Will You Win?
This problem was suggested by Brian Jacobs.
Your friend has a game for you to try. He rolls two n-sided fair dice and you roll one n-sided fair die. You win if the number on your die is between the numbers on his dice, inclusively. For example, if he rolls a double, then you win only if the number on your die matches the number on his dice. On the other hand, if he rolls a 1 and an n, then you win no matter what you roll. What is the probability that you will win if you choose to play this game once?
Please submit all your work for this problem to mathpotw@acu.edu no later than 5:00 PM on Thursday, October 10th to receive credit for your submission.
Solution to Will You Win?
Let’s focus on the “gap” between the two numbers rolled by our opponent. For example, if he rolls a double, then the gap is 0. And in the worst case (for him) he rolls a 1 and an n, making a gap of n-1. For a gap of size k, we have k+1 possible successful rolls. For example, if he rolls a double, then we have to roll that number and so we have only 1 possibility. The answer to the problem can now be obtained by calculating the probability that he rolls a gap of size k, then multiplying that probability by the probability that our roll is within that range given he has that gap, and then summing over k. The denominator of all the probabilities will be n3, so we compute the numerator.
There are n ways for him to roll a gap of 0 (roll a double) and then one way for us to match his number. There are 2(n-1) ways for him to roll a gap of 1, and then 2 ways for our roll to be between his numbers. There are 2(n-2) ways for him to roll a gap of 2, and then 3 ways for our roll to be between his. Continuing this reasoning leads to a numerator of
.
We will make the algebra that follows easier if we include the first term in the summation symbol, which we accomplish by adding an n at the beginning and then subtracting it out at the end. The resulting algebra is as follows:
= 
= 
.
Using well-known formulas for the sum of consecutive integers and the sum of squares of consecutive integers, we arrive at a numerator of
= (n2 + 3n – 1)/3. Therefore the probability of our winning if we play this game once is
(n2 + 3n – 1)/3n2.It may be worth observing that as n increases to infinity, the probability decreases to 1/3. If n is 2, the probability is 3/4. The probability of winning is more than 1/2 until n = 6, in which case the probability is 53/108.
