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ความคิดเห็นที่ 23 |
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The answer is e.
Proof:
Let N be the number of times the game is played before it is stopped.
From definition of expected value:
E[N] = Sum n p(n)
n runs from 2 to infinity, as the game will not stop at n = 1
p(n) is the probability that the game will be played exactly n times.
Designate the outcome of randomization as a1, a2, a3, a4, ...
The game will stop with n numbers when this condition is met:
a1 > a2 > a3 > ... > a(n-1) & a(n) > a(n-1) (*)
So, out of n! possible arrangement of n numbers (from a1...a(n-1)), n-1 are the possible places you can place a(n) so that it meets the conditions of (*) above. Hence we have,
p(n) = (n-1)/n!
so, the expected value is
E[N] = sum_from_n=2_to_infinity n * p(n)
= sum_from_n=2_to_inf n * (n-1) / n!
= sum_from_n=2_to_inf 1/(n-2)!
= e
Note: The assumption of uniform distribution of real number in range [0, 1] is used. So it is with probability 0 that we have a(k) = a(j)
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ม่ายด้ายทำมานาน อิอิ (kuzibimun) 
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27 ก.ค. 52 21:26:57
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