A Stats question?

Method A:

The outcomes are 2 through 7 with equal probability:

X | P(X)

2 | 1/6

3 | 1/6

4 | 1/6

5 | 1/6

6 | 1/6

7 | 1/6

E(X) = (2 + 3 + 4 + 5 + 6 + 7) / 6 = 27/6 = 9/2 = 4.5

Method B:

36 outcomes:

\ 1 2 3 4 5 6

..----------------

1| 1 2 3 4 5 6

2| 2 2 3 4 5 6

3| 3 3 3 4 5 6

4| 4 4 4 4 5 6

5| 5 5 5 5 5 6

6| 6 6 6 6 6 6

X | P(X)

1 | 1/36

2 | 3/36

3 | 5/36

4 | 7/36

5 | 9/36

6 | 11/36

E(X) = (1*1 + 2*3 + 3*5 + 4*7 + 5*9 + 6*11)/36

E(X) = 161/36

E(X) = 4.47222222222

Answer:

Though just slightly better, I'd choose method A with an expected outcome of 4.5 compared to 4.472222..

A's avg score = 4.5

B's avg score = 6*P(6) + 5*P(5) + ...

P(6) = 1 - (5/6)^2

P(5) = 2(1/6)(4/6) + (1/6)^2

P(4) = 2(1/6)(3/6) + (1/6)^2

...

B's avg score = 4.472

So if the averages are a true indicator, A will have the slightly higher chance of winning. But let's see.

P(A > B) = P(A=7) + P(A=6 & B<6) + ... = 1/6 * [1 + (5/6)^2 + (4/6)^2 + ... + (1/6)^2] = 91/216

P(A = B) = P(both 6) + ... + P(both 2) = 1/6 * [P(B=6) + ... + P(B=2)] = 35/216

P(A < B) = 1 - P(A>B) - P(A=B) = 90/216

91 > 90 so A is the better method.

If the players roll until there is no tie, then A will win 91/181 = 50.276% of the time.

The number of possible outcomes is small enough to analyze in a spreadsheet by listing them all, so that's what I did.

Method A gives you an average of 4.5. That's easy to see since the expectation for rolling one die is 3.5, and adding 1 gives you 4.5.

Method B gives you an average of 4.5 - 1/36 = 4.472222. That's a little more difficult to calculate theoretically. I haven't really thought about how to do that. I just listed all 36 rolls, took the larger value, and averaged it to get the result, so since it looked at the entire sample space, I know it's correct. Perhaps someone can contribute a theoretical method for calculating that.

So Method A gives you a slightly higher average score.

I also looked at a head-to-head match up. If two people play, and one uses Method A and the other uses Method B, then there are 216 possible outcomes (6 for A, 36 for B, 6x36=216 for both), and player A:

wins 91 out of 216

ties 35 out of 216

loses 90 out of 216

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