Every week, this site presents 2 new puzzles. This time we’re dealing with dice.

Reveal solution by highlighting. Note, there are 2 sets of solutions.

Læs mere: Fiddler on the proof, 20240112

I know I should be doing this with probabilities. The probability the first one is a d4 * 1/4 * the probability the other one is a d4 etc. But it feels like a lot of work! So I did a spreadsheet instead. That was fun too! I had to figure out some new functions. It seems to give the answer 9%.

Same site, second puzzle, same spreadsheet. Seems to give the answer 2.7.

Second set of solutions.

I changed my mind. I went back and did the probabilities. It’s satisfying, that these exact numbers are close to my estimates. This time I got the answers 9.37% and 2.729.

Rolling 2 dice, they have the same value, if the “biggest” die hits the “smallest” die’s value. Say my d4 rolls a 3. Then my d8 has to roll 3 as well, with probability 1/8. Given 2 dice dX and dY, the probability is 1/max(X,Y). Calculate this for all 36 die combinations, add them together, divide by 36, and we get our result.

The 3 dice situation is a little more complex, of course. The 3 dice are chosen: dX, dY and dZ.

1 distinct number means dX = dY = dZ. This happens with probability 1/median(X,Y,Z) * 1/max(X,Y,Z).

3 distinct numbers means all 3 dice are different from each other. Assume X <= Y <= Z. Then we get the probability (Y-1)/Y * (Z-2)/Z. This generalizes to (median(X,Y,Z)-1)/median(X,Y,Z) * (max(X,Y,Z)-2)/max(X,Y,Z).

Calculate the probabilities for each of the 6³ dice combinations, for 1 and 3 distinct numbers separately. Also separately, add all probabilities and divide by 6³. Now we have the separate probabilities for 1 distinct number and 3 distinct numbers (p(1) and p(3)). The probability for 2 distinct numbers is 1 – p(1) – p(3).

Get the average number of distinct numbers by 1*p(1)+2*p(2)+3*p(3).