throwing two dice

[hanshuckebein]

hi folks,

if I throw two dice twice what is the probability that the second throw shows at least one repeat?

example: first throw 1;2
so what is the probability that the second throw shows also a 1 or a 2?

I would calculate this way (probably all wrong  ):

the first dice has a 1/6 chance for a repeat of nr. 1 and a 1/6 chance of nr. 2. so does the second dice. it makes a total of 4/6 chance for at least one repeat.

but if the first dice shows a repeat for nr. 1 it doensn't have a chance to show nr. 2 and vice versa. the same goes for the second dice.

do I now have to subtract this so that the overall probability for a repeat is 4/6 -1/6 - 1/6 = 2/6?

thanks for your advice and cheers. 

hans

[GARNabby]

1/6 the of the time, 6 of 36 possibilities, the first roll is a double number.  Neither of the two second-roll dies will repeat 5/6 X 5/6 the of the time. Therefore, a repeat then will happen 1/6 X (1 - 25/36).
 
5/6 the of the time, 30 of 36 possibilities, the first roll is not a double number.  Neither of the two second-roll dies will repeat 4/6 X 4/6 the of the time. Therefore, a repeat then will happen 5/6 X (1 - 16/36).

Adds to 111/216, or 0.5138 repeat (the eight.)

[Colbster]

Actually, each die gives you a 1/3 chance, and the whole roll gives you the same 1/3 chance.  You have 2 chances to win out of 6 per die, meaning 4 out of 12 possibilities or 1/3.  Not as complicated as you might think it would be.

[hanshuckebein]

thanks a lot, guys. 

cheers

hans