Problem Of the Week: 4th March 2024
There are 25 red balls and 25 blue balls in a bag. Alice and Bob take turns drawing balls from the bag without replacement, with Alice going first. Interestingly, they notice that whenever a red ball was drawn, the next ball drawn (by any person) was never red. Suppose Bob drew m red balls. Given th
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Problem Of the week: 4th March 2024
Michael Wu
There are 25 red balls and 25 blue balls in a bag. Alice and Bob take turns drawing balls from the bag without replacement, with Alice going first. Interestingly, they notice that whenever a red ball was drawn, the next ball drawn (by any person) was never red. Suppose Bob drew m red balls. Given that the 33rd ball drawn was red, what is the sum of all possible values of m? Solution
At first it might appear that there are many possibilities to keep track of, but this ends up not being very true. First of all, imagine dividing the 50 drawings into 25 pairs. The two balls in each pair must be different colors, because otherwise some pair would have two red balls. Therefore, if we let an “X” represent a red draw followed by a blue draw and “Y” represent a blue draw followed by a red draw, we find that the sequence of draws can be represented by a sequence of 25 letters that are either X or Y. For example, “XXYXY” represents the sequence of draws “RBRBBRRBBR”.
Moreover, we know that a Y cannot be followed by an X, because “YX” translates to “BRRB”, which is impossible. Therefore, we are down to 26 possibilities: k X’s followed by 25-k Y’s, for any k value between 0 and 25 inclusive. It turns out that these all satisfy the conditions.
If the 33rd ball drawn was red, the 17th letter in the sequence must be X, so we know that k must be greater than or equal to 17. Bob draws a ball for every Y in the sequence, so m can be any number from 0 to 25-17 = 8. These add to 36.
