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James Fung
United States
Berkeley
California
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Here’s an example of using conditional probability to solve for a result in Can't Stop.
Q. What is the probability of rolling at least one 7?
A. Let A, B, C, and D represent the four dice rolls. Because of symmetry, it doesn’t matter what A is.
There are 3 cases to consider for B:
* B = 7 – A, a 1-in-6 probability, in which we have failed to not roll a 7.
* B = A, a 1-in-6 probability.
* Otherwise, a 4-in-6 probability.
Therefore, after 2 dice, we have 3 cases:
* 7 has been rolled: probability 1/6
* Only 1 unique number has been rolled: probability 1/6
* 2 unique numbers have been rolled (that don’t add up to 7, which I’ll leave implied): probability 2/6
Let’s add a third die:
* P(7 has been rolled after 3 dice) = P(7 has been rolled after 3 dice | 7 has been rolled after 2 dice) P(7 has been rolled after 2 dice) + P(7 has been rolled after 3 dice | 1 unique number after 2 dice) P(1 unique number after 2 dice) + P(7 has been rolled after 3 dice | 2 unique numbers after 2 dice) P(2 unique numbers after 2 dice) = (1)(1/6) + (1/6)(1/6) + (2/6)(4/6) = 15/36
* P(1 unique number after 3 dice) = P(1 unique number after 3 dice | 1 unique number after 2 dice) = (1/6)(1/6) = 1/36
* P(2 unique numbers after 3 dice) = P(2 unique numbers after 3 dice | 1 unique number after 2 dice) P(1 unique number after 2 dice) + P(2 unique numbers after 3 dice | 2 unique numbers after 2 dice) P(2 unique numbers after 2 dice) = (4/6)(1/6) + (2/6)(4/6) = 12/36
* P(3 unique numbers after 3 dice) = P(3 unique numbers after 3 dice | 2 unique numbers after 2 dice) P(2 unique numbers after 2 dice) = (2/6)(4/6) = 8/36
Sanity check: these add up to 1. Finally, the 4th die:
* P(7 has been rolled after 4 dice) = P(7 has been rolled after 4 dice | 7 has been rolled after 3 dice) P(7 has been rolled after 3 dice) + P(7 has been rolled after 4 dice | 1 unique number after 4 dice) P(1 unique number after 4 dice) + P(7 has been rolled after 4 dice | 2 unique numbers after 3 dice) P(2 unique numbers after 3 dice) + P(7 has been rolled after 4 dice | 3 unique numbers after 3 dice) P(3 unique numbers after 3 dice) = (1)(15/36) + (1/6)(1/36) + (2/6)(12/36) + (3/6)(8/36) = 139/216 = 59.7%
Of course, all this should be rewritten in the usual mathematical notation so it’s a bit easier to read.
However, I’m having trouble coming up with an elegant solution to what is the likelihood of busting if your columns are 6,7,8 or 5,7,8, etc. I think it becomes a rather tedious counting problem, not much better than just having a computer exhaustively check every possibility. Even the probability of rolling any other number (like 8 ) becomes complicated once we lose the symmetry. That may be an important lesson in itself.
(Games in the Classroom #2 is off-site in my wordpress blog.)
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