There were a few logic puzzles posed on the 6/5/2026 Fiddler on the Proof. Here is the first one verbatim.
From Alan Pierrot comes a great way to “cap” off your week:
Three game show contestants must work together to win a prize. They are shown a large bag that’s initially empty, and then they see three red hats and two white hats placed into the bag. At this point, the contestants are blindfolded. Each contestant then picks a hat at random from the bag and places it on their head.
One at a time, their blindfolds are removed and they can see the hats on the others’ heads—but not the hat on their own head. If they can, with absolute certainty, identify the color of the hat on their own head, then the game is over and all three contestants win a prize! Otherwise, they skip their turn, at which point the next contestant has their blindfold removed and the process continues. If all three contestants skip their turn, then no prize is won.
Will the contestants always win the prize? If so, why? If not, why not?
They can always win a prize, as somebody will always be wearing a red hat.
So, the following logic will always work. Contestant 1 looks at contestant 2 and 3’s hats. If they are both wearing white, 1 will know she’s wearing red. If any of the two are wearing red, 1 will pass.
Contestant 2 will ignore contestant 1’s hat, and only look at contestant 3’s hat. If 3 is wearing white, contestant 2 will know she is wearing red. If 3 is wearing red, 2 will pass, and then 3 will know if it ever becomes their turn, that 3 is wearing red.
So, in summary, for the RWW ordering, contestant 1 knows. For the xRW ordering, where x is R or W, contestant 2 knows. For the xyR ordering, where x and y are R or W, contestant 3 knows.
The extra credit problem is given below verbatim.
Now, there are four game show contestants who must work together. They are shown a large bag that’s initially empty, and then they see R red hats, W white hats, and B blue hats placed into the bag, such that R, W, and B are each less than or equal to 4. Otherwise, the game is played as before.
For some triples (R, W, B), the contestants can always win. Among these, what is the greatest value of R + W + B?
To go along with the original problem, let’s assume 4 red hats and 3 hats that are not red. This could be 3 white and 0 blue, 2 white and 1 blue, 1 white and 2 blue, or 0 white and 3 blue. The main thing is that there are 3 hats that are NOT RED.
The same logic applies of the original problem applies. Only pay attention to the remaining hats. If red hats remain, then pass. If all the remaining hats are not red, then you know you’re wearing red.
So, any permutation of (4,3,0) and (4,2,1) will result in the contestants always able to win.
I don’t see how having 8 or more hats could work.