Suppose we start at the number zero, and repeatedly roll a fair $6$-sided die, adding the resulting number to the total. We win if we reach the total $k$, and lose if we overshoot, for example by reaching a total of $k-1$ and then rolling a number greater than $1$. What is the probability to win?
Asked
Active
Viewed 64 times
1
-
1Well, there's a simple recursion ($P_n=\frac 16 \times \left(P_{n-6}+\cdots P_{n-1}\right)$). Were you looking for the limit as $k\to \infty$ or something like that? – lulu Jan 06 '17 at 02:01
-
k is a fixed number. I thought a dynamic programming solution but I am not sure it's the fastest and if it's correct at all. – MathGeek Jan 06 '17 at 02:03
-
Ok, but then the recursion works. It's linear so you can solve it in closed form (though the characteristic polynomial has degree $6$ so the closed form isn't very nice). – lulu Jan 06 '17 at 02:08
-
This question was technically about the case $k=100$ but the answers discuss the general case. The limiting probability is $2/7.$ – bof Jan 06 '17 at 02:15
-
How does one get such a recursion? – Kaynex Jan 06 '17 at 02:21
-
2@Kaynex In order to reach $k$ exactly one has to do exactly one of the following: for some $i\in [1,\cdots,6]$ you must reach $k-i$ and then throw an $i$. The recursion just restates that. – lulu Jan 06 '17 at 02:25
-
@Kaynex $P_n$ is your probability of hitting $n$ if you start from $0.$ On your first roll you are equally likely to roll $1,2,3,4,5,6.$ If you start by rolling $k$ your conditional probability of hitting $n$ is $P_{n-k},$ the probability of hitting $n-k$ with your subsequent rolls. $P_n$ is the average of those six conditional probabilities. – bof Jan 06 '17 at 02:27
-
@lulu Yes, that's another way to look at it. Reels the mind. – bof Jan 06 '17 at 02:31