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This is a variation on a common probability question: You deal cards from a well-shuffled pack until you deal the Two of Hearts. What is the probability you see at least one king, one queen, and one jack before you deal the Two of Hearts?

My Approach

We remove the four kings, four queens, four jacks and Two of Hearts from the pack and consider just these cards. The other 39 cards can be in any permutation.

Focusing on the 13 cards, we see that we can abstract away the actual permutation of the 13 cards, and focus just on the occurrences of the unique values (for eg. (King, Queen, Jack, Two) or (King, Two, Queen, Jack)). We care that at least one king, one queen, and one jack appear before the Two. So we want to find

$P($first type of card to appear is $i)\cdot{P(}$first non-i type of card to appear is $j)\cdot{P(}$first non-i and non-j type of card to appear is $k\cdot{P(}$Two of Hearts appears$)$

where $i, j, k$ can be King, Queen, or Jack.

This is equal to $\frac{12}{13}\times{\frac{8}{9}\times{\frac{4}{5}\times{1}}}=\frac{384}{585}$

Does this seem right to you?

Abhay Agarwal
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  • Note that this problem is not the same as the one in your prior question. There you say exactly and here you say at least. Was that your intent? – lulu May 10 '24 at 10:16
  • Hmm. Conditioned on first observing a king, the first non-king card will be out of the remaining 9, right? – Abhay Agarwal May 10 '24 at 10:17
  • I initially assumed that you were asking the same question as you did in your other post. Then I noticed that this one was different...was that your intent? – lulu May 10 '24 at 10:17
  • Yes, this one is different. I was trying to create a problem in which I could use a similar strategy to the candy question. – Abhay Agarwal May 10 '24 at 10:18
  • Ok, but did you intend for this problem to be the same as the one in your prior post? – lulu May 10 '24 at 10:19
  • Nope, this is intentionally different. Still not quite sure where I'm going wrong, but I do agree: my number is too high. – Abhay Agarwal May 10 '24 at 10:19
  • Let us continue this discussion in chat. – Abhay Agarwal May 10 '24 at 10:29
  • Your method works for this problem. As an alternative, you could use Inclusion-Exclusion. There is a $\frac 15$ chance that the $2$ precedes any given suit, a $\frac 19$ chance that it precedes any two given suits and a $\frac 1{13}$ chance it precedes all of them. Thus the answer is $1-\frac 35+\frac 39-\frac 1{13}$ which matches your result. – lulu May 10 '24 at 10:29
  • No need for a chat, your method is fine. My early comments were based on the bad reading of the question. – lulu May 10 '24 at 10:30
  • I agree with your analysis. A simpler, less elegant approach, that generalizes well for problems of this type is represented by the comment of lulu, that specifically refers to using Inclusion-Exclusion to attack the problem. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula. – user2661923 May 10 '24 at 13:59
  • Thank you very much! – Abhay Agarwal May 10 '24 at 18:52

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