# Practice Paper 3 Question 17

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: \(2,3,\ldots,10,\mathcal{J},\mathcal{Q},\mathcal{K},\mathcal{A}.\)

## Related topics

## Warm-up Questions

- 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?

## Hints

- Hint 1What is the probability you do not receive any \(\mathcal{J},\) \(\mathcal{Q}\) or \(\mathcal{K}\) after drawing \(k\) cards?
- Hint 2Drawing \(k\) cards at a time is analagous to drawing them one at a time without replacement.
- Hint 3The joint probability of \(n\) independent events is the product of probabilities of all of them.

## Solution

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: \[ \begin{align} P_l(n) &= \prod_{k=1}^{n}\prod_{j=0}^{k-1} \frac{40-j}{52-j} \\ &= \prod_{j=0}^{n-1}\prod_{k=j+1}^{n}\frac{40-j}{52-j} \\ &= \prod_{j=0}^{n-1}\left(\frac{40-j}{52-j}\right)^{n-j} \end{align} \] 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.

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