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发帖数: 129 | 1 Cant type pinyin now. my apology.
Q:
On average, how many times must a 6-sided die be rolled until a 6 turns up?
clearly, the answer is 6 times. However, is there a way to solve the problem
using martingale?
I assume X_i = 1, if you have 6, or 0 otherwise. S_n = sum of (X_1,... X_n)
It seems that (S_n)^2- n^2/36 - 5*n/36 is a martingale. But how come we end
up with n = 6?
Many thanks! | a**o
发帖数: 2 | 2 设计一个游戏, 投1-5就输1元, 投出6赢1元. 这样的游戏的收益是matingale. 因为E(X
) = 5/6* -1 + 1/6*5 =0
然后设投n次才出现6.那么前n-1次都投输了1元.
n次的总收益是 Xn= (n-1)*-1 + 1*5
X是matingale所以
E(Xn)=0 => (n-1)*-1 + 1*5 =0 => n=6 | c******0
发帖数: 59 | 3 Martingale?
i'd think about markov chain, so that's like a 2 state machine. | x******a
发帖数: 6336 | | G******n
发帖数: 572 | |
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