Finding Probability for Sum of Five Die Rolls Using Generating Functions

[tka2451]

Homework Statement

Suppose you throw a six-sided die five times. Find the probability that the sum of the outcomes of the throws is 15 using generating functions.

Homework Equations

Binomial theorem, generating functions

The Attempt at a Solution

Here's my attempt:

Okay, so I know so far that the generating function of a single throw is
G(x) = s/6 + (s^2)/6 + (s^3)/6 + (s^4)/6 + (s^5)/6) + (s^6)/6.

And that G(x) raised to the 5th power is the generating function for five throws.

Also, taking into account the independence of the x's:

G_x(t) = ((s+...+s^6)/6)^5 = s^5 * (1-s^6)^5 divided by 6^5 * (1-s)^5.

I get that:

1/(1-s)^5 = 1 + (5 choose 1) s + (6 choose 2) s^2 + (7 choose 3) s^3 + (8 choose 4) s^4 + (9 choose 5) s^5

and that

(1-s^6)^5 = 1 - (5 choose 1) s^6 + (5 choose 2) s^12 - (5 choose 3) s^18 + (5 choose 4) s^24 - (5 choose 5) s^30

However, I'm confused about how I use these to figure out the coefficient of s^15, which is the probability I'm looking for.

 

[mighty2000]

Can anyone help?



Hi there, your approach is on the right track, but there are a few errors in your calculations. Here's how you can use generating functions to find the probability of getting a sum of 15:

First, let's define the generating function for a single throw as G(x) = (x + x^2 + x^3 + x^4 + x^5 + x^6)/6. This is because each term represents the probability of getting a specific number on the die (e.g. x^3 represents the probability of getting a 3).

Next, we need to find the generating function for five throws, which is G(x)^5. This expands to (x + x^2 + x^3 + x^4 + x^5 + x^6)^5/6^5.

To find the coefficient of x^15 in this expansion, we need to look at the terms that contribute to x^15. These are: x^6 * x^6 * x^3, x^5 * x^4 * x^6, x^4 * x^5 * x^6, x^3 * x^6 * x^6, and x^6 * x^3 * x^6. The coefficient of each of these terms is (1/6)^3, since we need to get a 6 on each of the three throws that contribute to the sum of 15. Therefore, the total coefficient of x^15 is 5 * (1/6)^3 = 5/216.

Finally, to find the probability, we divide this coefficient by the total number of possible outcomes, which is 6^5. This gives us a probability of 5/6^8, or approximately 0.0000143.

I hope this helps clarify the process for finding probabilities using generating functions. Let me know if you have any further questions!