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
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
where each are iid copies of that above distribution.
Let the points you've accumulated after n plays.
since you tossed a 1, n times
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.
you can look at the vector
Maybe use http://en.wikipedia.org/wiki/Probabi...ating_function
You can use the trinomial theorem to calculate the probabilities
maybe you can use a markov chain here with an upper triangular matrix.
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 ....
Originally Posted by theodds
Okay. I'll at least post the answer so others can check it/against it, if they wish.
Originally Posted by mr fantastic
The probability that n occurs in a given infinite sequence of rolls, if my work is correct, is