# probability of a die

• Mar 15th 2011, 06:55 PM
turtlejacks
probability of a die
so I am doing some math that my friend asked me to do and for the life of me I cant figure it out... Here is the question. If anyone could solve this that would be amazing.

Suppose you have a fair 6-sided die. If you roll the die repeatedly, with 1 worth 1 point, and 6
worth 3 points, and all other outcomes (2, 3, 4 and 5) worth 2 points, and add up the points as you
go along (starting at 0), what is the probability pn that you’ll hit n? Check your formula with a
direct calculation of p
• Mar 15th 2011, 08:36 PM
matheagle
P(X=1)=1/6=P(X=3) while P(X=2)=2/3
where X is the points you accumulate in each play.
NOW, I doubt you want n twice.
I think you want

$\displaystyle P(X_1+\cdots +X_n=k)$ where each $\displaystyle X_i$ are iid copies of that above distribution.

Let $\displaystyle S_n=X_1+\cdots +X_n$ the points you've accumulated after n plays.

$\displaystyle P(S_n=n)=(1/6)^n$ since you tossed a 1, n times

$\displaystyle P(S_n=n+1)=n(1/6)^{n-1}(2/3)$
• Mar 15th 2011, 08:56 PM
turtlejacks
well the way we have been taught in our lin alg class is that we want to set it as a recursion formula and use eigenvectors/values I believe, I may be wrong though. Then he would set it as a linear combination.
• Mar 15th 2011, 09:13 PM
matheagle
you can look at the vector $\displaystyle (S_{n-1},S_n)$

Maybe use http://en.wikipedia.org/wiki/Probabi...ating_function

$\displaystyle G_X(z)=E(z^X)=(1/6)z+(2/3)z^2+(1/6)z^3$

Then $\displaystyle G_{S_n}(z)=\big(E(z^X)\bigr)^n=\big((1/6)z+(2/3)z^2+(1/6)z^3\big)^n$

So $\displaystyle P(S_n=k)={G^{(k)}_{S_n}(0)\over k!}$

You can use the trinomial theorem to calculate the probabilities
http://staff.spd.dcu.ie/breens/documents/BiTri2.pdf
• Mar 16th 2011, 05:48 PM
matheagle
maybe you can use a markov chain here with an upper triangular matrix.
• Mar 16th 2011, 07:37 PM
theodds
I think I've solved this. No time tonight to verify and post the answer, but I'll follow up tomorrow probably. It's based on a recursion and simple matrix stuff. All that's left to do is diagonalize an appropriate matrix to get the answer in a nice form.
• Mar 16th 2011, 11:05 PM
mr fantastic
Quote:

Originally Posted by theodds
I think I've solved this. No time tonight to verify and post the answer, but I'll follow up tomorrow probably. It's based on a recursion and simple matrix stuff. All that's left to do is diagonalize an appropriate matrix to get the answer in a nice form.

It would be better to wait a while before posting and see what progress the OP has made ....
• Mar 17th 2011, 07:59 AM
theodds
Quote:

Originally Posted by mr fantastic
It would be better to wait a while before posting and see what progress the OP has made ....

Okay. I'll at least post the answer so others can check it/against it, if they wish.

The probability that n occurs in a given infinite sequence of rolls, if my work is correct, is

$\displaystyle \frac 1 2 + \left(\frac {-1} 2\right)^n - \frac 1 2 \left(\frac {-1} 3 \right)^n$