- #1
- 1,650
- 246
Homework Statement
We have a game in which a player plays against a dealer. The goal is to get closest to 100 without going over.
The game goes like this:
A random integer is generated from 1 to 100 (including 1 and 100) which is given to the player. The player can either "hit" or "stay." If the player hits, they will get another random integer (from 1 to 100) which is added on to their number. The player can do this as many times as they wish, but they instantly lose ("bust") if they go over 100.
Once the player stays, the dealer will keep rolling a random integer (from 1 to 100) for their-self until they either beat the player (get closer to 100) or bust (go over 100).
I am curious to find what is the chance the player will win if the player chooses to stay at a value of "m"
Homework Equations
##\sum\limits_{n=1}^{m}\bigg (\frac{n}{100}\sum\limits_{s=1}^{n}\Big (100^{-s}{n-1 \choose s-1}\Big )\bigg )##
The Attempt at a Solution
The equation in "
Homework Equations
" is my guess at the answer.The reason I guessed this is because the problem reminded me of "pascal's triangle." So let us identify an entry in pascal's triangle by two numbers, n and s.
n is the row (starting from n=1 being "1"; n=2 being "1, 1"; n=3 being "1, 2, 1"; etc.) and s is the number of entries from the left (starting from s=1 being the leftmost entry, which is always 1; s=2 being n-1; etc.).
(So s takes values from 1 to n.)
The sth entry of the nth row will be given by ##{n-1\choose s-1}##
Now this is what I'm thinking (sort of a guess)... The sth entry of the nth row represents the number of paths for the dealer to get a value of n in s steps.
For example, there is only 1 way to get n in 1 step: roll an n; and there is only 1 way to get an n in n steps: roll all ones.
The probability for a specific path which takes s steps will be 100-s and so the probability that the dealer gets to a value of n in s steps will be ##100^{-s}{n-1\choose s-1}##
Therefore the chance of the dealer getting a value of n is ##\sum\limits_{s=1}^n100^{-s}{n-1\choose s-1}##
Now, if the dealer has a value of n, then the chance of you winning is n/100 on the next roll.
For example, if the dealer has a 2, there is a 0.02 chance you will win on the next roll, because the dealer can roll either 100 or 99.
So we should multiply the chance you will win if the dealer has n by the chance of the dealer having n, and then sum over all n from 1 to your value, m. That should be the probability of you winning by staying on m. Thus we get:
##\sum\limits_{n=1}^{m}\bigg (\frac{n}{100}\sum\limits_{s=1}^{n}\Big (100^{-s}{n-1 \choose s-1}\Big )\bigg )##
Can anyone confirm if this is correct or incorrect?