Probability of getting a sum of 14 when rolling 11

adocidasiaqxm

adocidasiaqxm

Answered question

2022-05-17

Probability of getting a sum of 14 when rolling 11 dice.
I assume that in P(A) = m/n, n is 6^11, but I don't know how to calculate the number of ways to get a sum of 14 when rolling 11 dice.

Answer & Explanation

Marquis Matthews

Marquis Matthews

Beginner2022-05-18Added 9 answers

The minimum sum is 11, which you can get only one way.
To get a sum of 12, you have to roll something greater than 1 on at least one die. But in fact if you roll 2 on one die then you have to roll 1 on every other die or you'll be over 12. You cannot roll anything over 2 or the sum will be too large. So the number of ways to do this is the number of ways to choose one die out of 11 (choose the die that will show a 2).
To get a sum of 13, you have more choices. You can roll one 3, but then all other rolls have to be 1. Alternatively, if you roll a 2 one one die, you must roll a 2 on one other die and 1 on every die in the remaining nine dice. You cannot roll anything over 3. So the total number of ways to add up to 13 is the number of ways to show exactly one 3 and ten 1s on 11 dice, plus the number of ways to show exactly two 2s and nine 1s on 11 dice.
The number of ways to roll a sum of 14 is just another counting exercise requiring a little more work than the sum 13.
othereyeshmt4l

othereyeshmt4l

Beginner2022-05-19Added 4 answers

1 + 1 + . . . + 1
And as noted, the sum is at least 11, so there are at least eleven 1s. Now the problem is to add three more 1s to the sum. Consider the following three cases:
(1) Number of one 4 and ten 1s =11
(2) Number of one 3, one 2 and nine 1 s = = ( 11 2 ) 2 !
(3) Number of three 2s and eight 1s = = ( 11 3 )
Thus the probability is
( 11 + 110 + 165 ) 6 11 = 286 362797056

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