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I'm working on a maths problem about finding the probability of getting '8' after throwing two fair 6-sided dice. I know the answer is 5/36 from looking at all possibilities of adding numbers 1-6.So, I have 2+6, 3+5, 4+4, 5+3 and 6+2.

The problem is I'm not sure how to work this out using permutations and combinations as this is the point of the exercise.

My first thought was to do $\frac {({6C1 * 6C1})*5} {12C2}$ . My thinking was the order doesn't matter for which the numbers are read but that we need to multiply by 5 as there are 5 possible true event. I wasn't confident that the denominator was accurate but I know the answer was wrong as it is 5/22

I Have had other ideas but because I know the answer I'm generally using ad hoc intuition. Is there a proper way to solve this?

Hussain
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  • Why do you think you need to use permutations and combinations? Particularly for such a problem? – David G. Stork Aug 07 '24 at 04:35
  • It was part of a exercise with other perms and combs questions so I assume that how it is meant to be solved – Hussain Aug 07 '24 at 04:39
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    This problem does not lend itself to solving by permutations or combinations. – user2661923 Aug 07 '24 at 04:40
  • Can i ask why that is the case? – Hussain Aug 07 '24 at 04:42
  • Your ordered appoach to counting looks good to me. If you take the sums with on dice you get 1 1 1 1 1 1 with sum to 1 to 6. Convolve with 1 1 1 1 1 1 and get one permutation with sum 2. The counts for sums of is 1 2 3 4 5 6 5 4 3 2 1. Starts with sum 2 and you get 5 answers sum to eight. Divide by 6^2. (Multinomial equivalent to Pascal's triangle.) You were effectively doing this with how you found sums to 8. – kirk beatty Aug 07 '24 at 10:36
  • For the denominator, we can use the Multiplication Principle to obtain $6 \cdot 6 = 36$ possible outcomes. For the numerator, simply listing the favorable cases is the easiest way to find the number of favorable cases. If you insist on a more formal solution, you could find the number of solutions of the equation $x_1 + x_2 = 8$ in the positive integers subject to the restriction that $x_1, x_2 \leq 6$. However, that is more work than simply listing the favorable cases. – N. F. Taussig Aug 08 '24 at 14:10
  • Expanding on the comment of @N.F.Taussig, see this answer for a blueprint of how to combine Inclusion-Exclusion with Stars-And-Bars to attack this generic type of problem. The original constraints are $$x_1 + x_2 = 8, ~~~x_1, x_2 \in \Bbb{Z_{\geq 0}}, ~~1 \leq x_1,x_2 \leq 6.$$ The final computation is $~\dfrac{N}{D} ~: ~D = 6^2,~$ and $$N = \left{ ~\binom{7}{1} - \left[ ~2 \times \binom{0}{0} ~\right] ~\right}.$$ – user2661923 Aug 14 '24 at 01:30

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The result $\frac{5}{36}$ is correct. In case of doubt I would always resort back to looking at the individual draws to decompose the problem into manageable sub problems. Decomposition is a good strategy for many difficult problems:

Let's look at the denominator (all possible throws) first: Throwing 2 dice is like drawing twice from an urn containing the numbers 1-6, putting the number back after each draw. Hence, there are $n^k_a=6^2=36$ possible outcomes/permutations that occur with equal probability. Looking at each individual draw: $n_{a1}^k*n_{a2}^k=6^1*6^1=36$

Now, the numerator (all throws that sum to 8): Observe that you cannot get the sum 8, if the first draw is a 1 (that you have to see). Hence, for the first draw there are only 5 numbers in the urn (2-6), which makes $n_{b1}^k=5^1$ permutations. Then for the second draw there is only a single feasible option left to draw a number so both numbers sum to 8, which makes $n_{b2}^k=1^1$ total possibilities. To get the number of ways to draw two numbers that equal 8, you compute $n_{b1}^k*n_{b2}^k$.

This yields: $\frac{n_{b1}^k*n_{b2}^k}{n_{a1}^k*n_{a2}^k}=\frac{5^1*1^1}{6^1*6^1}=\frac{5}{36}$

You can of course use the binomial operator to count options for each draw, which yields:

$\frac{5C1*1C1}{6C1*6C1}$

I can't rule out that there is some more elegant way for this specific problem to calculate probability.

ahi
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  • See my comment, following the posted question. It is very arguable whether my (overall) approach is more elegant than yours. However, it does represent an alternative approach. – user2661923 Aug 14 '24 at 01:58