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: $2,3,\ldots,10,\mathcal{J},\mathcal{Q},\mathcal{K},\mathcal{A}.$
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- 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?
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What is the probability you do not receive any
$\mathcal{J},$$\mathcal{Q}$or$\mathcal{K}$after drawing$k$cards? -
Drawing
$k$cards at a time is analagous to drawing them one at a time without replacement. -
The joint probability of
$n$independent events is the product of probabilities of all of them.
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We will derive a formula for the probability of losing. If you draw
$k$cards, the probability that none of them are a$\mathcal{J},$$\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.