Probability of getting a sum of 14 when rolling 11 dice

Question

So, this is a probability and combinatorics problem... I'm not sure how to solve this.

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.

Remember, I should be able to solve this with the knowledge of basic probability, because this is a high school-level problem.

Answer 1

I assume the dice are all standard six-sided dice.

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 $1$s on $11$ dice, plus the number of ways to show exactly two $2$s and nine $1$s 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 $13$.

Answer 2

First thing is to realize that a sum of 14 with 11 dice is quite low...

That's only possible with:

A) a subset of one die showing 4 and the other ten all showing 1

or

B) a subset consisting of one die showing 3 and one showing 2, and the other nine all showing 1

or

C) a subset of three dice showing 2, the other eight all showing 1

(To see how many ways each of these can happen may be bit easier if you imagine the dice are thrown one by one)

A can happen in 11 ways (1st die one, or second die one, or 3rd die one...)

To see how many ways you can get to B, note that the die that shows three can again be at 11 positions (1st, 2nd...), once the position for that die is "chosen", you have 10 positions left for the die that shows 2, so you have $11\cdot10$ possibilities.

This is nothing else but applying the formula for the number of $k$ permutations of $n$ (http://en.wikipedia.org/wiki/Permutation#k-permutations_of_n):

$_nP_k=\cfrac{n!}{(n-k)!}$

where $n$ is the total number of dice, and $k$ are the ones you "choose" or "label" as belonging to the subset.

For C, assume as an intermediate step, that the three dice showing 2 each had a different colour. Then the number of distinguishable outcomes would be analogous to B: $11\cdot 10\cdot9$.

With different colours, e.g. if the "first" three dice would show 2, and they were blue, red, green, and with chosing the position for the blue one first, then red, then green, there would be three options to place the blue one. Once that is done, two positions are left to place the red one and one place remains for the green one.

So there would be $k!$ ways to get three times 2 with the "first" three dice.

But as the dice do not have different colours here, these are not distinct outcomes (equivalent to the dice showing one in the remainder of the sequence not being distinguished) . In the total number of ("distinguishable") possible results, $6^{11}$, the result with three dice showing 2 at the three "first" positions is ("only") included once. So by calculating C with $_nP_k=\cfrac{n!}{(n-k)!}$, we would count every distinct outcome with three dice showing 2 not once, but $k!$ times - to correct for this, divide the result by $k!$ to get

$_nC_k=\binom{n}{k}=\cfrac{n!}{k!(n-k)!}$

which is the formula for the number of "$k$-combinations" in a set with a subset of $k$ distinct elements. (http://en.wikipedia.org/wiki/Combination)

Here: $_nC_k=\binom{11}{3}=\cfrac{11!}{3!(8)!}=\cfrac{11\cdot10\cdot9}{6}=165$

In summary, you have $11+110+165=286$ ways of getting 14 with 11 dice.

As you correctly stated, any specific distinguishable outcome has a probability of $\cfrac{1}{6^{11}}$.

Hence the probability to get 14 with eleven dice is $\cfrac{286}{6^{11}}$.

(=$0.00008\%$)

Answer 3

Reduce the problem to

$$1+1+ ... + 1$$

And as noted, the sum is at least 11, so there are at least eleven $1$s. Now the problem is to add three more $1$s to the sum. Consider the following three cases:

$(1)$ Number of one $4$ and ten $1$s $= 11$

$(2)$ Number of one $3$, one $2$ and nine $1$s $=\binom{11}{2}*2!$

$(3)$ Number of three $2$s and eight $1$s $= \binom{11}{3}$

Thus the probability is $$\frac{(11+110+165)}{6^{11}}=\frac{286}{362797056}$$