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Suppose I have a well shuffled standard deck of $52$ cards. I draw them one at a time (without replacement) and sort the chosen cards by rank so that it is easy for me to see pairs, triples, quads.... I want to know what is the probability of seeing exactly three triples (such as $5,5,5$, $3,3,3$, $K,K,K$) before seeing a quad (such as $10,10,10,10$). Note that the triples and quad need not be in any order, they can have other "irrelevant" cards drawn in between them.

Just to clarify, the triple and quads do not need to be consecutive cards. For example, if the first $6$ cards drawn are $3,5,6,5,K,5$, then that is a triple for the $5$s.

Note that when we either see $3$ triples or one quad, we are done with that hand. No other cards will be drawn. So for example, if you have $3,3,3,5,5,7,7,7$ so far (and all the other drawn cards are singles, then another $3$ or $5$ draw will immediately stop the hand but for different reasons. We will either have quad $3$s at that point or $3$ triples. A drawn $7$ would also stop this sample hand.

A quad winner would be something like $K,K,K,7,7,7,3,3,K$ with any "irrelevant" single or double (paired) cards interspersed such as $2,J,9...$.

Just for "extra" clarity, a quad does NOT also count as a triple. A triple becomes a quad and loses its triple status if the $4$th card of that same rank is drawn in the same hand.

For even more clarity (I didn't know this question would be ambiguous), I am $NOT$ asking for $3$ triples followed by a quad before seeing the $4$th triple. I am saying all you have to see is $3$ triples without seeing a quad and you can stop that hand and it is counted as a winner. Sorry for any confusion. I thought it was clearly stated but now that I reread the title it is a little confusing but not if you read the examples it shouldn't be.

David
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    My instincts would be to do this by Monte Carlo (I mean, just sample a few million shuffles). To do it analytically it might (possibly) help to look at the deck backwards. To describe the first completed quad you can start at the back of the deck and run until you have seen all $13$ ranks. Now, amongst those cards you went through you want the probability that at least three of the ranks occurred only once (four if you count the one you stopped at). Maybe that's easier (maybe not). – lulu Mar 29 '16 at 15:55
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    Just before completing the first quad, you obviously have at least one triple. So if you don't have three triples, you have at most one other, in which case you've drawn at most $3+3+11\cdot2=28$ cards. So an exact answer for the complementary probability could be computed with a frequency count for each of $26$ cases $4\le k\le29$, where $k$ is the card count that completes the first quad, and where what you're counting is the number of ways you can have $k-1$ cards with at most two triples. – Barry Cipra Mar 29 '16 at 16:10
  • Exactly three triples or at least three triples before a quad ? – true blue anil Mar 29 '16 at 16:10
  • To be clear: If the quad was $K,K,K,K$ I was not counting the first three $K's$ as a triple. Of course, I might be misinterpreting. – lulu Mar 29 '16 at 16:12
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    @lulu, and I'm clearly counting the first three $K$'s as a triple. I'm interpreting the "seeing three triples" and "seeing a quad" as "stopping" rules. But I agree, it would help for the OP to clarify. – Barry Cipra Mar 29 '16 at 16:14
  • @trueblueanil, my interpretation of "seeing three triples" as a "stopping" rule would render your question irrelevant. lulu's interpretation would not. – Barry Cipra Mar 29 '16 at 16:17
  • @BarryCipra Your reading certainly makes sense, and seems more tractable than mine. – lulu Mar 29 '16 at 16:18
  • @David: Your edit has rendered the question inconsistent. The only consistent interpretation of the original question was the one with "at least $3$ triples". In the current version, "exactly $3$ triples" is inconsistent with your statement "Note that when we either see $3$ triples or one quad, we are done with that hand.", which is only true if we don't need to continue in order to find out whether a fourth triple is completed before the first quad. – joriki Mar 29 '16 at 22:21
  • I don't see how my original question could have been misinterpreted this way. I stated what is the probability of getting 3 triples before getting a quad. This implies (quite clearly I might add) that any 3 triples ends the hand and any quad ends the hand both immediately. I even gave examples to clarify. I would say generally whenever anyone asks for 3 of something, it means exactly 3 of something, not 3 or more. If you interpret 3 to mean 3 or more then you are in trouble generally speaking. – David Mar 29 '16 at 23:30
  • The English language is "tricky" to word this problem clearly but it seems the natural interpretation would be if you see $3$ triples and haven't yet seen a quad, you are done with that hand cuz it is a winner. It doesn't mean you need to see if there is no $4$th triple and then you see a quad. The hand is terminated at either $3$ seen triples (with no quad seen) or a quad seen before $3$ triples. – David Mar 30 '16 at 00:48
  • @David: I think we're in agreement about the original question. I think you're right that it was unambiguous, and I don't think I misinterpreted it; I think my answer answers the question as you intended it. My problem is with the edit and with how you're using the expressions "exactly $3$" and "$3$ or more". It seems that you're using "exactly $3$" to mean what I would call "$3$ or more" (or "at least $3$", as true blue anil phrased it), and I don't understand what scenario you mean by "$3$ or more". I'll try to explicate how I'm using the expressions, and perhaps you can do the same. – joriki Mar 30 '16 at 08:41
  • @David: To my mind, "the probability of getting at least three triples before a single quad" means you count the number of triples you get before you get a quad, and if it's at least three, it's a success. Since you know that it will be at least three once you get the third, you can stop at that point. This is, if I understand correctly, what you're asking about. By contrast, "the probability of getting exactly three triples before a single quad" means you count the number of triples you get before you get a quad, and if it's exactly three, it's a success. – joriki Mar 30 '16 at 08:44
  • @David: In this case, you don't know whether it will be exactly three when you draw the third, so you have to keep going until you either draw a fourth triple (a failure) or a quad (a success). This, if I understand correctly, is not what you're asking about. You seem to be using "at least $3$" and "exactly $3$" to refer to the number of triples actually drawn in practice, rather than the number of triples seen before seeing a quad. (I see now that you removed "exactly" in the title, but it's still in the body of the question.) – joriki Mar 30 '16 at 08:54
  • This is why examples are sometimes needed cuz the wording can be difficult. As soon as you see 3 triples (and no quad) stop the hand. As soon as you see a quad (and not 3 triples) stop the hand. That pretty much clears it up. Any other interpretations are not what I intended. So the minimum number of cards to stop the hand is $4$. The minimum number of cards for a "winner" is $9$. The maximum number of cards needed for a winner is $29$. – David Mar 30 '16 at 13:51

