Probability Of A Royal Flush In Texas Holdem

In the 2008 World Series of Poker Motoyuki Mabuchi's quad aces were beaten by Justin Phillip's Royal flush. I have a simple question about the odds of this occurring. ESPN and others quoted it as 1 in approximately 2.7 billion. It appears to me that they simply took the published odds of quads occurring, and multiplied them by the odds of a royal flush occurring. Is this the correct method of calculation?

Odds of hitting a royal flush in texas holdem People also ask: ID Keyword Suggestions Search Volume Results Google Search CPC; 1: odds of hitting a royal flush in texas holdem: 94438: 964556551: odds of hitting a royal flush in texas holdem: $93.1USD: 2: texas holdem poker tables: 36895: 127179451: texas holdem poker tables: $53.2USD: 3: texas. An ace-high Straight flush is called Royal flush. A Royal Flush is the highest hand in poker. Between two Royal flushes, there can be no tie breaker. If two players have Royal Flushes, they split the pot. The odds of this happening though are very rare and almost impossible in texas holdem because the board requires three cards of one suit for. Odds of a Royal flush with 7 cards being involved (like in texas hold em): 4324 in 133784560. Which is 1 hand in every 30940 hands. Or if you like percentages. Quad Aces in Hold Em Ok, so the odds of me holding the Royal (and both cards playing in each hand) is about 878 million to one. Now of course I can calculate the probability of something happening with that odds in X number of chances.

The odds against making a royal flush are 649,739-to-1. By comparison, the odds of making a straight flush, poker’s second strongest hand, are 0.00139%, with the odds against at 72,192.3-to-1. Calculating the odds of royal flush for Texas Hold’em requires different mathematics, as Texas Hold’em hands are made by choosing the best five.

Probability Of A Royal Flush In Texas Holdem

I disagree with the 1 in 2.7 billion figure too. As you said, they seemed to calculate the probabilities independently for each player, for just the case where both players use both hole cards, and multiplied. Using this method I get a probability of 0.000000000341101, or about in 1 in 2.9 billion. Maybe the one in 2.7 billion also involves compounding a rounding error on both player probabilities. They also evidently forgot to multiply the probability by 2, for reasons I explain later.

There are three ways four aces could lose to a royal flush, as follows.

Case 1: One player has two to a royal flush, the other has two aces, and the board contains the other two aces, the other two cards to the royal, and any other card.

Example:

Player 1:
Player 2:
Board:

Probability Of A Royal Flush In Texas Holdem

In most poker rooms, to qualify for a bad-beat jackpot, both winning and losing player must make use of both hole cards. This was also the type of bad beat in the video; in fact, these were the exact cards.

Case 2: One player has two to a royal flush (T-K), the other has one ace and a 'blank' card, and the board contains the other three aces and the other two cards to the royal.

Example:

Player 1:
Player 2:
Board:

Case 3: One player has one to a royal flush (T-K) and a blank card, the other has two aces, and the board contains the other two aces and the other three cards to the royal flush.

Probability Of A Royal Flush In Texas Holdem

Example:

Player 1:
Player 2:
Board:

The following table shows the number of combinations for each case for both players and the board. The lower right cell shows the total number of combinations is 16,896.

Bad Beat Combinations

CasePlayer 1Player 2BoardProduct
1243443,168
22413213,168
3704312,112
Total8,448

However, we could reverse the cards of the two players, and still have a bad beat. So, we should multiply the number of combinations by 2. Adjusting for that, the total qualifying combinations is 2 × 8,448 = 16,896.

The total number of all combinations in two-player Texas Hold ’Em is combin(52,2) × combin(50,2) × combin(48,5) = 2,781,381,002,400. So, the probability of a four aces losing to a royal flush is 8,448/2,781,381,002,400 = 0.0000000060747, or about 1 in 165 million. The probability of just a case 1 bad beat is 1 in 439 million. The simple reason the odds are not as long as reported in that video is that the two hands overlap, with the shared ace. In other words, the two events are positively correlated.

