m*******s
发帖数: 758 | 1 T_k = first time to see consective k same values of one dice
what is the expection ?
It was discussed about consective k same value of "1" : T_k(1)
E[T_k(1)] = 1/6 (E[T_{k-1}(1)]+1) + 5/6(E[T_{k-1}(1)]+1+E[T_k(1)])
How about this question ?
hee T_k = min {T_k(1),T_k(2),...,T_k(6)}. | J****x
发帖数: 37 | 2 My two cents.
Let x = T_k(1). Then,
x = 1/6^k * k + sum_{i = 1}^k 5 * (x + i) / 6^k
Denote by y the expected number we want to have, then
y = x / 6 | m*******s
发帖数: 758 | 3
this result is trivial.
" y = x / 6
seems ok ? but reason ??? could you please the reason behind it ?
Does T_k(1), T_k(2), ... T_k(6) independent ?
What is the joint distribution of them ?
... seems not easy
【在 J****x 的大作中提到】
: My two cents.
: Let x = T_k(1). Then,
: x = 1/6^k * k + sum_{i = 1}^k 5 * (x + i) / 6^k
: Denote by y the expected number we want to have, then
: y = x / 6
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