Solution below.

Another week, another fiddler.

Reveal solution by highlighting.

15 minutes. As shown in this spreadsheet.

Now, I have to admit, that’s a guess. It’s based on, fastest child carries slowest child some of the way, slowest child walks the rest of the way, meanwhile middle child was walking, fastest child goes back and then carries that child the rest of the way. A solution, where everybody arrives at the same time, will be optimal.

I can’t really argue, why it has to be that way. And it’s complicated to do something similar for the extra credit, 4 child version.

Solution below.

Solution below.

Solution below.

Solution below.

Solution below.

Solution below.

Solution below.

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.

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).