It is my understanding that the 'racinos' at Monticello and Yonkers, New York, are known as 'Video Lottery Terminals.' I read that they are not true slot/video poker machines, because they do not use a random number generator, but are connected to a central computer in Albany, that controls the outcome of the game. For example, in video poker if you are initially dealt a four of a kind and you discard them all, it will reappear as a winner, since the central computer was programmed for your machine to get a four of a kind. Therefore, any strategy is useless. Is this correct?

You are absolutely right, according to the paper Telling the Truth about New York Video Poker. The player’s outcome is indeed predestined. Regardless of what cards the player keeps, he can not avoid his fate. If the player tries to deliberately avoid his fate, the game will make use of a guardian angel feature to correct the player's mistake. I completely agree with the author that such games should warn the player that they are not playing real video poker, and the pay table is a meaningless measure of the player's actual odds. It also also be noted these kinds of fake video poker machines are not confined to New York.

I use your great site quite often, thanks! I found a new pay table at the Borgata in Atlantic City, for the Three Card Bonus bet in Let It Ride. They implemented these very recently, to the point the dealers were struggling to remember the new odds. Here is the new pay table:

Mini Royal: 50 to 1
Straight flush: 40 to 1
Three of a kind: 30 to 1
Straight: 6 to 1
Flush: 4 to 1
Pair: 1 to 1

I am curious how it impacts the overall house edge.

That is not bad for a side bet. I show the house edge is 2.14%.

Hi Wizard, I came across a new online casino, and decided to give it a try. I was playing at their craps table and noticed that on 20 rolls of the dice, the field bet lost 16 times, and won only 4 times. The sequence went like this: L6,W1,L1,W1,L1,W1,L2,W1,L6. I realize this is a small sample, but is it enough to pass some sort of assessment as to whether this new casino is legit or not?

The probability of an event with probability p happening x times, out of a possible n, is combin(n,x) × px × (1-p)(n-x). In this case, p=4/9, x=4, and n=20. Here is the probability for all possible number of number of field rolls out of 20:

Bad Beat Combinations

WinsProbability
00.000008
10.000126
20.000954
30.004579
40.015567
50.039851
60.079703
70.127524
80.165782
90.176834
100.155614
110.113174
120.067904
130.033430
140.013372
150.004279
160.001070
170.000201
180.000027
190.000002
200.000000
Total1.000000

Taking the sum for 0 to 4, the probability is 2.12%. So, this could have easily happened in a fair game.

Thank you for your entertaining collection of math puzzles. My girlfriend and I came up with this variation on the pirate puzzle. What if all the pirates are of equal rank, and in each round the proposer of the division is chosen by lot? In this variation, assume that each pirate’s highest priority is to maximize his expected amount of coins received. I have what I think is the solution, but perhaps you’d like to try your hand at it first. Thanks again.

You’re welcome. If there are only two pirates left, then the one chosen to make a suggestion has no hope, because the other pirate will vote no. The one drawn will get zero, and the other all 1000. So, before the draw, the expected value with two pirates left is 500 coins.

At the three pirate stage, the drawn pirate should suggest giving one of the other pirates 501, and 499 to himself. The one getting 501 will vote yes, because it is more than the expected value of 500 by voting no. Before the draw, with three pirates left, you have a 1/3 chance each of getting 0, 499, or 501 coins, for an average of 333.33.

Chances Of Hitting A Royal Flush In Texas Holdem

At the four pirate stage the drawn pirate should choose to give 334 to any two of the other pirates, and 332 to himself. That will get him two ’yes’ votes from the pirates getting 334 coins, because they would rather have 334 than 333.33. Including your own vote, you will have 3 out of 4 votes. Before the draw, the expected value for each pirate is the average of 0, 334, 334, and 332, or 1000/4=250.

Probability Of A Royal Flush In Texas Holdemas Hold Em

By the same logic, at the five pirate stage, the drawn pirate should choose to give 251 to any two pirates, and 498 to himself. Unlike the original problem, it isn’t necessary to work backwards. Just divide the number of coins by the number of pirates, not including yourself. Then give half of them (rounding down) that average, plus one more coin.