The expected value is equal to 1 times the probability that 1 is the highest showing number plus 2 times the probability that 2 is the highest showing number, etc., up to 6 times the probability that 6 is the highest showing number.
Taking these probabilities in reverse order, the probability that 6 is the highest showing number is the same as the probability that 6 shows at least once. This is equal to 1 minus the probability that no 6's show up:
p(6)=1-(5/6)^n
The probability that 5 is the highest number showing is the probability that no 6's show up times the conditional probability that at least one 5 shows up:
p(5) = (5/6)^n * (1 - (4/5)^n)
The probability that 4 is the highest is the probability that no 5's or 6's show up and that at least one 4 shows up:
p(4) = (4/6)^n * (1 - (3/4)^n)
In general, the probability that i is the highest number showing on a die is
p(i) = (i/6)^n * (1 - ((i-1)/i)^n)
Multiplying each of these terms by the corresponding i and adding them together gives the expected value:
E(n) = Sigma{i = 1 to 6} i * (i/6)^n * (1 - ((i-1)/i)^n)
But written out it can be seen that this can be simplified:
6 - 5^n/6^(n-1) + 5^(n+1)/6^n * (1 - (4/5)^n) + 4^(n+1)/6^n * (1 - (3/4)^n) + 3^(n+1)/6^n * (1 - (2/3)^n) + 2^(n+1)/6^n * (1 - (1/2)^n) + 1^(n+1)/6^n * (1 - (0/1)^n)
Each term reduces, as for example, 5^(n+1)/6^n * (1 - (4/5)^n) is 5(5^n-4^n)/6^n. Reformulating the first two terms as well, we get
6*(6^n-5^n)/6^n + 5(5^n-4^n)/6^n + 4(4^n-3^n)/6^n + 3(3^n-2^n)/6^n + 2(2^n-1^n)/6^n + 1(1^n-0^n)/6^n
At this point we see that the 5*(5^n)/6^n portion of the second term can be combined with the -6*(5^n)/6^n to get just -5^n/6^n, and similarly down to the end, where, fortunately the last remaining unpaired portion is zero.
So ultimately we get:
6 - (5/6)^n - (4/6)^n - (3/6)^n - (2/6)^n - (1/6)^n
The following program evaluates this for various n:
DEFDBL A-Z
FOR i = 1 TO 20
t = 6 - (5 / 6) ^ i - (4 / 6) ^ i - (3 / 6) ^ i - (2 / 6) ^ i - (1 / 6) ^ i
PRINT i, t
NEXT
with the following results:
1 3.5
2 4.472222222222222
3 4.958333333333333
4 5.244598765432099
5 5.430941358024691
6 5.560292352537723
7 5.654117369684499
8 5.724354257163542
9 5.778176976165981
10 5.820159373068342
11 5.853357065278942
12 5.879889975825309
13 5.901275192312109
14 5.9186267025424
15 5.932780282948634
16 5.944374386207267
17 5.953904159524657
18 5.961758506478343
19 5.968246134835735
20 5.973614264074476
As expected, when n=1, the expected value is midway between the highest and lowest possible values for the one die. As more and more dice take part, the probability increases that there's at least one 6 (and other high-value numbers contribute of course).
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Posted by Charlie
on 2004-12-08 18:37:31 |
solution
