# What are the odds of getting the following hands in euchre?

I really like euchre, but I stink at statistics. Last time we were playing euchre, people would get some crazy hands, and we kept saying, "What are the odds?"

So now I ask you all, what are the odds?

Scenario 1:

What are the odds that you get dealt both Jacks of the same color, plus three other cards all of the same suit in the same color? (i.e. Jack of Hearts, Jack of Diamonds, plus Ace, Queen and Ten of Diamonds).

Scenario 2:

What are the odds you get dealt four of the cards asked in Scenario 1, and the fifth card you need is the face-up card in the pile of 4 remaining cards?

Scenario 3:

What are the odds you get dealt all 4 Jacks?

-

Surprisingly, the odds for Scenarios 1 and 3 are the same! The chances for each:

• Scenario 1: 0.0047%
• Scenario 2: 0.0535%
• Scenario 3: 0.0047%

Here are the calculations:

Note: The order you consider cards in these calculations doesn't matter, so I'll use the most convenient ordering possible.

Scenario 1:

First, there are 24 cards in a Euchre deck. To calculate the number of possible Euchre hands of five, you multiply the number of possibilities for each card: `24 × 23 × 22 × 21 × 20 = 5,100,480`.

Then we need to figure out how many unique ways you can get Scenario 1. We'll calculate the draw this way: (I) number of Jacks (4), (II) number of Jacks matching the color of the first Jack (1), (III) number of remaining cards (A, K, Q, 10, 9) matching the color of the Jack (10; 5 in each suit of that color), (IV) number of remaining cards matching the suit of card III (4), (V) number of remaining cards matching the suit of cards II and III (3). This calculates as `4 × 1 × 10 × 4 × 3 = 480`.

So the chance of getting Scenario 1 is `480 / 5,100,480 = 1 / 10,626`.

Scenario 2:

This one is more complicated.

Number of ways to get a mismatch by missing a Jack: (I) number of Jacks (4), (II) number of remaining cards (A, K, Q, 10, 9) matching the color of the Jack (10; 5 in each suit of that color), (III) number of remaining cards matching the suit of card II (4), (IV) number of remaining cards matching the suit of cards II and III (3), (V) number of cards that don't complete the set (19). This calculates as `4 × 10 × 4 × 3 × 19 = 9,120`.

Number of ways to get a mismatch by missing another card: (I) number of Jacks (4), (II) number of Jacks matching the color of the first Jack (1), (III) number of remaining cards (A, K, Q, 10, 9) matching the color of the Jack (10; 5 in each suit of that color), (IV) number of remaining cards matching the suit of card III (4), (V) number of cards that don't complete the set (17). This calculates as `4 × 10 × 4 × 3 × 17 = 8,160`.

So, the number of ways to end up with four of the five cards in Scenario 1 is `(9,120 + 8,160) / 5,100,480 = 6 / 1,771`.

Then the number of cards that would match (3) out of the number of possibilities for the face-up card (19) is `3 / 19`

So the chance of both occurring (and thus Scenario 2) is `(6 / 1,771) × (3 / 19) = 18 / 33,649`.

Scenario 3:

Number of ways of getting all four Jacks: (I) number of Jacks (4), (II) number of remaining Jacks (3), (III) number of remaining Jacks (2), (IV) number of remaining Jacks (1), (V) number of remaining cards (20). This calculates as `4 × 3 × 2 × 1 × 20 = 480`.

So the chances of getting Scenario 3 is `480 / 5,100,480 = 1 / 10,626`.

-
Hi Firefeather. I've been reading through some of the older posts recently, and I don't think the calculations above are quite right. I think you're right that you can ignore ordering, but in that case the total number of hands isn't 24*23*22*21*20, it's that divided by 5*4*3*2*1 (or 24C5). Also I didn't follow the logic in scenario 2 when it came to looking at the four remaining cards. – tttppp Oct 25 '12 at 8:20
For scenario 1 you need to multiply by 2 to account for suit choices. If you're dealt both red jacks, it doesn't matter if the rest of your cards are Hearts or Diamonds, but in your calculation you require them to be the suit of the first jack. – Gregor Nov 3 '12 at 19:13
And @tttppp is correct, 24*23*22*21*20 is the number of ordered Euchre hands, so your denominators all must be changed. – Gregor Nov 3 '12 at 19:34

## Solutions (TLDR)

The probabilities of a single player getting one of the above hands, from a single deal, are:

• Scenario 1: 0.09%-0.10%
• Scenario 2: 1.40%-1.54%
• Scenario 3: 0.04%-0.05%

The range of values in Scenario 2 is due to a discrepancy between my calculation and my simulation - the calculation gives the lower of these, and simulation gives the higher. Simulation of the other two scenarios backs up the calculations.

