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In 4-player trick-taking game where taking a trick constitute a penalty (e.g. Hearts), I've often seen the card 5 (the fourth-lowest rank in a suit, when 2 is the lowest and Ace is the highest) takes the trick because other players play 2, 3, 4, respectively.

Initially I thought playing the card 5 would be safe because the probability that 2, 3, 4, 5 each falls into exactly different hands must be very low, but it seems not. Given a standard 52-card deck shuffled randomly among 4 players, what is the probability that 2, 3, 4, 5 are split exactly among the four players, at least for one of the suits? What is the probability that this happens for two, three or four of the suits?

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  • $\begingroup$ Try math.stackexchange.com $\endgroup$ – Kun Apr 28 '16 at 20:04
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I find it easier to look at players being assigned to cards, rather than the other way around.

Probability of a specific suit

Lets start with a single suit, say ♣, of which we are interested in the lowest cards only (2♣, 3♣, 4♣, and 5♣).

A: The 2♣ gets assigned to someone (any player) with probability $\frac{52}{52}=1$

B: Given A occurred, the 3♣ gets assigned to a different player than the 2♣ with probability $\frac{39}{51} \approx 0.745$

C: Given B occurred, the 4♣ gets assigned to a different player than the 2♣ and 3♣ with probability $\frac{26}{50} = 0.48$

D: Given C occurred, the 5♣ gets assigned to a different player than the 2♣, 3♣, and 4♣ with probability $\frac{13}{49} \approx 0.204$

All of these events must occur for the ♣'s to be dispersed among four different players, and they are conditional on one another (and thus independent), so the probabilities can be multiplied.

The resulting answer for the lowest ♣'s ending up in four different players hands is $\frac{52}{52} * \frac{39}{51} * \frac{26}{50} * \frac{13}{49} \approx .1055$ or $10.55\%$

Probability of two specific suits

Given A, B, C, and D have occurred, this iterative process could be continued for the next suit, say ♠'s. Now we have 48 cards left to be assigned to players.

A: The 2♠ gets assigned to someone (any player) with probability $\frac{48}{48}=1$

B: Given A occurred, the 3♠ gets assigned to a different player than the 2♠ with probability $\frac{36}{47} \approx .766$

C: Given B occurred, the 4♠ gets assigned to a different player than the 2♠ and 3♠ with probability $\frac{24}{46} \approx 0.522$

D: Given C occurred, the 5♠ gets assigned to a different player than the 2♠, 3♠, and 4♠ with probability $\frac{12}{45} \approx 0.266$

The resulting answer for the lowest ♠'s ending up in four different players hands, given the lowest ♣'s are separated, is $\frac{48}{48} * \frac{36}{47} * \frac{24}{46} * \frac{12}{45} \approx .1067$ or $10.67\%$.

Probability two specific suits are separated is $0.1055*0.1067 = 0.0112$ or $1.12\%$

Probability of three specific suits

Given A, B, C, and D above have occurred, this iterative process could be continued for the next suit, say ♥'s. Now we have 44 cards left to be assigned to players.

A: The 2♥ gets assigned to someone (any player) with probability $\frac{44}{44}=1$

B: Given A occurred, the 3♥ gets assigned to a different player than the 2♥ with probability $\frac{33}{43} \approx .767$

C: Given B occurred, the 4♥ gets assigned to a different player than the 2♥ and 3♥ with probability $\frac{22}{42} \approx 0.524$

D: Given C occurred, the 5♥ gets assigned to a different player than the 2♥, 3♥, and 4♥ with probability $\frac{11}{41} \approx 0.268$

The resulting answer for the lowest ♥'s ending up in four different players hands, given the lowest ♣'s and ♠'s are separated, is $\frac{44}{44} * \frac{33}{43} * \frac{22}{42} * \frac{11}{41} \approx 0.1079$ or $10.79\%$.

Probability three specific suits are separated is $0.1055*0.1067*0.1079 = .0012$ or $0.12\%$

Probability of all four suits

Given A, B, C, and D above have occurred, this iterative process could be continued for the next suit, say ♦'s. Now we have 40 cards left to be assigned to players.

A: The 2♦ gets assigned to someone (any player) with probability $\frac{40}{40}=1$

B: Given A occurred, the 3♦ gets assigned to a different player than the 2♦ with probability $\frac{30}{39} \approx .769$

C: Given B occurred, the 4♦ gets assigned to a different player than the 2♦ and 3♦ with probability $\frac{20}{38} \approx 0.526$

D: Given C occurred, the 5♦ gets assigned to a different player than the 2♦, 3♦, and 4♦ with probability $\frac{10}{37} \approx 0.270$

The resulting answer for the lowest ♦'s ending up in four different players hands, given the lowest ♣'s, ♠'s, and ♥'s are separated, is $\frac{40}{40} * \frac{30}{39} * \frac{20}{38} * \frac{10}{37} \approx 0.1094$ or $10.94\%$.

