We’ve looked carefully at the Elchanan Mossel’s dice probability problem a few times – including last week. The problem goes like this:
If you have a fair 6-sided die with sides marked 1, 2, 3, 4, 5, and 6, how many rolls on average will it take to roll a 6 if any sequence of rolls containing an odd number prior to seeing a 6 doesn’t count. So, 2, 4, 4, 6 would count, for example, and 2, 4, 5, 6 would not count.
Here are a two of our prior looks at it:
I was looking for additional ways to discuss the problem in a slightly different way and found a fascinating idea in the comments (attributed to Paul Cuff) on Gil Kalai’s blog. We talked through that idea today and then looked at the problem on a 120-sided die.
Here’s now I introduced Paul Cuff’s idea:
Now that we had a series to sum, we had to talk about how to sum it! My younger son had a pretty clever idea and was able to find the sum with a little help. My older son found the sum using an idea from Calculus:
With the series summed, we could now calculate the answer to the original problem with a 6-sided die. We also looked at the same problem with a 120-sided die.
For the last part of our project we went to Mathematica to check the answer for the 120-sided die. There was one mysterious result in the histogram, but I think the simulation showed that our answer in the last video was correct.
Also, whoops . . . , sorry for forgetting to fix the focus on this one.