The way I understand it, you win on a 2 or 3; lose on a 7 or 11; push on a 12; go to point phase for 4-6 8-10. So that means you can win on 2 numbers, lose on 2, push on 1, and go to 50:50 odds with the others. Seems like the house doesn’t have an advantage. (As opposed to the Pass, where you lose on 3 and win on 2 in the come out.)
In: Mathematics
each number combination has a different probability of occurring.
A 2, for example, can ONLY occur when both dice show 1.
But a 7 can occur on a 1/6, 2/5, 3/4, 4/3, 5/2, or 6/1 combo. So a 7 is 6 times as likely to occur than a 2.
Similarly, an 8 can be a 2/6 or a 3/5, 4/4, 5/3, or 6/2. Less likely than a 7, but more likely than a 2.
7 is the most likely, 6 and 8 are equally likely, more than 5 and 9 which are equally likely, more than 4 and 10 which are equally likely. 2, 3, 11, and 12 are the least likely possibilities.
So it’s not a 1 in 12 chance of any number occurring. In fact there are 36 possible outcomes when rolling 2 dice (6 for die A and 6 for die B, 6*6=36)
a 7 can occur in 6 of those or 16.7% chance (1/6, 2/5, 3/4, 4/3, 5/2, 6/1)
an 8 in 5 of them or 13.9% chance (2/6, 3/5, 4/4, 5/3, 6/2)
a 9 in 4 or 11.1% chance (3/6, 4/5, 5/4, 6/3)
a 10 in 3 or 8.3% chance (4/6, 5/5, 6/4)
etc.
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