5
$\begingroup$

Suppose you have a pair of dice that have removable stickers for numbers on each of their 6 sides. Suppose that you unstick all 12 of the stickers from the dice and reapply them randomly to the 2 dice. You will still have 2 occurrences of each number 1 through 6. However, they may both occur on the same die. (For instance: after rearranging the stickers, you may have dice $d_1$ and $d_2$ with sides $d_1 = [1,2,2,4,4,6]$ and $d_2 = [1,3,3,5,5,6]$.)

Suppose now that you roll this randomized pair of dice. Is there a concise way to calculate the probability of each outcome? What is the probability of each possible outcome?

Just by working out some of the possible arrangements, it seems like $p(2)$ should be $\frac{1}{72}$ (which might not be correct), but the other probabilities are more difficult to compute this way.

$\endgroup$
  • $\begingroup$ How did you calculated $p(2)=\frac1{72}$ ? $\endgroup$ – callculus Jun 6 '17 at 18:03
  • $\begingroup$ I made an error. Starting with the fact that $p(2)=p(2|1$s on same die$)+p(2|1$s on different dice$)$, the first term is zero, so it reduces to $p(2)=p(2|1$s on different dice$)=\frac{1}{36}*p(1$s on different die$)$. My mistake was in error in the combinatorics $p(1$s on different die$)=\frac{6}{11}$ (not $\frac{1}{2}$). $\endgroup$ – DavidWayne Jun 6 '17 at 18:42
4
$\begingroup$

Here's a formula that generalizes to more digits.

Let's label the dice A and B, and write $(i,j)$ for the getting $i$ on A and $j$ on $B$. Consider the following two cases.

  1. $i=j$. This can only happen if we have exactly one $i$ on each die. The probability of this is $\frac{ 2\binom{10}{5}}{\binom{12}{6}}$ (chose one of the two $i$'s on A and the other on B, then choose $5$ more numbers from the remaining $10$ for die A). Once we have exactly one $i$ on each die, the probability of both landing $i$, is $\frac 16 \times\frac 16$. Therefore $$P( (i,i)) = \frac{ 2\binom{10}{5}}{\binom{12}{6}} \frac{1}{36}=\frac{1}{66}$$
  2. $i\ne j$. Since this is independent of the choice of $i$ and $j$, and there are exactly $6*5=30$ such pairs, the probability is equal to $$ \frac {1}{30} (1- \sum_{i} P( (i,i) )=\frac{1}{30}(1-\frac {1}{11})=\frac {1}{33}.$$

Now it remains to find the probability of a sum equal to $k$.

a. If $k\le 7$, there are exactly $k-1$ ways to write it: $ (1,k-1),(2,k-2),\dots, (k-1,1)$. Now if $k$ is odd, then in all of these $i\ne j$. Therefore the probability is $(k-1)/33$. If $k$ is even, then exactly one of these is of the form $i=j$, therefore the answer is $(k-2)/33 + 1/66$.

b. Finally if $k \ge 8$, and $(i,j)$ is such that $i+j=k$. Then $(7-i,7-j)$ has exactly the same probability, and has sum between $\{2,\dots,6\}$. Therefore the probability to get $k$ is the same as the probability to get $14-k$.

In other words, the distribution of the sum is symmetric about $7$.

$\endgroup$
4
$\begingroup$

The problem can be viewed as picking $2$ numbers from list of twelve numbers $1$,$1$,$2$,$2$,$3$,$3$,$4$,$4$,$5$,$5$,$6$,$6$ and looking at their sum.

Now there are $12 \cdot 11 = 132$ ways to pick two numbers from the list. In how many ways we can pick two same (specific) numbers? There are apparently exactly $2$ ways to do it (depends on which of the same two numbers you pick first). In how many ways we can pick two different numbers in given order (e.g. $1,2$)? There are exactly $2\cdot2 = 4$ ways to do it ($2$ two choices for first pick, $2$ choices for second pick).

Now for the sums, in how many ways we can get sum $2$? That is possible only as $1+1$, which by reasoning above can happen $2$ times (we have to pick two same numbers), so probability is $$p(2)=\frac{2}{132}=\frac{1}{66}$$

In how many ways can we get sum $3$? We have $3=1+2=2+1$ and again by reasoning above there is overall $4+4=8$ possibilities. This is because to get $2+1$ we need to pick two different numbers (and we know that is possible in $4$ ways), same for $1+2$. So probability is $$p(3)=\frac{8}{132}=\frac{2}{33}$$

Similarly $4=1+3=2+2=3+1$ with $4+2+4=10$ possibilities, so probability is $$p(4)=\frac{10}{132}=\frac{5}{66}$$

And so on...

$\endgroup$
  • $\begingroup$ Very nice answer; a very clever way of reducing it to a simpler problem. I have to give the best answer to @Fnacool, however, for providing a closed form solution for computing the probabilities. $\endgroup$ – DavidWayne Jun 6 '17 at 20:33
  • $\begingroup$ @DavidWayne That is alright, I would prefer closed form solution as well. $\endgroup$ – Sil Jun 7 '17 at 17:51
2
$\begingroup$

I wrote a basic Python program to find the distribution of the different values. There are ${12 \choose 6} = 924$ ways to select six values out of twelve, so the number of possible value distributions equals $\frac{924}{2} = 462$. For each of these combinations, I increased the probability of the 36 combinations of the two dice, resulting in the following probabilities:

 2 - 0.0152 -  1/66
 3 - 0.0606 -  4/66
 4 - 0.0758 -  5/66
 5 - 0.1212 -  8/66
 6 - 0.1364 -  9/66
 7 - 0.1818 - 12/66
 8 - 0.1364 -  9/66
 9 - 0.1212 -  8/66
10 - 0.0758 -  5/66
11 - 0.0606 -  4/66
12 - 0.0152 -  1/66

For instance, to get a 2 we need the two 1's to be on two different dice (probability $\frac{6}{11}$), in which case the probability of hitting a 2 equals $\frac{1}{36}$. The total probability of getting a 2 thus equals $\frac{5}{11} \cdot 0 + \frac{6}{11} \cdot \frac{1}{36}=\frac{1}{66} \approx 0.0152$. The Python program:

import itertools
import math
from collections import defaultdict

values = [x for x in range(1, 13)]
d = defaultdict(int)

for v in itertools.combinations(values, 6):
    w = [x for x in values if x not in v]
    for i in range(6):
        for j in range(6):
            d[math.ceil(v[i]/2) + math.ceil(w[j]/2)] += 1/36/924

for k in d:
    print(k, d[k])
$\endgroup$
  • 1
    $\begingroup$ It looks like all your probabilities are divisible by $1/66$ (alternatively, d[k] is always a multiple of 14.) It's not obvious why this should be. $\endgroup$ – Michael Lugo Jun 6 '17 at 18:14
  • $\begingroup$ @MichaelLugo That is correct. I updated the table to make this clear. $\endgroup$ – jvdhooft Jun 6 '17 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.