Then we need to look at the probability that the two dice will roll a number that matches.
Then we evaluate the probability of the rolls being this match outcome for each d12 case. Then sum.
| d12 | d12 prob | 2xd6's match prob | prob of this match |
| 1 | 1/12 | 0/36 | 0/432 |
| 2 | 1/12 | 1/36 | 1/432 |
| 3 | 1/12 | 2/36 | 2/432 |
| 4 | 1/12 | 3/36 | 3/432 |
| 5 | 1/12 | 4/36 | 4/432 |
| 6 | 1/12 | 5/36 | 5/432 |
| 7 | 1/12 | 6/36 | 6/432 |
| 8 | 1/12 | 5/36 | 5/432 |
| 9 | 1/12 | 4/36 | 4/432 |
| 10 | 1/12 | 3/36 | 3/432 |
| 11 | 1/12 | 2/36 | 2/432 |
| 12 | 1/12 | 1/36 | 1/432 |
| Probability of match | 36/432 |
The probability of a match is 8.333% - also 1/12
Hmm I think the thing we're missing is that dice 3 can roll a 1, but dice 1 plus dice 2 cannot equal 1. So the answer must be something less than 1/12.
I'll say it's 1/13, or 1 in 13 tries.
I'll say it's 1/13, or 1 in 13 tries.
Wait a minute. Did OP say there was a prize and then not pay it? wtf?
I sent you ! Thanks !
Oops...
