You draw $n$
cards at random from a standard deck of 52 playing cards. If the $n$
cards contain either a $\mathcal{J}$
, $\mathcal{Q}$
or $\mathcal{K}$
then you win. If not then you put all $n$
cards back in the deck, reshuffle and draw $n - 1$
new cards. You repeat until you win or until $n = 0$
, when you lose. What are the probabilities of winning and losing if you start with $\textit{(a)}$
$n = 1$
, $\textit{(b)}$
$n = 2$
, $\textit{(c)}$
$n = 8$
, or $\textit{(d)}$
$n=41$
?
$\textit{Note:}$
The standard deck of 52 cards has 4 cards of each of the following: $2,3,\ldots,10,\mathcal{J},\mathcal{Q},\mathcal{K},\mathcal{A}.$
-
- What is the probability of getting at least one heads in three coin flips?
- What is the probability of getting two cards of the same rank when you draw two cards?
-
What is the probability you do not receive any
$\mathcal{J},$
$\mathcal{Q}$
or$\mathcal{K}$
after drawing$k$
cards? -
Drawing
$k$
cards at a time is analagous to drawing them one at a time without replacement. -
The joint probability of
$n$
independent events is the product of probabilities of all of them.
-
We will derive a formula for the probability of losing. If you draw
$k$
cards, the probability that none of them are a$\mathcal{J},$
$\mathcal{Q}$
or$\mathcal{K}$
is${40\over52} \cdot {39\over51} \cdots {40-k+1\over52-k+1} = \prod_{j=0}^{k-1} \frac{40-j}{52-j}.$
For any given
$n,$
you need to not get a$\mathcal{J}$
,$\mathcal{Q}$
or$\mathcal{K}$
in$n$
different draws in order to lose. As these draws are independent, the probability of losing is the product of probabilities of not winning in each draw.For
$n>40$
winning is ensured. For$n \leq 40$
, the losing probability can be computed as follows:$$ \begin{aligned} P_l(n) &= \prod_{k=1}^{n}\prod_{j=0}^{k-1} \frac{40-j}{52-j} \\ &= \prod_{j=0}^{n-1}\prod_{k=j+1}^{n}\frac{40-j}{52-j} \\ &= \prod_{j=0}^{n-1}\left(\frac{40-j}{52-j}\right)^{n-j} \end{aligned} $$
We can now compute$P_l(1)=\frac{10}{13},$
$P_l(2)= \frac{100}{221},$
$P_l(8) = \prod_{j=0}^{7}\left(\frac{40-j}{52-j}\right)^{8-j}.$
The winning probabilities are found by subtracting the losing probabilities from 1.