(March 14, 2016 at 3:04 pm)Chas Wrote:(March 14, 2016 at 2:00 pm)robvalue Wrote: You're right Chas, but the problem is that the combinations don't all produce the same probability.
I think the solution involves both counting combinations and being able to count up how many there are for each probability. I got stuck into it at one point, but I didn't quite get there.
It's the "not your own name" that throws the wrench in, and the probability for each person changes depending on whether their number has already come up or not.
I did the example for 4 people to show this. There's 2 legal combos, but they have different probs:
For 4 players:
2, 3, 1, 4 = 1/3 * 1/3 * 1/2 = 1/18
or
3, 1, 2, 4 = 1/3 * 1/2 * 1/2 = 1/12
Total = 5/36
Every combination has the same probability of occurring.
It doesn't though, and here's why:
Consider the ABCD grouping. If A picks "C" out of the hat, then when B goes to pick he'll have "A", "B", and "D" left in the hat. B can't pick his own name, so B can either pick "A" or "D". Thus, the chance that the first two picks are, say, "CD" is 1/3*1/2 = 1/6.
But, if A picks "B" on the first pick, B has "A", "C" and "D" left in the hat. So, because none of these have to be put back, B has a 1/3 chance of picking each. Thus, the chance of the first two being "BD" is 1/3*1/3 = 1/9.
That's what Rob means when he says the probability changes depending on the previous pick.
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Don't worry, my friend. If this be the end, then so shall it be.