Hold Em: Flopping a Royal Flush vs 3-Aces vs 3-Jacks

Hello Wizard and other distinguished mathematicians,

What are the chances of this scenario:
6 people at the table

Dealer gets dealt Q10 hearts
Small Blind gets dealt AA
Under the Gun gets dealt JJ

Flop comes AKJ all hearts

First to act with AA shoves all in, Big Blind Folds, UTG calls with JJ, others fold, Dealer calls with royal flush.

This was a home game and dealer is relatively new to the game. I've played around 40k hands online and have never seen this before.

I think the odds of flopping a royal flush in Hold Em are somewhere in the neighborhood of 1 in 650,000 hands. What are the odds of flopping a royal flush against both trip Aces and trip Jacks in a 6-handed game?

I'm gong to take a few liberties with the problem description. First, I'll assume that the suit of the Royal doesn't matter. Second, to simplify the math, I'm going to cut out the other three players from the picture - their hands do actually matter, since they can't be dealt any of the key cards needed to complete the royal.

The lucky Royal recipient has to have a suited QT, KT, or KQ, since the flop has to come up AJ* (with * being the other card to complete the royal). Whatever suit he gets these cards in determines what every other player must hold for the hand to work. Since order doesn't matter, there are 4 ways of getting a suited QT, 4 ways of getting a suited KT, and 4 ways of getting a suited KQ, for a total of 12 combinations, times 4 suits = 48 ways of this hand being dealt. We'll assume KQ of hearts are chosen for the sake of example, but remember that it doesn't actually matter which of these combinations we get.

The rockets have to come up in the wrong suit - if the ace of hearts were in this hand, it couldn't come up on the flop to complete the Royal. Thus, there are three aces to choose from, which means there are 3 ways of picking two of them. The same is true for the jacks, so there are 48*3*3 = 432 ways for the initial hands to be dealt, and 3! ways of ordering the players, for a total of 2592 possibilities.

The flop has to come up AJT of hearts, but remember that order doesn't matter, so there's really only one way this can happen once we've chosen the two cards in the Royal hand. Thus, there are still only 2592 ways the situation you describe could occur.

We can calculate the total number of ways three hands and a flop could be dealt with Combin(52,2) * Combin(50,2) * Combin(48,2) * 3! * Combin(46,3), which is 166,882,860,144,000.

So, the probability of this hand occurring is 1 in 64,383,819,500.

Sometimes, if I want to bust chops of a floorman, I'll ask how the bad beat is split if it's a there are three or four qualifying hands.

FYI: A bad beat is what you're describing. Often the minimum hand is Aces full, sometimes it's quads. To qualify, all players involved must have a pocked pair, or use both cards for their best straight flush.

When it hits, the loser of the hand gets 50% of the jackpot, the winner gets 25%, and the other players split 25%.

I ask about the following scenario:

Player 1: Ks Qs
Player 2: 8s 7s
Player 3: Tc Th
Player 4: Jc Jd

Board: Js Ts 9s Td Jh

Player 1 has the high straight flush. Player 2 has the low. Players 3 and 4 have quads.

How is the money split?

The answer is usually something like: "Well, since that will happen about once in a million years, I'll just hope that it happens on my day off!"

Quote: DJTeddyBear

Sometimes, if I want to bust chops of a floorman, I'll ask how the bad beat is split if it's a there are three or four qualifying hands.

FYI: A bad beat is what you're describing. Often the minimum hand is Aces full, sometimes it's quads. To qualify, all players involved must have a pocked pair, or use both cards for their best straight flush.

When it hits, the loser of the hand gets 50% of the jackpot, the winner gets 25%, and the other players split 25%.

I ask about the following scenario:

Player 1: Ks Qs
Player 2: 8s 7s
Player 3: Tc Th
Player 4: Jc Jd

Board: Js Ts 9s Td Jh

Player 1 has the high straight flush. Player 2 has the low. Players 3 and 4 have quads.

How is the money split?

The answer is usually something like: "Well, since that will happen about once in a million years, I'll just hope that it happens on my day off!"

