Would anyone care to take a shot at a hard probability question?

What year/number AIME question was this? I thought AIME had to have integer answers...

Here's my best shot:

First, you can get 7 in 6 possible ways: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

Removing the faces and rearranging them and then rolling one die gives you the same probability of rolling any number as rolling a single standard die. Now suppose that we only look at the number that was rolled with the first die and didn't know where the rest were placed. This is where the rearrangements come in. Now, we have 11 possible numbers that we can roll with the second die. Of these, 2 are the corrosponding number needed to sum to 7. Thus, our probability of rolling the right number on the second roll is 2/11. Our total probability is thus 1/6*2/11, and since there are 6 ways we can do this the answer is 2/11.

First you must Calculate the probability of getting any given number (let's say 1) on the first dice. Of course it is 1/6, but let's get sure of that.

There are two different possibilities of getting 1 on the first dice:

1) After rearranging faces, two ones are on the first dice, which would give you a 2/6 probability.

2) After rearranging faces, only one of the ones got on the first dice, which would give you a 1/6 probability.

The probability of getting two ones on the first dice es 6/12*5/11, as there are 6 faces on the first dice of a total of twelve, and after putting the first one, only 5 out of 11 remain on the first dice.

The probability of getting only one one on the first dice is 6/12*6/11*2, applying the same principle, but in the second dice there are still 6 places remaining, and you have to multiply by two because you have two ones.

Then, the probability of getting two ones on the first dice and rolling a one with it would be 6*12*5/11*2/6=60/792

And the probability of getting one one on each dice and rolling a one with the first one would be

6*12*6/11*1/6*2=72/792

As both probabilities are mutually exclusive, you add them which gives you:

132/792=1/6.

Now we have established that yes, changing the faces does not affect the probability of getting any given number on the first dice, which will make the rest easier.

Now, with any given number on the first dice, we must calculate the probability of getting the number that adds up to 7 in the second.

If two sixes ended up in the second dice, the probability would be 2/6, but if only one ended up there, it would be 1/6.

Then, given the fact that the first one is a one, the probability of having two sixes in the second dice would be: 6/11*5/10 and the probability of having only one would be 5/11*6/10*2

Then we do the same thing we did previously:

6/11*5/10*2/6+5/11*6/10*2*1/6=120/660=2/11

As we have the same probability of getting any other number, we can conclude that the probability is 1/6*2/11*6=2/11

Which proves that the previous answer was right, I just did it in a more elaborate way, so it can be proven in a better way.

I hope I explained myself right.

in the system you describe, there are 12 faces and you will select any two of those faces at random. The means described is random, so there is no bias between any particular combination, because of the initial random mixing of the faces.

call the dice A and B

then you have

A1, A2, A3 .... A6

B1, B2 , B3 ....B6

so 7 can be made with 1 + 6

this could be

A1 + A6

A1 + B6

B1 + A6

B1 + B6

so 4 ways

this applies also to

2 + 5

and 3 + 4

so there are 12 ways of obtaining a 7

there are 12C2 = 12! / 10! / 2! = 12 x 11 / 2 = 66 ways of selecting 2 tiles from 12

the probability is

12/66 = 6/33 = 2 / 11