# Ask the Wizard #287

What is the probability of forming a Yahtzee with up to n rolls of the dice?

Anonymous

For the benefit of other readers, a Yahtzee is a five of a kind with five dice. In the game of Yahtzee the player may hold any dice he wishes and re-roll the rest. He can do this up to three rolls.

The player may re-roll previously held dice, if he wishes. For example, if the player's first roll is 3-3-4-5-6 and he holds the threes and then has 3-3-5-5-5 after the second roll he may keep the fives and re-roll the threes on his third roll.

The following table shows the maximum number of dice of the same face for 1 to 20 rolls. The table shows the probability of getting a Yahtzee within three rolls is about 4.6%.

### Yahtzee Probabilities

Rolls | Maximum Dice of Same Face | ||||
---|---|---|---|---|---|

One | Two | Three | Four | Five | |

1 | 0.092593 | 0.694444 | 0.192901 | 0.019290 | 0.000772 |

2 | 0.008573 | 0.450103 | 0.409022 | 0.119670 | 0.012631 |

3 | 0.000794 | 0.256011 | 0.452402 | 0.244765 | 0.046029 |

4 | 0.000074 | 0.142780 | 0.409140 | 0.347432 | 0.100575 |

5 | 0.000007 | 0.079373 | 0.337020 | 0.413093 | 0.170507 |

6 | 0.000001 | 0.044101 | 0.263441 | 0.443373 | 0.249085 |

7 | 0.000000 | 0.024501 | 0.199279 | 0.445718 | 0.330502 |

8 | 0.000000 | 0.013612 | 0.147462 | 0.428488 | 0.410438 |

9 | 0.000000 | 0.007562 | 0.107446 | 0.398981 | 0.486011 |

10 | 0.000000 | 0.004201 | 0.077416 | 0.362855 | 0.555528 |

11 | 0.000000 | 0.002334 | 0.055317 | 0.324175 | 0.618174 |

12 | 0.000000 | 0.001297 | 0.039279 | 0.285674 | 0.673750 |

13 | 0.000000 | 0.000720 | 0.027757 | 0.249063 | 0.722460 |

14 | 0.000000 | 0.000400 | 0.019543 | 0.215313 | 0.764744 |

15 | 0.000000 | 0.000222 | 0.013720 | 0.184883 | 0.801175 |

16 | 0.000000 | 0.000124 | 0.009610 | 0.157896 | 0.832371 |

17 | 0.000000 | 0.000069 | 0.006719 | 0.134258 | 0.858954 |

18 | 0.000000 | 0.000038 | 0.004692 | 0.113753 | 0.881517 |

19 | 0.000000 | 0.000021 | 0.003272 | 0.096100 | 0.900607 |

20 | 0.000000 | 0.000012 | 0.002280 | 0.080994 | 0.916714 |

This question is raised and discussed in my forum at Wizard of Vegas.

How many hands need to be played of a casino game for the **casino** to be confident of showing a profit?

cwwbjr

It depends on the game, naturally. The greater the house advantage and lesser the variance, the greater the chance is. It also depends on the level of confidence.

The following table shows how many bets are required, assuming the same bet amount, for the casino to be confident, at various levels of confidence, for some common games. For example, to have a 95% chance of showing a net profit on the Banker bet in baccarat, the casino would need to deal 20,791 hands.

This is based on the Normal Distribution in all cases except for Jacks or Better. That approximation becomes untrustworthy if the number of expected events of any one outcome is five or less. So, for video poker, I used the Poisson distribution for the royals and the Normal approximation otherwise.

For blackjack, the rules are: 6 decks, dealer stands on soft 17, double after split allowed, surrender allowed, re-splitting aces allowed.

This question is raised and discussed in my forum at Wizard of Vegas.

Why does the number 19,933,230,517,200 come up so often as the total combinations in video poker?

rjs357

For the benefit of other readers, here is the return table for 9-6 Jacks or Better.

### "9-6" Jacks or Better

Hand | Payoff | Combinations | Probability | Return |
---|---|---|---|---|

Royal flush | 800 | 493512264 | 0.00002476 | 0.01980661 |

Straight flush | 50 | 2178883296 | 0.00010931 | 0.00546545 |

Four of a kind | 25 | 47093167764 | 0.00236255 | 0.05906364 |

Full house | 9 | 229475482596 | 0.01151221 | 0.10360987 |

Flush | 6 | 219554786160 | 0.01101451 | 0.06608707 |

Straight | 4 | 223837565784 | 0.01122937 | 0.04491747 |

Three of a kind | 3 | 1484003070324 | 0.07444870 | 0.22334610 |

Two pair | 2 | 2576946164148 | 0.12927890 | 0.25855780 |

Jacks or Better | 1 | 4277372890968 | 0.21458503 | 0.21458503 |

Nothing | 0 | 10872274993896 | 0.54543467 | 0 |

Total | 19933230517200 | 1 | 0.99543904 |

Most of my video poker return tables for 52-card games have the same number of combinations of 19933230517200. The question is why?

First, there are combin(52,5) = 2,598,960 ways to choose five cards out of 52.