Probability all four specific suits are separated is $0.1055*0.1067*0.1079 = .00013$ or $0.013\%$


Putting it all together

Now we can employ the inclusion-exclusion principle on the specific suit combinations to compute the probability of this happening for any 1, any 2, any 3, or any/all 4. The probability of all four was already computed in the last step above.

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There are 4 (4 suits) * 4! (ways to split among players) * ${48\choose12} {36\choose12} {24\choose12}$ (ways to arrange the other 48 cards) $\approx2.2638e28$ ways for this to occur

There are ${52\choose13}{39\choose13}{26\choose13}\approx5.3645e28$ ways to arrange 52 cards into 4 hands of 13 cards

Dividing these numbers gives about a 42.2% chance of this occurring - pretty substantial!!!

The odds for this occurring for any particular suit are only 1/4 of that about 10.6%

As a quick check, consider an incorrect, but close approximation. If you take one of those cards - say the 2, somebody has to have it. The odds that the next card goes to someone else are a little better than 3/4 (since the one with the 2 only gets 12 chances to everyone elses 13), the third a bit better than 1/2, and the fourth a bit better than 1/4. So a good lower limit is $1\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{1}{4}\approx0.0938$ which is indeed just a little under 10.6%.

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This isn't a definitive answer, but I'm curious too and this might be helpful for others to check their work. Using random sampling:

library(parallel)
nums <- 1:13
suits <- LETTERS[1:4]
cards <- expand.grid(nums, suits)
x <- rep(1:4, each=13)

nperm <- 10000000
out <- mclapply(1:nperm, function(dummy) {
    deal <- sample(x)
    c(length(unique(deal[1:4]))==4, length(unique(deal[14:17]))==4, 
      length(unique(deal[27:30]))==4, length(unique(deal[40:43]))==4)
}, mc.cores=16)
out2 <- do.call(rbind,out)

The result out2 looks like this:

> head(out2)
      [,1]  [,2]  [,3]  [,4]
[1,] FALSE FALSE FALSE FALSE
[2,] FALSE FALSE FALSE FALSE
[3,] FALSE FALSE FALSE FALSE
[4,]  TRUE FALSE FALSE FALSE
[5,] FALSE FALSE FALSE FALSE
[6,] FALSE FALSE FALSE FALSE

where each row is one trial, and each column is whether or not the four lowest cards of that suit are distributed one per player.

$P(one\ suit)$ is then:

> table(out2[,1]==1)

  FALSE    TRUE 
8945197 1054803 
> 1054803/nperm
[1] 0.1054803

$P(two\ suits)$ is:

> table(rowSums(out2[,1:2])==2)

  FALSE    TRUE 
9888034  111966 
> 111966/nperm
[1] 0.0111966

$P(three\ suits)$ is:

> table(rowSums(out2[,1:3])==3)

  FALSE    TRUE 
9988022   11978 
> 11978/nperm
[1] 0.0011978

and finally all four suits:

> table(rowSums(out2)==4)

  FALSE    TRUE 
9998680    1320 
> 1320/nperm
[1] 0.000132
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  • $\begingroup$ This is strange way of dealing cards! Most of the time more than 13 of at least one suit appears in your sample... . The results have no evident connection to the situation described in the question. Doesn't it seem strange that they differ substantially from the two theoretical answers already offered? $\endgroup$ – whuber Apr 28 '16 at 20:13
  • $\begingroup$ Wow, I somehow got it in my head that the sampling would "deal" the 52 cards evenly across 4 players. Editing now. Not sure I get your second comment - you go through each trial and check to see if all four of the lowest cards are distributed evenly... $\endgroup$ – fanli Apr 28 '16 at 20:42
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There are ${52}\choose{13,13,13,13}$ ways to distribution 52 cards among 4 people.

There are 4! ways to distribute 4 lowest cards of a particular suit to 4 people, and there are 4 suits. On top of this, there are ${48}\choose{12,12,12,12}$ ways to organize the rest of the cards. In total, there are $4*4!*$${48}\choose{12,12,12,12}$ ways to distribute the cards such that the 4 lowest cards of a suit are distributed among 4 players.

Therefore the probability in question is

$\frac{$4*4!*{48}\choose{12,12,12,12}}{{52}\choose{13,13,13,13}}$

I get the same answer as MikeP but I'm not sure why this is inconsistent with the first answer and the numeric simulation...

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