This is probably a pretty easy problem for those who know how to calculate it, but here goes:
Say you got the opportunity to throw a die an unlimited amount of times. You had to pay $1 for each throw, but every time you threw a six you got payed $7.
So you have an expected roi of 16,66%, but what are the chances you are a losing player after for example 1000 throws, and how do you calculate it?
Say you throw the dice n times and it shows a six k times. For you to loose money you need k to be smaller than n/7. (For example you'd break even if after 70 throws you only got 10 sixes and lose money with less sixes.)
Since the number of sixes you throw is binomially distributed after n throws you will have thrown k sixes with the probability B(k,n,1/6) = C(n,k) * (1/6)^k * (1-1/6)^(n-k).
Now to calculate the probability of throwing less than n/7 sixes you 'simply' need to add up the probabilities - i.e. sum( B(k,n,1/6); 0<=k<n/7 ).
Fortunately computers can calculate those long sums for us (for example Calc has a function BINOMDIST).
In your case the probability of losing money is 1.8%.
Sixhigh has mentioned one way of doing this but you need the assistance of a computer to solve for the sum of a bunch of binomials. You may approximate the solution using the normal distribution if you know your win-rate and standard deviation.
EV = 1/6
Compute var(x) where x represents the amount won or lost when a fair die is rolled.
Var(x) = E(x^2) - [E(x)]^2
E(x^2) = 5*(-1)^2*1/6 + (+6)^2*1/6
E(x) = 1/6
var(x) = 6.80555...
s.d(x) = sqrt(var(x))
s.d(x) ~ 2.6087
s.d over 1000 trials is ~ 2.6087*sqrt(1000) ~ 82.494
After 1000 trials, you expect to be up $166.66... in profits.
We have z = (0-166.66)/82.494 ~ -2.02
We can do better than this since we can never win exactly 0 at the end of the 1000 rolls. We can be down $6 which gives us a better approximation.
z= (-6-166.66)/82.494
z ~ 2.093
This means that you are 2.093 standard deviations below the mean which corresponds to a 1.83 % chance that you will show a negative profit after 1000 trials.
Saying the same thing in a less high fallutin style:
In 1000 throws you need to get 143 sixes to be ahead. You ought to get 167. To figure out the chances you will fall 24 or more short, you see what one standard deviation works out to. The formula is the square root of npq. The square root 1000x5/6x1/6 is slightly below 12. So you would have to have a bit more than two standard deviations of unluckiness to be behind.
@op: there's a nice little weird thing about this problem: although the probability of losing converges to zero with the number of throws, this probability does some unexpected jumps. For example the probability of being behind behind after 175 throws (which have an expectation of 29.2) is greater than the probability of being behind after after 125 throws (which have an expectation of only 20.8).
@op: there's a nice little weird thing about this problem: although the probability of losing converges to zero with the number of throws, this probability does some unexpected jumps. For example the probability of being behind behind after 175 throws (which have an expectation of 29.2) is greater than the probability of being behind after after 125 throws (which have an expectation of only 20.8).
I don't think i believe this. Can u show the math please.
Then f(n) is the probability of being behind after n throws.
using calc or excel we get
f(125) = 0.2146
f(175) = 0.2316
This is so because you need to throw more sixes to be ahead after 175 throws (25) than after 125 (18).
An easier example: after 6 throws you need to have thrown at least one six to be ahead, this will happen with probability 0.67. But to be ahead after 7 throws you need to have thrown at least two sixes which will only happen with probability 0.33.
My first guess was that if it was true, it was sort of a modular arithmetic thing..
I have excel, but its in thai, so i can't check your work.. its interesting tho.. the ev would def be higher, but your saying the chance of being ahead is less... interesting..
