# what is the probability that if 5 fair 6-sided dice are rolled the product is 4

what is the probability that if 5 fair 6-sided dice are rolled the product of the numbers is 4

 ebaines Posts: 10,033, Reputation: 5529 Expert #11 Jul 2, 2012, 08:53 AM
You're missing quite a few combinations for the 1,1,1,2,2 set: such as 1,1,2,1,2. Don't forget that the 2's don't have to be next to each other.

Yes, there is a formula for calculating this: for the 1,1,1,2,2 set you can calculate the number of ways that the two 2's can be arranged in a set of using using combinations C(5,2), where

$
C(a,b)= \frac {a!}{b!(a-b)!}
$

Try that and tell us what you get. You should also try calculating the number of ways that the three 1's can be arranged in a set of 5 - the answer should be the same! I would also suggest that you try writing out all the combinations by hand (don't forget to include all the cases where the 2's aren't together) and see if it all agrees.
 833student Posts: 9, Reputation: 1 New Member #12 Jul 2, 2012, 09:01 AM
Quote:
 Originally Posted by ebaines You're missing quite a few combinations for the 1,1,1,2,2 set: such as 1,1,2,1,2. Don't forget that the 2's don't have to be next to each other. Yes, there is a formula for calculating this: for the 1,1,1,2,2 set you can calculate the number of ways that the two 2's can be arranged in a set of using using combinations C(5,2), where $ C(a,b)= \frac {a!}{b!(a-b)!}$ Try that and tell us what you get. You should also try calculating the number of ways that the three 1's can be arranged in a set of 5 - the answer should be the same! I would also suggest that you try writing out all the combinations by hand (don't forget to include all the cases where the 2's aren't together) and see if it all agrees.
so would it be 5C2= 10 for the 2's and 5C1= 5 for the 1's and 5C4=5 for the 4's and then add together?
 ebaines Posts: 10,033, Reputation: 5529 Expert #13 Jul 2, 2012, 09:06 AM
Getting close. Both 5C2 and 5C3 = 10, which both are the number of ways that 1,1,1,2,2 can be arranged - so you only need one of them. And ether 5C1 or 5C4 =5 gives the number of ways that 4,1,1,1,1 can be arranged. So all you need is to add together 5C2 + 5C1. Again - I suggest you write out all combinations to see if it's correct.
 833student Posts: 9, Reputation: 1 New Member #14 Jul 2, 2012, 11:47 AM
so the answer would be 15. What if i changed it to be 4 fair dice with a product of 12? The numbers involved are
3*2*2*1
3*4*1*1
2*6*1*1
would the combination formula be as follows?
4C3+4C2=10?
 ebaines Posts: 10,033, Reputation: 5529 Expert #15 Jul 2, 2012, 11:59 AM
No. The problem is that once you try to address how three different outcomes can be arranged it's a bit more complicated. For example: the set 3, 2, 2, 1 can be arranged as follows:

3,2,2,1
3,2,1,2
3,1,2,2
2,3,2,1
2,3,1,2
2,2,3,1
2,2,1,3
2,1,3,2
2,1,2,3
1,2,2,3
1,2,3,2
1,3,2,2

That's 12 possible ways, which is not the same as 4C2 or 4C3. If you repeat this for sets 3,4,1,1 and 2,6,1,1 you'll see that each of these has 12 possible arrangements. So that's a total of 36 possible "winning" combinations of rolls out of 6^4.
 833student Posts: 9, Reputation: 1 New Member #16 Jul 2, 2012, 12:20 PM
so what would be an equation to use instead of writing everything out?
 ebaines Posts: 10,033, Reputation: 5529 Expert #17 Jul 2, 2012, 12:32 PM
There's no single equation. But there are a couple of different ways to think it through.

Given a set (a, b, c, c) where one element is repeated (the 'c') and the others occur once:

1. Consider what happens if the sequence starts with 'a' - that leaves one b and two c's for the last three positions, so there are 3C1 = 3 ways the b and c's could be arranged.
2. Same thing if the sequence starts with 'b' - there are 3 ways that the remaining a and 2 c's can be arranged.
3. Suppose the series starts with a 'c' - that leaves one a, one b and one c. There are 3! = 6 ways they can be arranged.

So the total is 3 + 3 + 6 = 12.

Another way to approach this is to consider that these 4 could be arranged with the a in one of 4 places (either first, second, third or fourth in sequence), which leaves any one of three places that the b could be. The c's then just fill in whatever is left. So the total number of ways to arrange (a,b,c,,c) is 4 x 3 = 12. Or alternatively you could think of the two c's as distinct items - let's call them c1 and c2. The number of ways that a, b, c1, and c2 can be arranged is 4! = 24. Now since in fact c1=c2 you have to divide by 2! to take into account the duplicate c's, So 24/2 = 12. This same approach can be used if there are more c's - for example the number of ways that (a, b, c, c, c) can be arranged is 5!/3! = 20. It's also a handy technique if there are multiple a's and/or b's as well - for example the set (a,a,b,b,b,c,c,c,c,) can be arranged in 9!/(2! x 3! x 4!) = 1260 ways.

As you can see there are various ways to approach these types of problems.
 833student Posts: 9, Reputation: 1 New Member #18 Jul 2, 2012, 12:47 PM
Ok! thanks so much for all your help

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