction
[ QUOTE ]
and P(5,n-2) is permutations of 5 things taken n-2 at a time = 5!/(n-2)! [/ QUOTE ] P(5,n-2) = 5!/[5-(n-2)]! = 5!/(7-n)! |
#8
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Re: correction
thank you once again Bruce.
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