PeroK's latest activity

PS note that we can simply add the expected value for each colour: $$E = E(N, R) + E(N, B) + E(N, G)$$

Inspired by the posts by @Gavran we can show that for ##R## red balls and ##N## balls in total, the expected number of red balls at the...

## P(X=g) ## - probability that we have g green cards remaining ## P(X=b) ## - probability that we have b blue cards remaining ## P(X=r)...

I also calculated it to be 1.515 using a different but related method. What the OP is asking for is modeled by the negative...

That would be the same as drawing N cards without replacement that match the first color.

That is very far off. The correct answer is 1.515 Your concept of “green survives” appears to be the problem.

I put the precise calculations into a spreadsheet (it only took a few minutes) and got 1.514988104

Alternatively, you could note that the probabilities above are related to each other by: Probability that precisely the last (r+1)...

Here's some pseudo-code: Specify number of balls of each colour R = 40, B = 35, G = 25, N = R + B + G Initialize expectation value E =...

If you can revolutionise physics in your spare time, why are you wasting your talents on software development?

Where did you get that idea from?

PS multiply the probability in each case by the relevant number to get the contribution to the expected value.

You could crunch a precise answer using a program. Calculate the probability of the last card being 1-40 red, then 1-35 blue and 1-25...

If you lay the 100 cards out in a row, there is a statistical symmetry going first to last or last to first.

Highly recommended, but very math-heavy. Fine for math nerds like me, but others may not enjoy it as much as I did. Thanks Bill