1 Answers1

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Note: This is an answer to the original question, which seemed to be asking for the probability of getting at least three triples before a quad, not exactly three triples.

The probability of completing a third triple without having completed a quad on draw $k$ is

$$ \frac{\sum_{m=0}^{\min(10,\left\lfloor\frac{k-9}2\right\rfloor)}\binom{12}2\binom43^2\binom32\binom{10}m\binom{10-m}{k-9-2m}\binom42^m\binom41^{k-9-2m}}{\binom{51}{k-1}}\;, $$

since, given the card drawn, we can select $k-1$ from the remaining $51$ by selecting $2$ of $12$ ranks, and in each $3$ of $4$ suits, for the triples already completed, $2$ of $3$ suits for the double being completed to a triple, $m$ of $10$ ranks for which doubles have been drawn and $k-9-2m$ of the remaining $10-m$ ranks for which singles have been drawn, along with the corresponding choices of suits. Summing this for the possible values $8\le k\le29$ yields

$$ \frac{2556075271418192}{3583273791308225}\approx71\%\;. $$

Here's code that computes the sum and checks it with a simulation.

joriki
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  • Wow what a "mess" of a formula! More involved than I thought it would be. – David Mar 29 '16 at 16:31
  • @David: The relatively low denominator (compared to $52!$) might indicate that there's a simpler way to do this that I didn't think of. – joriki Mar 29 '16 at 16:38
  • Yes there will be a winner way before the deck is exhausted normally so that may help some. $29$ cards drawn will guarantee at least $3$ triples and $40$ cards drawn will guarantee at least one quad. There is no need to analyze all $52$ cards drawn. A Monto Carlo simulation can not only tell you the probability, but also the average # of cards drawn per hand which would be informative and interesting. Perhaps less than half of the deck need be drawn on average but that is a guess. – David Mar 29 '16 at 16:43
  • @David: I added that to the simulation. The average number of cards drawn until either a third triple or a quad is completed is about $21$. I'm not sure what you mean by "There is no need to analyze all $52$ cards drawn." I didn't analyze all $52$ cards drawn, only the $k$ that are drawn, but I have to take into account that they could be any of the $52$. – joriki Mar 29 '16 at 17:16
  • $21$ seems about right for average number of cards. I guess what makes this problem more challenging than a "typical" counting problem is that there is a "competition" between $2$ events. The $3$ triples are a "winner" and the single quad is "loser". The not having to analyze all $52$ cards will help with simulation since we can stop at card $40$ but likely much sooner. – David Mar 29 '16 at 17:18
  • I might try a computer simulation and see what I get and then post the results. – David Mar 29 '16 at 23:37
  • @David: Note that the code I posted includes a simulation, whose results match the result of the calculation. – joriki Mar 30 '16 at 08:56