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Roswell
04-22-2005, 02:07 PM
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.

Roswell
04-22-2005, 02:09 PM
and I mean the chance of rolling AT LEAST one double.

BruceZ
04-22-2005, 02:35 PM
[ 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 &lt; 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.

Roswell
04-22-2005, 02:36 PM
thank you for your timely response, Bruce.

Roswell
04-22-2005, 03:52 PM
Sorry to bring this up again: What about the probability of making EXACTLY ONE double when rolling n dice?

BruceZ
04-23-2005, 02:16 AM
[ 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 &lt; 8)

6*C(n,2)*P(5,n-2)/6^n

For n &gt;=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)!

BruceZ
04-23-2005, 11:12 PM
[ 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)!

Roswell
04-25-2005, 10:24 AM
thank you once again Bruce.