Is the probability of rolling 5 dice and getting 55566 = 5! * (1/6)^5 (1.5%)?

(order doesn't matter)

C(5,2) * (1/6)^5

= 5*4/2 * (1/6)^5

=~ 0.1286%

There are C(5,2) ways to choose the 2 dice with the sixes.

As one that mostly has to at least triple check my permutation and combination formulas, I can do makeup so much easier...

I say (5*4/2) / 6^5
because I think of the favorable outcomes and total possible outcomes (n/N)

I know /6^5
is the same as *1/6^5

My question is why use *1/6^5
and what does that actually mean?

Ah, it is raining outside, I go and play with the rain drops
Sally

The dice are independent, so (1/6)^5 is the probability of getting a specific number on each die. But since we don't care which of the dice have the sixes, we multiply that by C(5,2).

Note that if all of the dice were different, like 12345, then we would multiply by 5! as the OP did.

OK. That is clear.
I do understand the combination part. That is easy.

I really like to see 10/7776 as, to me, it also makes it clear but maybe in a different question it might mess me up.

I'll try to give a better example if I think of one
Thanks!
Sally

Got it, thanks

To me BruceZ is sense on this forum.

But I think I can add a little from an instructive perspective.

Permutations and combination are counting techniques that "can be used" to derive probabilities.

Using combinations to arrive at the probability of rolling 66655 (order makes no difference), you simply count the total number of ways it can happen and divide by the total number of ways anything can happen.

So, C(5, 2)/6^5 is the probability. But one might argue that if order makes no difference than 66655 is the same as 55666 and there's only 1 way to get 66655. But we need to think in terms of sample space. The sample space has C(5, 2) = C(5, 3) elements that equate to 66655. And P of each of every element in the sample space is (1/6)^5. So you just add them up and walah, you get C(5, 2)*(1/6)^5 = C(5,2)/6^5.

I hope that helps and doesn't confuse. Probability and counting are not the same but counting is very useful in getting to the probability of an event.

another way of writing the same thing:

Imagine you roll one dice 5 times and write down the numbers in the order they come, then the options to get what you want are:

66555
65655
65565
65556
56655
56565
56556
55665
55656
55566

There are 10 of those, but when rolling a dice 5 times you had 6^5 possible sequences, so the answer you want is 10 / (6^5).

The shortcut to getting the figure "10" without having to write them all out, is the C(5,2) formula - the number of ways you can select 2 unordered objects out of a set of 5. (i.e. how many ways you can change two of the 5's to 6's in the string "55555").

I think I'm preaching to the choir here

Hopefully I'm helping jmark.

From an instructive point of view: There are (at least) two ways to look at it.

1) Consider the sample space. It consists of every possible outcome when you toss 5 dice. Each outcome has equal probability and the probability of each single outcome is 1/6^5. Now using the rule of sum simply add up the probability of all the outcomes you're interested in. There are 10 outcomes that contain three 5's and two 6's. So the probability of getting three 5's and two 6's is 10/6^5. You can use counting techniques (combinations) to get the numerator but the key here is that you're adding up the probabilities of each outcome you're interested in.

2) Use combinations (counting techniques) to arrive at the probability of getting three 5's and two 6's when you roll five dice. In this method you count up the number of ways it's possible to get three 5's and two 6's. (i.e. 55566, 55656, 55665, etc.) Using fast counting this will be C(5,2) OR C(5,3) just as kittens pointed out. This is the number of outcomes of interest. To get the probability you divide that by the total number of ways anything could happen = 6^5. So, the probability is C(5,2)/6^5 = 10/6^5. In this case we didn't add up any probabilities - we computed the value of a fraction where the numerator is the number of outcomes we're interested in and the denominator is the total number of outcomes possible. No rule of sum.

When you think about more than one way and you get the same result, this is a strong indication you're thinking about it correctly in all cases.

Thanks DP and K.
Excellent explanations for a thick headed gurl like me.
Hope it helps many others also!

From a friend using combinations.
Try using "we have" and "we want" in your sentence.
This is a common way to teach permutations and combinations.

55566
we have 5 and we want 3
C(5,3)
AND we are left with 2
we have 2 and we want 2
C(2,2)
The AND means we multiply
C(5,3)*C(2,2) = 10 combinations
I see the C(2,2) is really not needed here, but it hurts no one.

Say 66551
we have 5 and we want 2
C(5,2)=10
AND we are left with 3
we have 3 and we want 2
C(3,2)=3
AND we are left with 1
we have 1 and we want 1
C(1,1)=1

C(5,2)*C(3,2)*C(1,1)

and 55621
we have 5 and we want 2
C(5,2)=10
we have 3 with no pairs
3!=6
10*6=60
or
C(5,2)*C(3,1)*C(2,1)*C(1,1) it still works.

Yeah! for all those who struggle with combinations