The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field.

The Computer Laboratory

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

The above links are provided as is. They are not affiliated with the Climb Foundation unless otherwise specified.

Warm-up Questions

  1. What is the probability of getting at least one heads in three coin flips?
  2. What is the probability of getting two cards of the same rank when you draw two cards?


  • Hint 1
    What is the probability you do not receive any \(\mathcal{J},\) \(\mathcal{Q}\) or \(\mathcal{K}\) after drawing \(k\) cards?
  • Hint 2
    Drawing \(k\) cards at a time is analagous to drawing them one at a time without replacement.
  • Hint 3
    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: \[ \begin{align} P_l(n) &= \prod_{k=1}^{n}\prod_{j=0}^{k-1} \frac{40-j}{52-j} \\ &= \prod_{j=0}^{n-1}\prod_{k=j+1}^{n}\frac{40-j}{52-j} \\ &= \prod_{j=0}^{n-1}\left(\frac{40-j}{52-j}\right)^{n-j} \end{align} \] 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.

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at[email protected]. Please do not write to this address regarding general admissions or course queries.