Though they are probably being cute, I would bet everything I've got that they (at least the card room) know(s) exactly what to do in those situations. If you do want to know, I'm sure it would be in the rules. Not just because the casino cares, but the Gaming Boards are very specific in their rules about nearly all possible possibilities. There is little to nothing up for debate. Maybe the losers with qualifying hands split the 50, winner still gets 25, and the rest of the table still splits 25. That would stink because the winner gets more than the losers, but I'm sure you could find the answers pretty easily.

The odds of getting the three remaining cards in a royal given that you have 2 cards to the royal and you don't know any of the other cards dealt is (50 x 49 x 48) / 6 = 19,600:1.

The odds that you get dealt two cards to a royal is 20/42 * 4/49 * = 31.85:1
The odds that any player is dealth AA = 4/52 * 3/51 = 221:1
The odds that any player is dealt JJ is also 221:1.

(all odds based on independent events)...

The odds that three of six people at the table have that combination: 12,963.21:1

The odds then of the entire combination = 12,963.21 x 19,600 = 254,079,022:1

While that may seem astronomical, what you are really looking for are the odds of two people getting a high pair (10-A) and another getting two to a royal and then have the remainder of the royal come up in the flush.

Quote: curtmack

I'm going to cut out the other three players from the picture - their hands do actually matter, since they can't be dealt any of the key cards needed to complete the royal.

I do not agree.

I think the other players cards matter about as much as the unseen burn cards, and cards not dealt.

The only time they matter is while the hand is in play.

For example, often when players are all-in before the river, someone who is on a draw will ask if someone folded the card they need.

As long as nobody responds, the odds are unaffected by their hand.

Note, if they respond, then the odds DO change - regardless if the response is yes or no. (Assuming it was a truthful reply.)

I disagree with all of the above math, although there are different correct probabilities based on which probability one is seeking. My calculations assumed the following:

1) The suit of the royal doesn't matter
2) The player with the royal could have been dealt any 2 royal cards as long as neither was an Ace or Jack
3) The flop was the remaining three royal cards
4) Neither of the players with a pocket pair ended up with quads at the end
5) It doesn't matter which player gets which hand

Bearing those assumptions in mind, here's what I came up with:

1) The royal can be in any suit, so the numerator starts at 4.

2) The player that gets the royal must hold QT, QK, or TK. That's 3 possibilities. The numerator is now 12.

3) The player with the Aces must have 2 of the 3 non-royal Aces, so combin(3,2) = 3. The numerator is now 36.

4) The player with the Jacks must have 2 of the 3 non-royal Jacks, so combin(3,2) = 3. The numerator is now 108.

5) The flop must be the remaining 3 royal cards. There is only one way to do this (their order doesn't matter since they all appear at the same time). The numerator is still 108.

6) The turn card can be anything except the last Jack or Ace, so there are 41 possibilities. The numerator is now 4428.

7) The river card must also not be the last Jack or Ace, so there are 40 possibilities. The numerator is now 177120.

8) Lastly, any of the three players could have ended up with the royal, and either of the remaining 2 could have ended up with the Aces. That's 3 * 2 * 1 = 3! = 6 possibilities, so the numerator is 1062720.

The denominator is calculated as follows:

1) Number of combinations for player A = combin(52,2) = 1326
2) Number of combinations for player B = combin(50,2) = 1225
3) Number of combinations for player C = combin(48,2) = 1128
4) Number of combinations for the flop = combin(46,3) = 15180
5) Number of combinations for the turn = combin(43,1) = 43
6) Number of combinations for the river = combin(42,1) = 42
7) The denominator is 1326 * 1225 * 1128 * 15180 * 43 * 42 = 50,231,740,903,344,000

The total number of ways the situation could happen (using the assumptions I made) is 1,062,720 out of 50,231,740,903,344,000, which reduces to approximately 1 in 47,267,145,535.

I don't think he mentioned quads, though. I just calculated the odds of getting that particular flop - the Royal vs. three aces vs. three jacks.

Quote: curtmack

I don't think he mentioned quads, though. I just calculated the odds of getting that particular flop - the Royal vs. three aces vs. three jacks.

You're absolutely right. I was under the impression that the situation was AAAA vs. JJJJ vs. Royal Flush. I guess I should have read it more carefully.

I edited my earlier post to reflect my revised math.

deleted