4 dice are rolled. what is the chance of gettin two 5's and two 6's? i've done the exhaustive approach and found it to be 1/216. how do i do it using a proper method?

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- Mar 14th 2007, 01:46 PMShaun GillRolling 4 dice
4 dice are rolled. what is the chance of gettin two 5's and two 6's? i've done the exhaustive approach and found it to be 1/216. how do i do it using a proper method?

- Mar 14th 2007, 02:03 PMPlato
The string “5566” can occur in (4!)/[(2!)^2] = 6 ways.

So 6/[6^4]=1/[6^3]=1/(216). - Mar 14th 2007, 02:11 PMShaun Gill
thanx a lot, but could you just tell me the theory behind how u can say that?

- Mar 14th 2007, 02:31 PMPlato
If one rearranges the letter string “SOUTHERN” there are (8!) ways to do that. That is because all 8 letters are distinct. Now take the string “UNUSUAL” there are seven letters but 3 are the same so there are (7!)/(3!) ways to rearrange the string. There are (11!)/[(4!)(4!)(2!)] ways to rearrange the string “MISSISSIPPI”.

To toss two 5’s and two 6’s is to toss the string “5566” is some order.

How many ways are there to rearrange that string?

There are 6^4 outcomes in tossing four dice. - Mar 14th 2007, 02:35 PMSoroban
Hello, Shaun!

Quote:

Four dice are rolled.

What is the chance of getting two 5's and two 6's?

We know that there are: 6^4 = 1296 possible outcomes, right?

You said you did an exhaustive approach.

You must have found that there were__six__outcomes with two 5's and two 6's

. . 5566, 5656, 5665, 6556, 6565, 6655

Can we find that number (6) without making a list? . . . Yes!

There are four "slots: . _ _ _ _

Pick two of them to place the 5's (the 6's will go in the other two slotd).

There are: .C(4,2) .= .(4!)/(2!2!) .= .6 ways that this can be done.

Got it?