Data Challenge
Today I walked into combinatorics, games and puzzles day, with a challenge. Aleeza (a student) and I alternated three throws each of some special dice I brought in, and read out the results for the rest of the class. We ended up rolling over 100 times.
At the end, I challenged the students singly, or in groups, to write down what they thought was going on. I unkindly gave them just a bit more than 5 minutes, and no hints or help, other than to tell them that neither die was weighted. Take a look at the data (beneath the fold) and see if you can figure it out. Do you have an interesting approach or a useful hint?
Many of my students came close, but only one group got a complete and accurate description.
Click here to see the rolls: —>
We agreed to always list the lower number first
- 56, 33, 25, 35, 25, 16
- 33, 34, 56, 35, 16, 56
- 33, 55, 11, 26, 55, 15
- 36, 14, 45, 55, 13, 23
- 33, 26, 36, 34, 13, 25
- 55, 26, 56, 23, 26, 34
- 12, 55, 35, 35, 13, 11
- 25, 11, 33, 35, 23, 35
- 56, 36, 35, 16, 26, 23
- 13, 25, 45, 45, 36, 25
- 13, 25, 45, 34, 35, 45
- 35, 25, 15, 34, 25, 23
- 15, 45, 34, 45, 15, 13
- 25, 46, 35, 35, 55, 56
- 13, 45, 55, 55, 16, 26
- 56, 23, 15, 23, 36, 36
- 66, 36, 55, 12, 45, 35
JD2718 
Well, there’s a shortage of 2’s and 4’s, and a complete absence of 22, 44 or 24, so I think on die was missing a 2 and the other was missing a 4. There’s an excess of 3’s and 5’s, so I’m thinking one die an extra 3 and the other an extra 5.
But I’m not convinced the dice were 6-sided…
The dice were both 6-sided, and a bunch of my kids thought the same as you. Same answer, too: good thinking, you’ve got the right ideas in play, but not quite there yet.
Is it safe to say the two dice are not identical?
Were you each throwing one die at a time, or did she throw both 3 time, then you threw both 3 times? Were you reading the true result from both dice?
And if the two dice were not identical, where the two rolls independent, or did the roll you got on the first die determine which die you used for the second roll?
You don’t have any 22 24 44 66 BUT do have 26 & 46
So (unless you’ve just been incredibly unlucky in rolling), 6 is missing on one die. 2 and 4 missing from the other die. Still need to work out which numbers the 2 and 4 are replaced with and which the 6 is replaced with.
Efrique, there is a 66 in the last line of data. (actually caused a little murmur when it came up)
The dice were not identical. We threw them together. But when we read the result we always named the lower number first.
you say the dice are not weighted but special. from mere observation it looks like the dice favor 3 & 5, perhaps 6 as well. are you sure they are fair dice ?
They are not weighted. Nor are they fair.
Hmm, unweighted and not fair? Is one or both not a cube?
Okay, thanks. I searched and searched for it. I guess I just get a bit blind after a while. So existence of 66 says 6 is on both dice, but still
You don’t have any 22 24 44 so 2 and 4 are missing from one die.
The results are fairly consistent with one normal die and the other one being numbered 1 3 3 5 5 6. The outcome “56” occurs somewhat too often but there’s not enough deviation to suggest that it’s anything but random chance operating on dice with those faces.
bravo!
did you organize the data, or just read it?
I suspected from reading it that it was like this (after your earlier correction), but I had to organize it to see clearly that it really was the 3 rather than the 1 that was doubled (the 5 was clear enough beforehand).
So just looking at the numbers was enough to generate the two possibilities 1 1 3 5 5 6 and 1 3 3 5 5 6 in my mind. But to take it further I had to count the results, which clearly indicate that the second is more likely.
So I had my conjecture.
Then, given a conjecture, in order to see that the data really was consistent with that second arrangement on the nonstandard die, I went further and computed a residual and a residual standard error for the counts for each outcome pair based off a multinomial with the relevant probabilities – which showed no consistent bias in any of the numbers (5 occurs more than it should, for example, but not in EVERY cell involving 5), and no combinations really stand far enough out to cause concern either.
So my conjecture was about as consistent with the data as I would expect it to be if it were true. That doesn’t establish the result of course, but (taken together with the other information you provided, like that both dice were six-sided) it’s both a simple and plausible explanation for the data.
I’m a statistician by training, so my impulse was to collate and analyze the data form the beginning; I avoided it only so long as it was possible to make good progress without doing so.