Which is more likely, rolling three normal dice, and having one be the sum of the other two? or rolling three normal dice, and having one be the product of the other two?

In addition to your regular work, try to find a (good!) explanation that appeals to intuition.

1. July 28, 2008 pm31 11:04 pm 11:04 pm

Ooh, good puzzle! Let’s see…

Since the dice rolls are independent, in either case the probability can be computed by determining all pairs of outcomes for the first two dice that yield a sum/product of 1…6 (and their probabilities) and then multiplying by 1/6 (the probability of getting that outcome on the third die). Thus for the purpose of comparison we can ignore the factor of 1/6 and just count the valid first pairs.

1st die 2nd die
1 1, 2, 3, 4, 5
2 1, 2, 3, 4
3 1, 2, 3
4 1, 2
5 1

For multiplication:
1st die 2nd die
1 1, 2, 3, 4, 5, 6
2 1, 2, 3
3 1, 2
4 1
5 1
6 1

So in the case of addition there are 5+4+3+2+1=15 possible valid outcomes, but for multiplication there are 14. The first scenario is slightly more likely.

Extending this to dice with an arbitrary number of sides (all dice the same, of course), I believe the gap between the two results would get wider, because multiplication “grows faster” than addition. More pairs of numbers would have a sum that is within the bounds than a product. But I haven’t sat down and worked out a formula yet because you’ve only just posted the problem!

2. July 28, 2008 pm31 11:25 pm 11:25 pm

Oops, I mistakenly assumed one specific die had to be the sum/product of the other two. If we allow any die to be the “third”, then there are exactly three times as many possibilities for addition (so a total of 45), whereas for multiplication there are fewer than three times as many possibilities. (Some outcomes would be counted more than once; for instance, 1-2-2 would be counted when die #2 is the product and also when die #3 is the product.) So addition wins.

3. July 29, 2008 am31 2:16 am 2:16 am

*SPOILER* full answer (though I’d like to see a better intuitive explanation)

Yeah, the simple combinatorial approach gives 25 “product” triples and 45 “sums”, so the sum is nearly twice as likely. 224 and its two permutations count in both camps, of course.

Simple arguments from intuition: firstly, the “result” must be 6 or less, and products grow faster than sums. Sadly this is nonsense, as if you neglect order there are 8 disting product triples and 9 distinct sums – much too close for this argument to apply. The disparity is mostly because almost all of the “product” triples contain a double number, so they only contribute three outcomes, whereas almost all of the “sum” triples don’t and so contribute 6.

Detail of the combinatorial approach: note that the two smallest dice must combine to make the largest (ties permissible as we’ll see). We consider only increasing orders.

“sum” combinations:
1 1 2, 3 orders – 3
1 n (n+1), n from 2 to 5, 6 orderings of each – 24
2 2 4, 3 orders – 3
2 n (n+2), n = 3 or 4, 6 orders – 12
3 3 6, 3 orders – 3 This is clearly the last possibility.
total 45

“product” combinations:
1 1 1, only one ordering – 1
1 n n, n=2 to 6, three orders – 15
2 2 4, three orders – 3
2 3 6, six orders – 6 This is clearly the last possibility.
total 25