Even 4 dice can create havoc
Four regular non-identical dice are rolled. The probability that the outcome is more than
1
4
is of the form
B
A
, where
A
and
B
are co-prime integers.
Find the value of B − A .
Details and Assumptions
-
Outcome means
n = 1 ∑ 4 ( number appearing on n th die )
-
The four dice are regular ones, numbered 1 , 2 , ⋯ , 6 .
-
The four dice are not identical. You can consider them to be of four different colors.
The answer is 721.
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3 solutions
You're too Smart! It is difficult to solve the problem without noticing that the outcomes are evenly distributed about 1 4 ..
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Thanks. Yes, without that initial observation this problem would involve a great deal of 'brute force'.
If there were an odd number of dice then the distribution would be even around two central values. For example, if there were 5 dice then there would be the same number of permutations for sums of 1 7 and 1 8 , and then the same number of permutations for sums less than 1 7 as there are for sums greater than 1 8 .
This is much easier than what I was doing. Ended up missing by 12 on my final try (563/1296). I'm still looking for where that missing 12 could have gone.
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Haha. "The Missing Dozen" sounds like a good title for a murder mystery. :) Yes, it's easy to miss a case or two with this type of question. I was off by 4 on my first attempt at counting just those permutations giving an outcome of 1 4 .
this is havoc
I did this a very long winded way. The table below shows the possible rolls of two dice on both the x axis and y axis, and then the appropriate probability of getting those rolls. Then inside the table itself is all the outcomes where the score is greater than 14 (14 or less is marked with a -). Then in the boxes available is the numerators of their probability (out of 6 4 = 1 2 9 6 ): ↘ P r o b 3 6 1 3 6 2 3 6 3 3 6 4 3 6 5 3 6 6 3 6 5 3 6 4 3 6 3 3 6 2 3 6 1 P r o b ↘ 2 3 4 5 6 7 8 9 1 0 1 1 1 2 3 6 1 2 − − − − − − − − − − − 3 6 2 3 − − − − − − − − − − 2 3 6 3 4 − − − − − − − − − 6 3 3 6 4 5 − − − − − − − − 1 2 8 4 3 6 5 6 − − − − − − − 2 0 1 5 1 0 5 3 6 6 7 − − − − − − 3 0 2 4 1 8 1 2 6 3 6 5 8 − − − − − 3 0 2 5 2 0 1 5 1 0 5 3 6 4 9 − − − − 2 0 2 4 2 0 1 6 1 2 8 4 3 6 3 1 0 − − − 1 2 1 5 1 8 1 5 1 2 9 6 3 3 6 2 1 1 − − 6 8 1 0 1 2 1 0 8 6 4 2 3 6 1 1 2 − 2 3 4 5 6 5 4 3 2 1 Then adding the numbers up = 575. As 1 2 9 6 5 7 5 is in it's simplest form, the answer is 1 2 9 6 − 5 7 5 = 7 2 1
We can solve this using generating functions. Find the coefficients of x^15 to X^24. Add all of them and divide it by 6^4
That's not feasible to do by hand, is it?
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No we can do it using multi nominal theorem, though I used wolfram alpha for it.
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Yes we could do it, but calculating just one coefficient is cumbersome enough, let alone 10..
A quick outline of one possible solution.....
Without restrictions there are 6 4 = 1 2 9 6 possible permutations of the 4 (distinct) dice.
Now there are the same number of permutations leading to outcomes of less than 1 4 as there are permutations leading to outcomes of more than 1 4 . (Note that 1 4 is the distribution midpoint.) So all we need to calculate is the number of permutations leading to an outcome of 1 4 . These will be all permutations of the following sets of integers:
( 6 , 6 , 1 , 1 ) , ( 6 , 5 , 2 , 1 ) , ( 6 , 4 , 3 , 1 ) , ( 6 , 4 , 2 , 2 ) , ( 6 , 3 , 3 , 2 ) , ( 5 , 5 , 3 , 1 ) ,
( 5 , 5 , 2 , 2 ) , ( 5 , 4 , 4 , 1 ) , ( 5 , 4 , 3 , 2 ) , ( 5 , 3 , 3 , 3 ) , ( 4 , 4 , 4 , 2 ) , ( 4 , 4 , 3 , 3 ) .
The number of permutations is then
6 + 2 4 + 2 4 + 1 2 + 1 2 + 1 2 + 6 + 1 2 + 2 4 + 4 + 4 + 6 = 1 4 6 .
Thus the probability that the outcome is more than 1 4 is
1 2 9 6 2 1 2 9 6 − 1 4 6 = 1 2 9 6 5 7 5 .
Thus A = 5 7 5 , B = 1 2 9 6 and B − A = 7 2 1 .