Edited by Richard Walker, Thursday 19 February 2026 at 15:05
This puzzle asked
In a litter of mice 3 are white and the others brown.
If 4 of the mice are chosen at random, the probability that the sample contains all 3 of the white mice exactly equals the probability that it contains none of them.
How many mice are in the litter altogether?
I'll give two solution: the first from an AI, correct and not too hard to follow, but long-winded, the second shorter and more insightful
Solution 1
I asked 'AI Overview and it reasoned essentially as follows (I've abridged its answer but not altered the logic).
Suppose there are in the litter. Then 3 are white and brown We can choose 3 white mice from 3 in 1 way and 1 brown from in ways. So there are ways to include all 3 white mice.
On the other hand we can pick 4 brown mice from in .
We are told the two probabilities are the same and so these two numbers must be equal
Cancelling amd multiplying both sides by 24 we obtain
So we seek three consecutive numbers whose product is 24 and this is satisfied by . So and .
Solution 2
If the 4 mice selected include all three white mice, the mice remaining must include none of the white mice.
Conversely, if the 4 mice selected include none of the white mice, the mice remaining must include all three white mice.
So the situation is symmetrical with respect to the location of the white mice and from the information that the two cases have the same probability we can deduce the two groups must be the same size anf thus there are altogether mice in the litter.
I put this to AI Overview and it gave me a pat on the back!
That is a brilliant and elegant way to solve it!
I actually adapted this from another question I saw, in which one probability was twice the other and I wondered if there were other numbers that gave a nice answer, for example, if the probabilities were equal. So I used the long method to get an equation and found that this would be the case if the litter size were twice the sample size. I thought that was rather neat, but then the penny dropped and I saw it was obvious!
PS This was another chance to extend my LaTeX, I now know how to do binomial coefficients.
Solution To 'Three White Mice'
This puzzle asked
I'll give two solution: the first from an AI, correct and not too hard to follow, but long-winded, the second shorter and more insightful
Solution 1
I asked 'AI Overview and it reasoned essentially as follows (I've abridged its answer but not altered the logic).
Suppose there are in the litter. Then 3 are white and brown We can choose 3 white mice from 3 in 1 way and 1 brown from in ways. So there are ways to include all 3 white mice.
On the other hand we can pick 4 brown mice from in .
We are told the two probabilities are the same and so these two numbers must be equal
Cancelling amd multiplying both sides by 24 we obtain
So we seek three consecutive numbers whose product is 24 and this is satisfied by . So and .
Solution 2
If the 4 mice selected include all three white mice, the mice remaining must include none of the white mice.
Conversely, if the 4 mice selected include none of the white mice, the mice remaining must include all three white mice.
So the situation is symmetrical with respect to the location of the white mice and from the information that the two cases have the same probability we can deduce the two groups must be the same size anf thus there are altogether mice in the litter.
I put this to AI Overview and it gave me a pat on the back!
I actually adapted this from another question I saw, in which one probability was twice the other and I wondered if there were other numbers that gave a nice answer, for example, if the probabilities were equal. So I used the long method to get an equation and found that this would be the case if the litter size were twice the sample size. I thought that was rather neat, but then the penny dropped and I saw it was obvious!
PS This was another chance to extend my LaTeX, I now know how to do binomial coefficients.