I have got a formula, but I do think it is the smartest way to do ... I
state as following:
for a single case: like 3351, for example, it is
probability = 0.2*0.2*0.2*0.1
prize = 3+3+5+1= 12
weight of 3351 = probability * prize = 12 * (0.2*0.2*0.2*0.1)
So the forumula of weight for a single case is: weight =
(w_i+w_j+w_k+w_l)*(p_i*p_j*p_k*p_l)
Enumerate all the case, from 1111 to 6666, get all the weight, and add
them together, then get the average:
for i=1~6
for j=1~6
for k=1~6
for l=1~6
W_{avg}=\sum_{l=1}^{6}\sum_{k=1}^{6}\sum_{j=1}^{6}\sum_{i=1}^{6}
(w_{i}+w_{j}+w_{k}+w_{l}) *(p_{i} * p_{j} * p_{k} * p_{l})
But this algorithm will end up with a high complexity, because there is
redudency: e.g.
1121 and 1112, in the enumeration, have the same sum, but were counted
twice, similarly, 2111, 1211.
So I guess that there will be a smarter way than mine... Any suggestion
is highly appreciated!
.
.
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