The “u-number” of a number x is defined as the ones digit of 2^x. Anto

the units digit of 2^x for individual values of the dice is 2,4,8,6,2,4.

corresponding 3^ for the units digits is 9,81,6561,729,9,81.

If 9 is chosen, 4 values are greater than 9
similarly if 81 is chosen, 2 values are greater than 81.
similarly for all other numbers.

Favourables Outcomes = 4+2+0+1+2+4=13
Total outcomes = 6x6=36

Hence probability=13/36.

If you calculate 3^u for 1st roll, all 6 results will be 9, 81, 6561, 729, 9, 81. This result is the same for 2nd roll. 9, 81, 6561, 729, 9, 81.

About distance: If the first result is 9 and the second is also 9, the distance is 9-9=0 which is smaller than 9. If the first result is 9 and the second is 81, the distance is 81-9=72 which is also smaller than B which has the value of 81. If the first result is 81 and the second is 9, the distance will be greater than B. Distance 81-9=72>9.

On the first roll, the probability of getting result 9 is 2/6. In this case no other alternative values for second roll which would make the distance greater than B. So probability is 0. So next estimations are:

probability of getting 81 on the first roll (2/6) * probability of getting 9 on the second roll (2/6) = 1/9
probability of getting 729 on the first roll (1/6) * probability of getting 9, 81 on the second roll (4/6) = 1/9
probability of getting 6561 on the first roll (1/6) * probability of getting 9, 81, 729 on the first roll (5/6) = 5/36

All together: 1/9 + 1/9 + 5/36 = 13/36

we can get 1,2,3,4,5,6 on rolling a dice.
corresponding value of u for each number will be 2,4,8,6,2,4
thus 3^u can be 3^2, 3^4 , 3^8, 3^6, 3^2, 3^4

now,
we want that A-B > B or A > 2B

Case 1:
when B is 3^2
there are 4 possible values which A can take- 3^4 , 3^8, 3^6, 3^4
so the probability will be
(1/3)(1/6) + (1/3)(1/6) + (1/3)(1/3)

Case 2:
when B is 3^4
there are 2 possible values which A can take- 3^8, 3^6
so the probability will be
(1/3)(1/6) + (1/3)(1/6)

Case 3:
when B is 3^6
there is only 1 possible value of A- 3^8
so the probability will be
(1/6)(1/6)

we will get final probability when we add 3 cases

(1/3)(1/6) + (1/3)(1/6) + (1/3)(1/3) + (1/3)(1/6) + (1/3)(1/6) + (1/6)(1/6)

= 13/36

Answer:- B

Any faster way of solving this question?

Possible 3^n s = 3^2 (if he rolls a 1), 3^4 (if he rolls a 2), 3^8 (if he rolls a 3), 3^6 (if he rolls a 4), 3^2 (if he rolls a 5), 3^4 (if he rolls a 6). There are 6 possible differences that could be made. B will absolutely be 3^2 if A is 3^4, 3^6, or 3^8. B will absolutely be 3^4 if A is 3^6 or 3^8. And B will absolutely be 3^6 if A is 3^8. However, since the result of 3^4 can be obtained if Antony rolls either a 6 or a 2, and also 3^2 can be obtained if Antony rolls either a 1 or a 5. So for every possible differences one has to multiply by 2 if there is either 3^4 or 3^2. If both were to appear, then one should multiply by 4 to get all the possible outcomes. Therefore the sample space of rolling 2 dice is 36 and after some calculation the event we desire has 13 outcomes.

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