Probability of Losing
This problem is from Grant Fikes.
In a certain game played with four dice, a player wins points if the player rolls the four dice and gets at least two of the same number ( for example, {1,2,2,5}, {2,2,3,3}, {1,3,3,3}, etc.) or has at least three consecutive numbers ( for example {1,2,3,6}, {2,3,4,5}, {1,3,4,5}, etc.). Find the probability that a player will not get any points with a single roll of the four dice.
Please submit all solutions to mathpotw@acu.edu by Thursday, Nov 5.
Solution to Probability of Losing
There were no completely correct solutions, though several students got close.
We will assume that the dice are all a different color so that “order matters” when we count. Then there are 64 different possible outcomes for a roll of four dice, each outcome being equally likely.
Now we need to count how many of these outcomes are not winners. Let’s count how many ways there are of rolling four dice so that no number is repeated, and then subtract out of those all that have at least three consecutive numbers. There are 6*5*4*3 ways of rolling no repeated numbers. Among these, there are just a few ways that at least three can be consecutive. Let’s list (and count) them, first with exactly three consecutive, then exactly four consecutive:
- First, those with exactly three consecutive numbers :
- 1 2 3 X — There are 2 choices for X, and 4! permutations to account for the fact that order matters. The result is 4!*2 = 48.
- 2 3 4 X — In this case, the X has to be a 6, so there are 4!*1 = 24 of these.
- 3 4 5 X — As in the previous case, the X has to be a 1, so there are also 24 of these.
- 4 5 6 X — In this case, the X has two choices (1 or 2), resulting in 48 possible.
- 1 2 3 4 — 24 of these.
- 2 3 4 5 — 24 more.
- 3 4 5 6 — another 24.
Next, those with four consecutive numbers:
Therefore the numerator of our probability is 6*5*4*3 – 24*(2+1+1+2+1+1+1) = 144. So the probability of not scoring points is
144/1296 = 1/9
