Puzzles: Counting to 6 Quickies
July 5, 2007 pm31 11:58 pm
Assume that n regular (six-sided) dice are thrown. In terms of n, find:
- The minimum and maximum sums possible.
- The number of sums possible.
- The most likely sum.
Image is from “think again! seductive math problems for the modern mind” Click on the dice to go there… I have not explored yet.
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JD2718 
Hmmm, here’s my thinking on the problem (which is a very nice problem):
1. Min is n
Max is 6n
2. Number of all sums possible:
If one considers 3 = 1+ 2 as different from 3 = 1+2 (i.e. order matters), then the possible sums are 6^n.
If order doesn’t matter, then I think the number of possible sums is 5n +1
3. Most likely sum: 3.5n (when n is odd, plus/minus 0.5 of this result are equally likely).
Anyone find anything similar? I’m not totally confident in all of the above, but what the heck, I’m posting it anyway!
Dang I wish I had gotten on earlier! I actually did this one, and I agree mostly with Jackie. The only one I don’t is on the if the sum orders do matter, if you have 1 die, then there are 6 combinations, 2 dies have 21 combinations compared to the 36 possibilities with order not mattering, and 3 dice have 56 combinations (if I did my figuring right) versus 216 orginial possibilities. So I am not sure what the actual formula for order not mattering. but I think I will work a bit longer and figure it out. Thanks for the fun puzzle.
hmm.. off to double check… guess I’ll try to write up may thinking (which may change my thinking!)
Ahh.. I think we’re defining things differently. What I meant by 5n +1 applied to distinct sums.
When rolling two dice, the possible sums are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, & 12 ( so I considered 1 + 3, 3+1, and 2+2 as one result – a sum of 4).
When rolling three dice, the possible sums are 3, 4, 5, …, 18.
So, JD, what did you mean in the original question?
(btw, when I posted the last comment it was 5:41 pm central time… to what time zone is the blog set?)
Jackie, as you assumed, that’s what I meant. With 2 dice there are 11 possible sums: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
And in response to your answer, why does each extra die only increase the number of possible sums by 5?
Mathnerd, I understand your confusion. However, sum is the operative word here – we can have sums of 4, 8, 11. But that doesn’t mean I couldn’t word this more clearly. I will work on that.
Argh, you had to ask why, didn’t you? I can’t articulate my thinking on this clearly (which tells me I don’t fully understand it). Maybe it’ll come to me in my sleep tonight.
–in comment #3, should have been “my” thinking, not “may” –sorry.
JD, thanks for the great problem. It provided a nice, guilt free break from planning for class next week (I was doing math, so my rationale is that it will someday apply to some planning I have to do).
For why each new die adds 5 new possible sums: once you’ve realized that the numbers you can get with
dice form a string of consecutive integers, adding one new die makes six new numbers accessible (the largest roll with 
dice plus 1, the largest roll with 
dice plus 2, …, the largest roll with 
dice plus 6), but you can no longer get the smallest possible roll, and 6 – 1 = 5.
Another way to look at it is to see that rolling dice labelled 0 to 5 gives you the same number of possibilities as rolling dice labelled 1 to 6 — then you always add 5 numbers at the large end of the scale.