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

Fixed number of successes without replacemement: Negative Hypergeometric.

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 $NHGeom(4,48,1)$

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

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

$$P(X=5-1)={\dbinom{4}{0}\dbinom{48}{5-1} \over \dbinom{52}{5-1}}\cdot{4 \over 52-(5-1)}=$$ $${ \dfrac{48!}{4!(48-4)!}\over \dfrac{52!}{4!(52-4)!}}\cdot{4 \over 48}=$$ $$={48(47)(46)(45) \over 52(51)(50)(49)}\cdot{4 \over 48}={3243 \over 54145}\approx0.0599$$

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

$$E(X)={1\cdot48\over 4+1}={48\over 5}=9.6$$