Suppose you throw 4 identical dice at once. What is the probability that at least one pair will be add to 7 or 11?
An exact calculation or proof rather than a simulation is preferred but I welcome either.

Larry

I was looking for an elegant solution and couldn't come up with one. By brute force I came up with a probability of 56%:

There are 1296 possible combinations.
Looking first at getting one or more sevens:

Die 1 Die 2 Die 3 Die 4 combinations
1 6 - - 25
1 - 6 - 25
1 - - 6 25
1 6 6 - 5
1 6 - 6 5
1 - 6 6 5
1 6 6 6 1

Total 91

Total for die 1 having values from 1-6, where die 2-4 are chosen to add to 7 : 1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, and 6 and 1.
91(6) = 546

For total to equal 11, the dice must be 6 and 5 or 5 and 6;
91(2) = 182

546 + 182 = 728

Total is 728

728/1296 = 0.5617 or about 56%

I think that's right but these probability problems can be elusive.

Larry

Larry, when you say at least one pair adding up to 7 or 11, does that mean that 3 or 4 dice adding up to 7 or 11 is also acceptable? It will make a difference in the final calculation.

Scott,
I mean there needs to be at least one pair that adds to 7 or 11
So 2,3,6,2 would not be accepted even though 2,3,and 2 add to 7 and 3,6, and 2 add to 11
Only pairs of numbers would count

acceptable example: 1,6,5,5
2,5,6,1
3,4,6,6
5,6,4,4

I see...so we're focusing on pairs of numbers exclusively, and it's alright if we have a second pair which adds up to 7 and 11. Thanks for the clarification! Let me ponder this...

Larry, I know it's not a very mathematical way of doing it, but a quick 10 minutes with a spreadsheet gives me:

834/1296 combos that yield at least one SEVEN pair (64.4%)
302/1296 combos that yield at least one ELEVEN pair (23.3%)
920/1296 combos that yield at least one SEVEN or ELEVEN pair

so that's about 71% chance of some combination adding up to 7 or 11.

The probability calculations got too comlicated quickly, so I opted for brute force: http://jsfiddle.net/68zk4Lhk/

In that run, any time there's one or more pair of numbers, that only counts as 1 positive result. For example, (2, 6, 5, 1) = 7: [2+5], [6+1], 11: [6+5], means that when 2, 6, 5, and 1 is rolled, you could use 2+5 to make 7, 6+1 to make 7, or 6+5 to make 11, but all 3 together only count as 1 additional result (since they all derive from a single roll).

By that count, 920 of the 1,296 possible rolls contain a pair of dice which can sum to either 7 or 11. That means there's a 70.98765432% (call it a "just under 71%" chance) that you'll have at least 1 pair of dice which sum to 7 or 11.

WilburrUK and Scott Cram. Thanks so much for all your efforts. This confirms my experience that these combination problems are hard to attack by the usual methods.

Larry