Does a player having one hand increase the likelihood of other players also having that hand?

by alex   Last Updated October 10, 2019 04:20 AM

Say a deck of cards is dealt out equally to four players (each player receives 13 cards).

A friend of mine said he believed that if one player is dealt four-of-a-kind (for instance), then the likelihood of another player having four-of-a-kind is increased - compared to if no other players had received a four-of-a-kind.

Statistics isn't my strong point but this sort of makes sense given the pigeonhole principle - if one player gets AAAAKKKKQQQQJ, then I would think other players would have a higher likelihood of having four-of-a-kinds in their hand compared to if the one player was dealt AQKJ1098765432.

I wrote a Python program that performs a Monte Carlo evaluation to validate this theory, which found:

  • The odds of exactly one player having a four-of-a-kind are ~12%.
  • The odds of exactly two players having a four-of-a-kind are ~0.77%.
  • The odds of exactly three players having a four-of-a-kind are ~0.03%.
  • The odds of all four players having a four-of-a-kind are ~0.001%.

But counter-intuitively, four-of-a-kind frequencies appear to decrease as more players are dealt those hands:

  • The odds of two or more players having a four-of-a-kind when at least one player has four-of-a-kind are ~6.24%.
  • The odds of three or more players having a four-of-a-kind when at least two players have four-of-a-kind are ~3.9%.
  • The odds of all four players having a four-of-a-kind when at least three players have four-of-a-kind are ~1.39%.

The result is non-intuitive and I'm all sorts of confused - not sure if my friend's hypothesis was incorrect, or if I'm asking my program the wrong questions.



Answers 3


First off: there are

$$\frac{1}{4!} \cdot \left(\begin{array}{c}52\\13\end{array}\right) \cdot \left(\begin{array}{c}39\\13\end{array}\right) \cdot \left(\begin{array}{c}26\\13\end{array}\right) = 22,351,974,068,953,663,683,015,600,000$$

distinct combinations of four thirteen-card hands drawn from a deck of 52 cards. If each combination takes just one nanosecond to examine, it would take thousands of years to examine any significant fraction of the possibilities -- and about 700 billion years to examine every combination. It seems likely that your Monte Carlo simulation just didn't have enough time to get an accurate statistic here. Did you get the same values from running the simulation multiple times?

Since the probability space is so large, we'll probably need to approach this based on theory. Inconveniently, analyzing four-of-a-kinds with thirteen-card hands is complicated, so let's simplify to a smaller deck. Consider a deck of cards with just two ranks (A and 2) and two players. In that case, if one player has a four-of-a-kind, then the other player automatically does as well. This extreme case suggests that your friend's hypothesis is correct.

To test the idea, let's step it up a little: three ranks (A, 2, and 3) but still two players. Then the probability of one player having a four-of-a-kind is given by

$$\frac{3}{11}\cdot\frac{2}{10}\cdot\frac19 \cdot \left(\begin{array}{c}6\\4\end{array}\right) = \frac{1}{11}$$

but the probability of the second player having a four-of-a-kind given that the first player does is $1/2$ (as long as the leftover two cards in the first player's hand match, the second player automatically has the other four-of-a-kind).

This pretty clearly highlights the trend: one player having a four-of-a-kind does increase the likelihood of other players having four-of-a-kinds. It's just a difficult phenomenon to see in a large deck.

Reese
Reese
October 09, 2019 17:29 PM

You have to be careful what question you are asking. What your friend seems to be claiming is if you draw a hand of $13$ cards for one player and it has four of a kind, the chance of a second hand you draw from the rest of the deck having four of a kind is increased. My intuition agrees with that. Distributing all the cards and asking the chance of two four of a kinds given there is one four of a kind is a different question. It seems likely you could modify your program to ask the first question.

For a comparison, I imagined four players each flipping a coin with $0.05$ chance to come up heads. Clearly here your friend's question has player $2$ getting heads with chance $0.05$ regardless of what player $1$ does. The chance at least one player gets heads is about $0.1714$. The chance at least two players get heads, given that at least one did, is about $0.07558$, which is less than the chance of at least one getting heads.

Ross Millikan
Ross Millikan
October 09, 2019 18:02 PM

Without even doing the calculations you can see intuitively that this must be true. If one player has four-of-a-kind then they have taken all four cards with the same number/face out of the deck, which means that the remaining deck now has heavier concentrations of each of the remaining numbers/faces. Conditioning on this event clearly increases the probability that another player will have four-of-a-kind. The increase in probability will not be huge, and it will still be a rare event, but your friend is correct.

Ben
Ben
October 10, 2019 03:50 AM

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