Possible outcomes for rolling 2 dice twice with same sum.

[bodensee9]

Hello:

I am wondering if I am doing the following correctly. This is the problem:

(a) how many different outcomes are possible when a pair of dice are rolled 2 successive times?

roll once with 2 dice, 36, so roll twice, 36*36.

(b) what's the probability that each die shows same value on second roll as first row?

for 1 dice, there are 6 cases of the same value, so it's 1/6. For both die, it's (1/6)*(1/6) = 1/36.

(c) what's the probability that the sume of the two dice is the same on both rolls?

For this you have the following possibilities:

(1 1) (2 2) (3 3) (4 4) (5 5) (6 6)
(1 2) (1 3) (1 4) (1 5) (1 6)
(2 3 ) (2 4) (2 5) (2 6)
(3 4) (3 5) (3 6)
(4 5) (4 6)
(5 6)

The possible sums are : 2 (1 1)
4: (2 2) (1 3)
5: (1 4) (2 3)
6: (3 3) (1 5) (2 4)
7: (3 4) (2 5)
8: (4 4) (2 6) (3 5)
9: (3 6) (4 5)
10: (5 5) (4 6)
11: (5 6)
12 (6 6)

So, the probability of rolling a (1 1) both times is (1/6)^4.
Rolling a 4 both times is (1/6)^4 + ((1/6)^4))*2 (because you can have 1 3 or 3 1)
Rolling a 5 is (1/6)^4 * (2) * 2
and so on?

 

[tiny-tim]

Hello bodensee9!

(a) and (b) look good.

So, the probability of rolling a (1 1) both times is (1/6)^4.
Rolling a 4 both times is (1/6)^4 + ((1/6)^4))*2 (because you can have 1 3 or 3 1) …

That's basically the correct method …

but what happened to 2 2?