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Sagot :
If we roll a dice three times and we can stop in the first time only when it come 6. The expected payoff is 4.557.
According to the question:
The number of times you can get 06 are:
(i). If the third toss rolls a 6: 6
In this case her first two digits can be any of the five numbers. No. The number of cases is 5 × 5 = 25.
(ii). If you get a 6 on the second throw. The number of cases is -5 (no third throw in this case)
(iii). If the first roll rolls a 6. The number of cases is -1.
2. Number of times you can get 5 Are-
(i). If there is only one 5 in 3 rolls . The number of cases will simply be
3 × 4 × 4
(Because there are 5 possibilities in 3 places and 4 in the remaining places.)
(ii). If there are two 5s, the number of cases is 3×4. (Because there are 3 places where 5 cannot exist, and the places are filled with 4 possibilities.)
(iii). If there are three 5s, the number of possibilities is -1.
Similarly, for 4,3,2,1, the number of possibilities is (3*3*3 + 3*3 + 1), (3*2*2 + 3*2 + 1), (3 × 1×1 + 3×1 + 1), 1 each.
So the number of possibilities for 1,2,3,4,5,6 are 1,7,19,37,61,31 respectively .
You can now get the expected profit by taking the weighted average of the numbers with the possible outcome odds.
So the result is (6*31 + 5*61 + 4*37 + 3*19 + 2*7 + 1)/(31+61+37+19+7+1)
which is 711/ 156 = 4.557
Therefore, the expected payout is 4.557.
Learn more about Expected payout:
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