Games in the Classroom #3: Can’t Stop

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:

  1. B = 7 – A, a 1-in-6 probability, in which we have failed to not roll a 7.
  2. B = A, a 1-in-6 probability.
  3. Otherwise, a 4-in-6 probability.

Therefore, after 2 dice, we have 3 cases:

  1. 7 has been rolled: probability 1/6
  2. Only 1 unique number has been rolled: probability 1/6
  3. 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:

  1. 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
  2. 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
  3. 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
  4. 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.

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