# Drawing cards from a deck with multiple success states

Given a standard 52-card deck, you make draws (with replacement) except you keep Kings and Jacks (they are not replaced). After each draw / replacement the deck is shuffled. How would you calculate the expected number of draws until you have exactly two Kings and two Jacks? (If you have two Kings already and you draw a third King, it is put back into the deck).

Thank you!

• I think some casework is necessary due to the lack of replacement after getting a desired card. Commented Jun 1, 2021 at 18:10

The number of draws until keeping four cards is $$T=T_1+T_2+T_3+T_4$$ where $$T_i$$ is the number of draws after obtaining the $$(i-1)$$th card until obtaining the $$i$$th card.

The expected number of draws until getting the first card is $$E[T_1] = \frac{52}{8}$$.

The expected number of draws until getting the next card is $$E[T_2] = \frac{51}{7}$$.

From here, some casework is necessary. There is a $$\frac{3}{7}$$ chance that the second kept card is the same rank as the first kept card (call this event $$A$$), and a $$\frac{4}{7}$$ chance that this second kept card is the other rank (J if the first kept card was K, and vice versa). In the first case, you are now only pursuing cards of the missing rank; in the second case you are still open to receiving either rank.

So, $$E[T_3] = E[T_3 \mid A] P(A) + E[T_3 \mid A^c] P(A^c) = \frac{50}{4} \cdot \frac{3}{7} + \frac{50}{6} \cdot \frac{4}{7}.$$

Finally $$E[T_4] = \frac{49}{3}$$.

My quickest idea is this approach with Markov Chains. Below the matrix has the states with the number of cards that you hold in your hand and the lines have the transition probabilities. The first state is that you have no King or Jack, so the chance of after this draw to not have a King or Jack is $$44/52$$ whilst the chance that you end now with one of these is $$8/52$$. The second state as mentioned is that you have one of them, the third line gives that you have two of the same type, the fourth that you have one of each and the last that you have two of a type and one of the other. We only have here the transient states of the Markov Chain. $$A =\begin{bmatrix} 44/52 & 8/52 & 0 & 0 & 0 \\ 0 & 44/51 & 3/51 & 4/51 & 0 \\ 0 & 0 & 46/50 & 0 & 4/50 \\ 0 & 0 & 0 & 44/50 & 6/50\\0 & 0 & 0 & 0 & 46/49 \end{bmatrix}$$ After this we can calculate $$B= (I-A)^{-1}$$. This matrix entries gives the expected amount of time reaching a state from a starting state. The answer that you are looking for will be the first entry of $$BV$$, where $$V$$ is the column vector made of ones, since we are simply summing all the expected time in the transient states. I found this value to be $$845/21$$.

• I think your $A$ should be the $4 \times 4$ submatrix of the full transition matrix you've written? Commented Jun 1, 2021 at 19:29
• Nah, I have a state with no cards, one card of a type, two of the same type, 1 of each type and one with 2 of one type and none of the other type. So in total 5 states... And my solution matches with the sum given by you... Commented Jun 1, 2021 at 19:31
• Thanks; sorry I had misread something. Commented Jun 1, 2021 at 19:32