Consider a collection of 4 Fermi dice That is no two die may

Consider a collection of 4 \"Fermi\" dice. That is, no two die may have the same face up. (This may be accomplished by sequentially tossing each die, but always re-throwing any die that comes up with a face value already present on the table.) Answer the following two questions. What is the least probable value(s) of the sum of the dice? What is the average value of the sum of the dice? You may provide a logical argument for your answer to the last question, rather than a detailed listing of states.

Solution

The minimum possible sum will be 1+2+3+4 = 10 ( consideriing the least values on all fermi dice)

The maximum possible sum will be 6+5 +4+3 = 18 ( considering the maximum values on fermi dice)

Note: ignoring the rearrangement of dice ( i.e every posiibility can be multiplied with 4!)

Possible sums will be from 10 to 18.

No. of possible die combinations will be 6C4 = 15 different possible combinations.

But sums possible are only 10 to 18 which is 18 -10 +1 = 9

odd number and we can expect a symmetrical distribution .

So Sum = 14 has several die configurations and will be the maximum.

i.e( 14 = 2,3,4,5 and 1,3,4,6 and 1,2,5,6) So 14 has 3 possible configurations ( which is 1 +2) i.e extra 2 configs added.

Similarly by symmetry 15 and 13 will have anothr 1 configuration ( 15 = 2,3,4,6 and 1,3,5,6)

Symilarly 12 and 16 has another 1 additional configuration

So total of extra configurations has become 6.

So probabilities will be

10 - 1/15

11 -1/15

12 -2/15

13 - 2/15

14 - 3/15

15 - 2/15

16 - 2/15

17 - 1/15

18 - 1/15

Now For (a) least probable values for the sum will be 10, 11 ,17,18

For(b) average value = 10(1) + 11(1) + 12(2) +13(2) +14(3) + 15(2)+16(2) + 17(1) + 18(1) /15 = 14