# Ask the Wizard #215

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?

Wade

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:

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.

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

Case | Player 1 | Player 2 | Board | Product |
---|---|---|---|---|

1 | 24 | 3 | 44 | 3,168 |

2 | 24 | 132 | 1 | 3,168 |

3 | 704 | 3 | 1 | 2,112 |

Total | 8,448 |

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.

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.

Kyle from Leesburg, VA

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?

Mark from Ottawa, Ontario

The probability of an event with probability p happening x times, out of a possible n, is combin(n,x) × p^{x} × (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

Wins | Probability |
---|---|

0 | 0.000008 |

1 | 0.000126 |

2 | 0.000954 |

3 | 0.004579 |

4 | 0.015567 |

5 | 0.039851 |

6 | 0.079703 |

7 | 0.127524 |

8 | 0.165782 |

9 | 0.176834 |

10 | 0.155614 |

11 | 0.113174 |

12 | 0.067904 |

13 | 0.033430 |

14 | 0.013372 |

15 | 0.004279 |

16 | 0.001070 |

17 | 0.000201 |

18 | 0.000027 |

19 | 0.000002 |

20 | 0.000000 |

Total | 1.000000 |

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.

Jon S

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.

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.

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.