I have also simulated the scenarios to find the probabilities of any player getting one of the above hands, from a single deal. I think this more closely matches the observations described in the question. These probabilities are:

• Scenario 1 for any player: 0.38%
• Scenario 2 for any player: 5.98%
• Scenario 3 for any player: 0.19%

## Introduction

A standard euchre deck contains the 9, 10, J, Q, K and A in each of the four suits, making a total of 24 cards. Each of the four players are dealt a hand of five cards, and there are four cards left over.

We will split the deck, `D`, into the four Jacks `J`, and the other twenty non-Jack cards `N`. We refer to the player's hand as `H`, and the table cards as `T`.

We use the notation `nCk` to refer to the Combinations function; the number of ways of choosing `k` elements from a set of size `n`.

We refer to the probability of an event using the notation `P(Event)`. We will refer to the conditional probability of event 2 happening, given that event 1 happens, with the notation `P(Event 2 | Event 1)`. In several places we do not explicitly state that an event is conditional on another event, but hopefully the context makes it obvious when this is the case.

## Scenario 1: What are the odds that you get dealt both Jacks of the same color, plus three other cards all of the same suit in the same color? (i.e. Jack of Hearts, Jack of Diamonds, plus Ace, Queen and Ten of Diamonds).

``````P(Scenario1) = P(H contains exactly 2 Jacks) * P(The Jacks are the same colour)
* P(3 other cards are same suit) * P(The suit is the same colour as the Jacks)
P(H contains exactly 2 Jacks) = (Ways to pick 2 from J) * (Ways to pick 3 from N)
/ (Ways to pick 5 from D)
= (4C2)(20C3) / (24C5)
= 6*1140 / 42504
P(The Jacks are the same colour) = 1/3
P(3 other cards are same suit) = P(The second card is the same suit as the first)
* P(The third is the same suit as the other two)
= 4/19 * 3/18
P(The suit is the same colour as the Jacks) = 1/2
∴ P(Scenario1) = 6*1140 / 42504 * 1/3 * 4/19 * 3/18 * 1/2
= 5/5313
~= 0.09%
``````

## Scenario 3: What are the odds you get dealt all 4 Jacks?

This calculation is slightly more straightforward than for Scenario 1, as we no longer need to worry about suits.

``````P(Scenario3) = P(H contains exactly 4 Jacks)
= (4C4)(20C1)/(24C5)
= 1*20 / 42504
= 5/10626
~= 0.04%
``````

## Scenario 2: What are the odds you get dealt four of the cards asked in Scenario 1, and the fifth card you need is the face-up card in the pile of 4 remaining cards?

This is harder to compute, and so I wouldn't be surprised if there's a mistake in my calculations.

First we split the calculation down by distribution of `H`. I've written `d(H)=(3,2)` to mean that `H` contains 3 Jacks and 2 non-Jacks. Note that it is only possible to get Scenario 2 if `H` contains 1,2 or 3 Jacks. We refer to "Scenario 2" as `S2` for brevity.

``````P(S2) = P(d(H)=(3,2))P(S2 | d(H)=(3,2)) + P(d(H)=(2,3))P(S2 | d(H)=(2,3))
+ P(d(H)=(1,4))P(S2 | d(H)=(1,4))
P(d(H)=(j,n)) = (4Cj)(20Cn) / (24C5)
``````

### Scenario 2, Part 1: Three Jacks in the hand

Consider the case where `H` contains 3 Jacks and 2 non-Jacks.

``````P(S2 | d(H)=(3,2)) = P(2 of the Jacks are same colour) * P(2 non-Jacks are same suit)
* P(Suit of non-Jacks is same colour as 2 Jacks)
* P(T contains at least one non-Jack of same suit)
P(2 of the Jacks are same colour) = 1
P(2 non-Jacks are same suit) = 4/19
P(Suit of non-Jacks is same colour as 2 Jacks) = 1/2
P(T contains at least one non-Jack of same suit) = 1 - P(T contains 0 non-Jacks of same suit)
= 1 - 16/19 * 15/18 * 14/17 * 13/16
= 514/969
∴ P(S2 | d(H)=(3,2)) = 1 * 4/19 * 1/2 * 514/969 = 1028/18411
``````

### Scenario 2, Part 2: Two Jacks in the hand

Next consider the case where `H` contains 2 Jacks and 3 non-Jacks. We treat the cases where the Jacks are the same colour separately from the cases where they are different.

``````P(S2 | d(H)=(2,3)) = P(Jacks are same colour)P(S2 | d(H)=(2,3) and Jacks are same colour)
+ P(Jacks not same colour)P(S2 | d(H)=(2,3) and Jacks not same colour)
= 1/3 P(S2 | d(H)=(2,3) and Jacks are same colour)
+ 2/3 P(S2 | d(H)=(2,3) and Jacks not same colour)
``````

