If I rolled n fair six-sided dice rolled simultaneously, what's the expected value (in terms of n) of the highest valued die?

(Note: "expected value" refers to the number you would expect to get if you ran this simulation many times and averaged the results.)

(In reply to re: solution by charlie by Ady TZIDON)

The formula

6 - (5/6)^n - (4/6)^n - (3/6)^n - (2/6)^n - (1/6)^n

just didn't come to me directly.  I think the explanation you give is not quite adequate, in that it seems to imply that the probabilities of other values are merely subtracted.

Looking back I can see how the above works, somewhat more directly than I had: for example, the (1/6)^n is the probability that nothing higher than a 1 shows up.  In taking an expected value one would expect to subtract 5 times this probability, as 1 is 5 less than 6.  But this is actually accomplished as, for example, the (2/6)^6 incorporates all those circumstances in which there is nothing higher than a 2, including those circumstances again in which nothing higher than a 1 appeared.  So ultimately, those situations are indeed counted 5 times, while those in which 2 is definitely the highest are counted 4 times, etc.

With this explanation, I can understand the direct going to this formula.

Posted by Charlie on 2004-12-08 19:48:33