Quickly calculate the probability of 12 dice having sum less than 30

Question

I try to quickly answer the question of whether one should play a game: roll 12 fair dice and sum up the face values; if the sum is less than 30, win 10 dollars, otherwise lose 1 dollar.

Let $S_{12}=\sum\limits_{i=1}^{12} x_i$ be the sum where $x_i$ is the value for the ith die.

I think if the expected gain is greater than 0, we should play the game, and the gain is $$E = 10 \times P(S_{12}<30) - 1\times [1-P(S_{12}< 30)] >0 ?$$

So I tried to quickly calculate or approximate $P(S_{12}<30)$. My way is through CLT, $$\frac{S_{12} - 12E[x_i]}{\sqrt{12 var(x_i)} } \xrightarrow{d} N(0,1)$$ $$\Rightarrow P(S_{12} < 30) \approx p(z < \frac{30-42}{\sqrt{35}}) = \Phi \left( \frac{30-42}{\sqrt{35}} \right) $$

But this approach still seems to require a certain amount of computation. Especially, it requires evaluating the normal cdf $\Phi(z)$.

So I was wondering if there is a better/quicker way without referring to computers/calculators.

Thanks in advance.

Answer 1

Well, your approximation is correct though you can do things in a quicker way without requiring paper. One fact is that the variance of a uniform variable interval of length $L$ is $L^2/12$, so it’s a quick approximation to get that the variance of a die is around $6^2/12=3$. Combining this with a total mean of $12\cdot 7/2$ gives a threshold of $(42-30)/\sqrt{12*3}=2$ standard deviations so $(1-.95)/2=.025$. That is there’s approximately a 2.5% chance of the sum being less than 30.

If it hadn’t luckily been an integer number, you could still approximate the number with linear approximation between 68,95,99.7%, rounding upwards for concavity.

Answer 2

I read that you are looking for easier method , so the suggestion is to use generating functions .We know that the die (standard) can have only the values of $1,2,3,4,5,6$. Then , the generating function for die is :$$x+x^2+x^3+x^4+x^5+x^6 = \frac{x-x^7}{1-x}$$

Now , we have $12$ dice , so if we find the product of these $12$ dice such that $$\bigg(\frac{x-x^7}{1-x}\bigg)^{12}$$ the sum of coefficient ranging from $x^{12}$ to $x^{29}$ will give us the number of possible occasions such that the summation over the dice less than $30$.Because of summing the values in the link is cumbersome , i did not write the exact solution , but if you write a computer program that can calulcate sum the coefficent of the values whose exponents from $x^{12}$ to $x^{29}$ and sum the coefficent of the values whose exponents from $x^{12}$ to $x^{72}$ , then when you divide them it will give you the probability

Answer 3

You have $E[x]= 3.5$ and $ \sigma_x^2 \approx 3 $. Then $E[Z]= 42$ , $ \sigma_Z^2 \approx 36 $, $ \sigma_Z \approx 6 $

By Chebyshev's inequality (one sided) $$P(Z<30) \approx P(\mu -Z \ge 12 ) = P(\mu -Z \ge 2 \sigma_Z) \le \frac12 \frac14 = \frac18 $$

This is non conclusive (you want $P(Z<30)>\frac{1}{10}$ to play the game), but knowing that the Chebyshev's inequality is very loose, you should not play.