Answer to Question #105344 in Statistics and Probability for This Website is Legend

Question

What is the probability that you will first cut an ace
a) on the 5th cut ?
b) in fewer than 4 cuts?
c) what is the expected waiting time before you cut an ace ?

Answer 1

The negative hypergeometric distribution describes probabilities for when sampling from a finite population without replacement in which each sample can be classified into two mutually exclusive categories: $NHGeom(w,b,r)$ (Introduction to Probability, Second EditionJoseph K. Blitzstein, Jessica Hwang pages 168-169)

$$P(X=k)={\dbinom{w}{r-1}\dbinom{b}{k} \over\dbinom{w+b}{r+k-1}}\cdot{w-r+1\over w+b-r-k+1}$$

The expected value of $NHGeom(w,b,r)$ is

$$E(X)={rb \over w+1}$$

If we shuffle a deck of cards and deal them one at a time, the number of cards dealt before uncoveeering the first ace is $NHGeom(4,48,1).$

$$P(X=k)={\dbinom{4}{1-1}\dbinom{48}{k} \over\dbinom{4+48}{1+k-1}}\cdot{4-1+1\over 4+48-1-k+1}=$$ $$={\dbinom{4}{0}\dbinom{48}{k} \over\dbinom{52}{k}}\cdot{4\over 52-k}$$

(a) The probability that you will first cut an ace on the 5th cut is 

$$P(X=5-1)={\dbinom{4}{0}\dbinom{48}{4} \over\dbinom{52}{4}}\cdot{4\over 52-4}=$$ $$={1\cdot\dfrac{48!}{4!(48-4)!} \over\dfrac{52!}{4!(52-4)!}}\cdot{4\over 48}={3243 \over 54145}\approx0.0599$$

Or

$${48\over 52}\cdot{47\over 51}\cdot{46\over 50}\cdot{45\over 49}\cdot{4\over 48}={3243 \over 54145}\approx0.0599$$

(b) The probability that you will first cut an ace in fewer than 4 cuts is 

$$P(X=1-1)+P(X=2-1)+P(X=3-1)=$$ $$={4\over 52}+{48\over 52}\cdot{4\over 51}+{48\over 52}\cdot{47\over 51}\cdot{4\over 50}={1201\over 5525}\approx0.2174$$

(c) The expected waiting time before you cut an ace is

$$E(X)={1\cdot48 \over 4+1}={48 \over5}=9.6$$