# What are odds of 2 players having same 4 hole cards in omaha?

What are odds of 2 players having same 4 cards in Omaha?

• 0% all 52 cars in a deck are different. Rephrase your question please.
– Jon
Commented Nov 11, 2017 at 12:56

Very interesting question. I assume that you mean that all 4 cards have the same rank and suits are ignored. This is not going to be a simple or short answer. Please bear with me.

To answer this question we need the probability of getting dealt quad, trips, double pairs, pairs and no pairs. We will need these numbers later on. Why? Because when player 1 gets dealt quads or trips, player 2 can't possibly get 4 cards with the same values, because they are not in the deck. Also it is more difficult to make the same 4 cards when player 1 has pair(s) in his hand, which we will see later.

(3/51) * (2/50) * (1/49) = 0.000048019

This is correct since the first card doesn't matter. The second card needs to have the same value as the first one and there are 3 cards left for which this will be the case. And so on for cards 3 and 4.

Probability of trips:

4 * (3/51) * (2/50) * (48/49) = 0.009219688

We multiply by 4, since there are 4 unique ways to get trips:

XXXO
XXOX
XOXX
OXXX

A 'X' represents a card that is part of the three of a kind. A 'O' represents a card that is not. If we calculate the top one we will find that the first card again doesn't matter. The second card needs to pair, which are 3 cards. Third card same story, but there are 2 left and then the fourth card needs to be any card except the one card that would make us quads. The probability of the 3 other ways to make trips are equal to the probability of the first way.

Probability of double pair:

3 * (3/51) * (48/50) * (3/49) = 0.010372149

We multiply by 3, since there are 3 unique ways to get a double pair:

XXOO
XOXO
XOOX

'X' represents the first pair. 'O' represents the second pair. Probability of each of these unique ways are equal. If we look at the top one, then the first card doesn't matter. The second card needs to be paired, which are 3 cards. The third card needs to not make trips, which are all remaining cards except 2. Finally the fourth card needs to pair with card 3, which are 3 cards.

Probability of single pair:

6 * (3/51) * (48/50) * (44/49) = 0.3042497

We multiply by 6, since there are 6 unique ways to make a single pair:

XXOO
XOXO
XOOX
OXXO
OXOX
OOXX

'X' represents the pair. 'O' represents any other card that doesn't pair its partner or 'X'. Again the odds of each of them happening are the same. If we take a look at the top one, we will find that the first one doesn't matter. The second card needs to pair with the first card, which are 3 cards. Then the third card needs to not make trips, which are all cards except for 2. Finally the fourth card needs to not make trips and not make a pair, which are all cards except 2 plus 3 equals 5.

Probability of no pair:

(48/51) * (44/50) * (40/49) = 0.676110444

This is correct, since the first card doesn't matter. The second card needs to not make a pair, so any card except for 3. Then the third card needs to not make any pairs, so any card except 6. And so on.

Since there exists no other starting hands these numbers should equal one added together and frankly, they do. Now that we got these numbers, let's move on.

0

This is not possible, because there are no 8 cards with the same rank in a deck of cards.

Probability of same trips:

0

This is not possible, because there are no 6 cards with the same rank in a deck of cards.

Probablity of same double pair:

(4/48) * (3/47) * (2/46) * (1/45) = 0.000005139

This is correct since we need to get four particular cards, but order of course, does not matter.

Probability of same single pair:

6 * (2/48) * (1/47) * (6/46) * (3/45) = 0.000046254

We can again write down all the possible ways this can happen:

XXOO
XOXO
XOOX
OXXO
OXOX
OOXX

'X' represents the card that is equal to the rank of the pair and 'O' represents the two other cards. The odds of all the possible ways are again equal. Looking at the top one: The first card needs to be the same rank of the pair, which are 2 cards. The second card needs to be the same rank again, which is now only 1 card. Then we need the rank of one of the two remaining cards, which are 6 cards. And finally we need to last rank, which are 3 cards.

Probability of same no pairs:

(12/48) * (9/47) * (6/46) * (3/45) = 0.000416281

The first card needs to be any rank that is equal to the 4 cards of player 1, which are 4 times 3 equals 12 cards, and so on.

Now that we got these numbers we can finally get to our answer. To get to the answer we need to multiply the probability of each of the categories of hands that player 1 has by the chance of player 2 getting the exact same ranks and add them all together for all categories. Since the probability of getting the same quads or trips are 0 we will get:

(0.010372149 * 0.000005139)
+ (0.3042497 * 0.000046254)
+ (0.676110444 * 0.000416281)

= 0.000295578

So the answer is 0.000295578 or 0.0295578% or 1 : 3383.2