Firstly apologies for the vague post title. Apart from probability I don't really know what sub category this question falls within.

If I have a pack of cards and I draw one card from it the chance of pulling a Jack, Queen, King or Ace is 16/52. Then pulling a second is obviously 15/51.

Now if I have a 5 pack of cards the chance of me pulling a Jack, queen, king or ace from any one pack 16/52 * 5. Let's assume pulling one card from each pack is considered a "round".

So for the first round you pull one card from each pack. If you do not pull one of the jack, queen, king or ace card from a pack them you return the card. If however you do pull one of these cards from a pack then you do not return the card to that pack. Instead you play another round which means you pull one more card from each pack. If one of the pack yields one of the cards you are interested in you play another round. The game ends when one round does not give one of those cards.

So in summary 1. Pull one card from all 5 decks 2. Return any cards that you are not interested in. 3. If one or more of the packs gives a card you are interested in do not return it to pack and repeat. 4. Stop when no pack gives interested card.

I want to know how calculate the average number of rounds until the game stops.

Thanks for taking the time to read and for any help you may have.

  • 2
    $\begingroup$ Didn't finish reading, but your calculation for 5 decks is wrong. You can't just add the probabilities like that. $\endgroup$ – lulu Jul 23 '15 at 19:19
  • $\begingroup$ The game you describe is unclear. If, say, in your first draw of 5 cards you get no High cards...do you put all 5 back in their respective decks? If you get, say, one K and no other High Cards, do you retain the K and replace the other 4? $\endgroup$ – lulu Jul 23 '15 at 19:21
  • $\begingroup$ I don't get the point of replacing the cards if you have drawn a High Card. The game ends the moment you get the High Card, right? So what difference does it make what you do next? $\endgroup$ – lulu Jul 23 '15 at 19:23
  • $\begingroup$ Hi lulu, the game ends if you do NOT pull a high card from ANY of the 5 packs. $\endgroup$ – codetemplar Jul 23 '15 at 19:30
  • $\begingroup$ Yes you would retain the k and replace other 4 to their respective packs. Then repeats until you don't pull a single high card $\endgroup$ – codetemplar Jul 23 '15 at 19:31

This gets quite messy, though it should be easy to automate. To illustrate the idea I'll start with a simpler version of the game: You draw one card from each of 5 ordinary decks. If you get no High Cards (one of J, Q, K, A) the game ends. Otherwise you replace all the drawn cards in their respective hand and play another round.

Given that I have the game right, the question is: What's the expected number of rounds it takes to play a complete game.

To answer it, we really only need to know the probability of the game ending on a single round, call it P. As you miss with a given deck with probability $\frac {36}{52}$ you miss on all five with probability $$P = \left(\frac {36}{52}\right)^5\sim.159$$.

Now, let E be the expected number of rounds you play (counting the final round as a round). Starting at the beginning, you play one round. That ends it with probability P and returns you to the beginning with probability (1-P). In the first case you only needed 1 round. In the second, you'll need E + 1 on average. Thus $$E = P*1 + (1-P)*(E+1)\;\;\;\Rightarrow\;\;\;PE = 1$$ Using the value of P we computed, we see that $E=\left(\frac {52}{36}\right)^5\sim6.288$.

Now, the actual game is harder to analyze because you keep removing High Cards, thereby making your chances better on the next round. it is clear that the answer has to be less than 6.288 but I don't see an easy way to be much more precise. To do it properly (on a computer) I would introduce states labeled by 5 integers $(a_1,...a_5)$. These integers reflect the number of High Cards you have removed from each deck. At the start, of course, you are in state (0,0,0,0,0). On any round any of the indices $a_i$ might either stay the same or ratchet up by 1. It is easy to compute the probability of either occurrence so we know the transition probabilities between all the states. We know, too, that the $a_i$ are all bounded by 16 so from the state (16,16,16,16,16) you are guaranteed to win on the next round. We (or, better, our computers) could then easily work backwards...computing the expected number of rounds from states like (15,16,16,16,16) and so on. Too many steps to do by hand (at least as far as I can see) but a good computing project otherwise.

  • $\begingroup$ Wonderful, thank you very much for your help. I am going to go away and try it out! $\endgroup$ – codetemplar Jul 23 '15 at 21:24

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