Discussion about this post

User's avatar
Paul Fischer's avatar

B is the following:

M=red balls

N=total balls

K=number of draws before the K+1 draw

L=at least one red ball drawn in previous k draws

Rk+1=red ball on k+1 draw

P(no red balls in k draws)=(N-M choose K)/(N choose K)

=(N-M)(N-M-1)…(N-M-K+1)/N(N-1)…(N-K+1)

P(L|B)=1-P(no red balls drawn)=1 - (N-M)(N-M-1)…(N-M-K+1)/N(N-1)…(N-K+1)

Now Bayes gives P(Rk+1|BL)=P(BL|Rk+1)P(Rk+1)/P(BL)

We know P(BL)= from the above

but P(BL|Rk+1) = P(BL) if we know we draw a red ball on the k+1 and we don’t change the previous draws so we have

P(BL|Rk+1)=P(BL) then P(Rk+1|BL)=[P(BL)(M/N)]/P(BL)=M/N

Expand full comment
Paul Fischer's avatar

Great stuff William! Love every second. It compliments my 5 decades of misunderstanding (frequentist) probability well. HAHA. Honestly, I had a solid year of graduate level probability theory and not once heard about any of this.

Expand full comment
6 more comments...

No posts

Ready for more?