Find graphically or mathematically the probability of the sum of two dice will be at most 9

Question

In a single toss of fair (evenly-weighted) -sided dice, I want to find the probability of that their sum will be at most 9.

So at first I did:

$$\frac{\sum_{i=2}^{9}\binom{12}{i}}{\sum_{i=2}^{12}\binom{12}{i}}$$

But I'm so prone to do mistakes in these kind of equations, and that looked to complicated to compute so I did a boolean array of whether the sum was <= 9

  1 2 3 4 5 6 
1 y y y y y y
2 y y y y y y
3 y y y y y y
4 y y y y y n
5 y y y y n n
6 y y y n n n

I thought the answer would have been the number of n (30) divided by all combinations (36). But it was the wrong answer.

The second solution is correct. Can you explain why you tried the first solution? — Asinomás, Apr 11 '21 at 14:30
@HereToRelax I tried the first solution to have on the top the number of combinations that sum to 9 and on the denominator all the solutions — Revolucion for Monica, Apr 11 '21 at 14:46
@lulu the website made me think the second approach was incorrect! They wanted 5/6. — Revolucion for Monica, Apr 11 '21 at 14:46
Your second approach is correct. I do not even know what is first one. In your second approach, it is easier to count sum $\gt 9$. There are only $6$ arrangements out of $36$ that do not have sum $ \leq 9$. — Math Lover, Apr 11 '21 at 14:46
@RevolucionforMonica but the number of possibilities is just $6\times 6$ and the number of combinations that sum $9$ or less is $30$. I don't see where those formulas come from — Asinomás, Apr 11 '21 at 14:47
Are you interested in solving it for arbitrary sides of dice and for arbitrary rolls of dice and for arbitrary sums? Otherwise I don't think the first approach is really very sensible. — Asinomás, Apr 11 '21 at 15:06

score 1 · Answer 1 · answered Apr 11 '21 at 14:56

I think this is the closest thing I can get to the current first solution.

We can use stars and bars for the numerator. The number of solutions to $a_1+\dots+a_k = n$ in positive integers is $\binom{n-(k-1)}{k-1}$.

So we get the formula:

$$\frac{\sum\limits_{i=2}^9 \binom{i-1}{1}}{\sum\limits_{i=2}^{12} \binom{i-1}{1}}$$

You can then recalculate the top and bottom using the hockey stick identity to get:

$$\frac{\binom{10}{2}}{\binom{12}{2}} = \frac{90}{132} = \frac{30}{43}$$

But this doesn't work. Why? Because we are also counting solutions like $7+0 = 7$ etc. If you want to get a nice formula you are going to have to do something similar to this:

Stars and bars with restriction of size between bars via generating functions.

Find graphically or mathematically the probability of the sum of two dice will be at most 9

1 Answers1