Dice question?

Question

A game is played as follows:  Ten dice are thrown simultaneously and the dice that show the most common outcome are set aside [for example, three sixes are observed and those three dice are set aside].  The remaining dice are again thrown simultaneously and those that show the same number as the dice already set aside are also set aside.  This continues until all the dice have shown the original number.  How many total throws of the dice would this take, on average?

Anonymous · Accepted Answer

Corrected results
I did a simulation with 10 million trials and the result was 15.3393393.  That's a little different than I had before. There was a bug in the program.

I also figured out the theoretical answer, which agrees closely with that.

The ways of getting a max of N on the first roll of the 10 dice is
N	 WAYS	PROB
10	 6 	 0.000000 
9	 300 	 0.000005 
8	 6,750 	 0.000112 
7	 90,000 	 0.001488 
6	 787,500 	 0.013024 
5	 4,721,220 	 0.078080 
4	 18,774,000 	 0.310488 
3	 32,004,000 	 0.529288 
2	 4,082,400 	 0.067515 
TOTAL	 60,466,176 	 1.000000

I calculated the ways of getting a max of N on the first roll by looking at the ways of getting certain combinations. I'd start with, say, a certain maximum, and then looking at all the ways that could happen. E.g., with a max of 5, the number of ways of getting that is found by adding up the ways of getting every possible pattern. I've indicated the details on a couple of them:

AAAAA BBBBB	3780 = 6C2 x 10! / 5!5!
AAAAA BBBB C	151200
AAAAA BBB CC	302400
AAAAA BBB CD	907200 = 6 x 5 x 4C2 x 10! / (5! 3!)
AAAAA BB CC D	1360800
AAAAA BB CDE	1814400
AAAAA BCDEF	181440
total	4721220

The expectations for each maximum on the first roll are in this table. I show both what I got within the simulation, and the theoretical result. This also calculates the overall expectation.

n	Simul'n	Theory	P(n on 1st)
10	n/a . .  . . 	1.000000	0.0000001
9	6.991520	7.000000	0.0000050
8	9.725211	9.727273	0.0001116
7	11.55716	11.55544	0.0014884
6	12.91882 12.92670	0.0130238
5	14.03108 14.02366	0.0780803
4	14.93655 14.93780	0.3104876
3	15.74158 15.72134	0.5292876
2	16.40160 16.40694	0.0675154
Expectation = ...... 15.34848239
Simulation gave 15.3393393

I calculated the Expectations for each of those recursively.
If you get a max of 10 on the first roll, it's 1.
If you get a max of 9 on the first, then the expectation is
E = (1+E)5/6 + 1(1/6) = 6/6 + 5/6E
1/6 E = 1
E = 6.00

Then if you get 8 on the first roll the expectation is
E = (1+E)(5/6)^2 + (1 + 6.00)(2)(5/6)(1/6) + (1+0)(1/6)^2
E = (1+E)0.69444 + (7)(0.27778) + 0.027778
0.30556 E = 2.666666667
E= 8.7272727
etc.

I trust that's enough detail of the method for anyone that wants to reproduce this. I couldn't have done the calculations without Excel. Even with Excel, there was a lot of work to do, but it was interesting.

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