Second, there are up to combin(47,5) = 1,533,939 combinations on the draw, depending on how many cards the player discards. The following table shows the number of draw combinations in the second column according to the number of discards.

### Draw Combinations

Discard | Combinations | Weight | Product |
---|---|---|---|

0 | 1 | 7,669,695 | 7,669,695 |

1 | 47 | 163,185 | 7,669,695 |

2 | 1,081 | 7,095 | 7,669,695 |

3 | 16,215 | 473 | 7,669,695 |

4 | 178,365 | 43 | 7,669,695 |

5 | 1,533,939 | 5 | 7,669,695 |

The least common multiple of the numbers in the second column is 7,669,695. This number can be expressed as 5×combin(47,5). To keep the total number of combinations the same for every hand on the deal I weight the draw combinations so that the total combinations on the draw is 7,669,695.

So, 19,933,230,517,200 = combin(52,5)×combin(47,5)×5. Some of my 52-card video poker tables have a smaller number of combinations. This is because sometimes the total number of combinations for each hand in the final return table has a greatest common divisor greater than one. In this case I sometimes divide each total by whatever the greatest common divisor is. My video poker analyzer does this automatically.

This question was raised and discussed at my Wizard of Vegas forum.

Two players each are dealt a random number in (0,1). First player decides to either stand pat or discard and draw a new number. Second player then does the same. High number wins. What is the optimal strategy for each player? Assuming optimal strategy, what is the probability each player wins?

Joe Shipman from New Jersey

I was playing Pai Gow Poker at an Indian casino and received a hand of Kings and Queens and was counting my winnings in my head. Then the dealer flipped over TTTJK*A (*=wild), with no flush possibility. She set her hand as TTTKJ, A*. I asked to see the house way since I thought that TJ*KA, TT would be a better play and I had a financial interest in the outcome.

They showed me the house way and it did not clearly address this hand. They ended up taking my money and giving me double my bet in non-negotiable chips so that the game could continue, which was okay. There was a pow wow in the pit with 4+ suits about how to set the hand and it lasted well over half an hour. They finally decided that TTTJK/A* was the proper way to set it and would change the house way as such.

My question is what is the house way for this hand and how do you interpret the house way for confusing hands like this?

bigfoot66

Pai gow poker was a badly designed game from day one and decades later nobody has bothered to clean it up. Flaws to pai gow poker include:

- An overly confusing and complicated house way.
- The ridiculous rule that the A2345 straight is the second highest.
- Option for player banking and co-banking, which almost nobody, except me, invokes.

When I briefly worked for a major Strip casino, that shall remain nameless, I offered to create a house way that would have been shorter, more powerful, and clearly covered every possible situation. Not just for pai gow poker, but tiles as well. Of course, that suggestion was shot down without comment.

Now that my rant is out of the way, I'll try to answer your question. While there are some differences from one house way to another, they are all more or less organized the same way. The Foxwoods house way is typical. Using your hand as an example, it isn't clear whether to treat is as a straight hand and play it as AKQJT/TT or a full house hand and play it as TTTJK/AA.

Not that you asked, but let's look at what would be the mathematical better play for the house. Using my pai gow poker appendix 1 we get the following, assuming the dealer is banking:

By playing the hand as a straight the dealer can expect to win 96.46% of the money, as opposed to 87.49% as a full house. So, going by the straight rule is significantly better.

Personally, the way I code pai gow poker is to start with the highest ranked hand (five aces) and move my way up the page in an if/else if/else if/else if/end loop. In other words, I classify the hand according to the highest possible five-card hand that can be made and then follow those rules.

I think that this is how the house way is intended to be interpreted. As evidence why, consider the Canterbury Park house way. In particular the rules for flush hands, which reads as follows.

Six and seven card flushes play the highest two cards possible in the Low Hand.

(Exception: If two pair are with the flush, the two pair rule will apply.)

As with all other house ways, the lowest hands are listed first. If the hand was supposed to be played by the first set of rules encountered, then they would have followed the two pair rule anyway. There would be no reason to explicitly make an exception to follow the two pair rule, since it was listed before the flush rule.

For practical purposes, I asked two dealers how they would handle this hand. They both pretty much said that the written house way is just a guideline and that in the event of a confusing hand, ask the floor what to do, and then just do as you're told without asking why. One of the dealers said that an unspoken policy is to go by the rule that seems to fit that situation best. In the case of Foxwoods, there is a rule for straight hands with a three of a kind. Since that most specifically describes the hand in question, follow that rule, despite the fact that the full house rule covers it too.

To make a long story short, no casino house way I have ever seen, in either pai gow poker or pai gow tiles, clearly explains how to play every possible hand. They are full of contradicting rules. Until somebody cares to do anything about it, you'll have to go by the interpretation of whoever is working the floor. In my experience, these confusing cases seem to usually get ruled against the player.

I hope one of these days a player will get fed up with this and make a complaint to the appropriate Gaming authority the next time an ambiguous hand is set in a way that favors the dealer.

This question is raised and discussed in my forum at Wizard of Vegas.