When the Jacks are the same colour, we need exactly two of the three non-Jacks to share a suit, and that suit to be the same colour as the Jacks.

``````P(S2 | d(H)=(2,3) and Jacks are same colour) = P(Exactly 2 non-Jacks share a suit)
* P(Suit is same colour as Jacks) * P(T contains at least one non-Jack of same suit)
= (3*(4*15)/(18*19)) * (1/2) * (514/969)
= 2570/18411
``````

When the Jacks are different suits, we need all three non-Jacks to share a suit.

``````P(S2 | d(H)=(2,3) and Jacks not same colour)
= P(All 3 non-Jacks share a suit) * P(T contains other Jack of same colour)
= (4*3)/(19*18) * 4/19
= 8/1083
``````

Putting these two cases together we get:

``````P(S2 | d(H)=(2,3)) = 1/3 * 2570/18411 + 2/3 * 8/1083 = 2842/55233
``````

### Scenario 2, Part 3: One Jack in the hand

The final distribution of `H` to consider is 1 Jack and 4 non-Jacks.

``````P(S2 | d(H)=(1,4)) = P(At least 3 non-Jacks share a suit) * P(The Jack is the same colour)
* P(T contains other Jack of same colour)
P(At least 3 non-Jacks share a suit) = (4*4*3*15 + 4*3*2) / (19*18*17) = 124/969
P(The Jack is the same colour) = 1/2
P(T contains other Jack of same colour) = 4/19
∴ P(S2 | d(H)=(1,4)) = 124/969 * 1/2 * 4/19 = 248/18411
``````

### Scenario 2: Conclusion

So putting all of that together.

``````P(S2) = (4C3)(20C2)/(24C5) * 1028/18411 + (4C2)(20C3)/(24C5) * 2842/55233
+ (4C1)(20C4)/(24C5) * 248/18411
= 4*190/42504 * 1028/18411 + 6*1140/42504 * 2842/55233 + 4*4845/55233 * 248/18411
= 26029130/1858535217
~= 1.40%
``````

## Simulation

To check my calculations I have written a Python script to evaluate ten million deals, to see how frequently each of the scenarios occurs. The script is here.

I also used it to compute the probability of any of the four players satisfying each of the three scenarios for a given deal. I thought this would be interesting, as this is what was observed in the question, although not actually the probability being asked for.

``````Probability for a single player   | Probability across all four players
Scenario1 | Scenario2 | Scenario3 | Scenario1 | Scenario2 | Scenario3
0.0972%  |  1.5319%  |  0.05  %  |  0.3703%  |  5.9367%  |  0.1748%
0.0927%  |  1.519 %  |  0.0442%  |  0.3716%  |  5.9698%  |  0.1914%
0.0911%  |  1.5607%  |  0.0466%  |  0.3896%  |  5.9523%  |  0.187 %
0.096 %  |  1.5207%  |  0.0442%  |  0.3753%  |  5.9583%  |  0.1853%
0.0946%  |  1.5594%  |  0.0457%  |  0.3711%  |  5.9709%  |  0.1844%
0.1004%  |  1.561 %  |  0.0466%  |  0.3824%  |  6.0065%  |  0.1916%
0.0964%  |  1.5367%  |  0.0476%  |  0.3776%  |  5.9651%  |  0.1882%
0.098 %  |  1.563 %  |  0.0452%  |  0.3775%  |  6.0131%  |  0.188 %
0.0942%  |  1.5385%  |  0.0469%  |  0.3849%  |  6.0205%  |  0.1852%
0.0983%  |  1.535 %  |  0.0418%  |  0.3641%  |  5.9913%  |  0.1931%
-----------------------------------------------------------------------
0.0959%  |  1.5426%  |  0.0459%  |  0.3764%  |  5.9785%  |  0.1869%
``````

The simulation backs up the calculation for Scenarios 1 and 3, but differs by about 0.14% for Scenario 2. This indicates that there is an error in either the calculation or the programme, but at least provides a ball park figure of roughly one occurrence every seven hundred hands.

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Very nice and clear. Since you define all your other notation so well, you might want to mention that P(X) is the probability of X. – Gregor Nov 3 '12 at 19:27
@shujaa Thanks! I've made an edit to include that now. – tttppp Nov 5 '12 at 17:01
@tttppp This is excellent work. Is the discrepancy between your Scenario2 simulation and calculation due to the fact that you checked for success in the simulation without subtracting the success of S1 and S3? ie. S1&S3 fulfills requirements of S2. It looks like the difference between sim and calculation is about the sum of S1+S3. – ghoppe Nov 7 '12 at 23:57
@ghoppe Thanks! I don't think that's the issue, but it's an interesting anomaly. The python script definitely excludes scenario 1 from scenario 2. My guess is that my arithmetic is wrong somewhere in the calculation, but I haven't spotted where yet! – tttppp Nov 8 '12 at 8:26