# Ask the Wizard #296

relaxmax

The answer is 50/50. This will be true for any number of dice rolled, not just two.

A bit off-topic, but I've always thought an odd/even set of bets would be a good way to replace the dreaded big 6/8 bets in craps. To give the house an advantage, here are my proposed pay tables and analysis.

### Odd Bet

Event | Pays | Combinations | Probability | Return |
---|---|---|---|---|

3 or 11 | 1.5 | 4 | 0.111111 | 0.166667 |

5 or 9 | 1 | 8 | 0.222222 | 0.222222 |

7 | 0.5 | 6 | 0.166667 | 0.083333 |

Even | -1 | 18 | 0.500000 | -0.500000 |

Total | 36 | 1.000000 | -0.027778 |

### Even Bet

Event | Pays | Combinations | Probability | Return |
---|---|---|---|---|

2 or 12 | 3 | 2 | 0.055556 | 0.166667 |

4 or 10 | 1 | 6 | 0.166667 | 0.166667 |

6 or 8 | 0.5 | 10 | 0.277778 | 0.138889 |

Odd | -1 | 18 | 0.500000 | -0.500000 |

Total | 36 | 1.000000 | -0.027778 |

Please note that I claim all rights with this publication.

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

"Anonymous" .

For those unfamiliar with the rules, the player is given four price tags and must place them on four items. When he is done he pulls a lever which gives the number of correct matches. If the player has less than four correct, then he may rearrange the tags and try again. The player may try as many times as he can within 45 seconds.

My advice is to always submit a selection that has a chance of winning, given the previous history of selections and scores. If the first score is 0, then don't reverse two sets of two tags, but instead move everything by one position in either direction.

If you're not able to do the logic on the spot, then I spell it out for you below. To use this strategy, assign the different tags the letters A, B, C, and D. Then place them in the order shown, from left to right on the stage. Always start with ABCD. Then look up the score history below and choose the sequence of tags indicated for that score sequence.

If 0, then BCDA

If 0-0, then CDAB

If 0-0-0, then DABC (must win)

If 0-1, then BDAC

If 0-1-0, then CADB (must win)

If 0-1-1, then CDBA

If 0-1-1-0, then DCAB (must win)

If 0-2, then BADC

If 0-2-0, then DCBA (must win)

If 1, then ACDB

If 1-0, then BDCA

If 1-0-0, then CABD

If 1-0-0-1, then CBAC (must win)

If 1-1, then BDCA

If 1-1-0, then CABD

If 1-1-0-1, then CBAC (must win)

If 1-1-1, then BCAD (must win)

If 2, then ABDC

If 2-0, then BACD (must win)

If 2-1, then ACBD

If 2-1-0, then DBCA

If 2-1-1, then ADCB

If 2-1-1-0, then CBAD (must win)

The following table shows the probability of each number of total turns. The bottom right cell shows an expected number of turns of 10/3.

### Race Game

Turns | Combinations | Probability | Return |
---|---|---|---|

1 | 1 | 0.041667 | 0.041667 |

2 | 4 | 0.166667 | 0.333333 |

3 | 8 | 0.333333 | 1.000000 |

4 | 8 | 0.333333 | 1.333333 |

5 | 3 | 0.125000 | 0.625000 |

Total | 24 | 1.000000 | 3.333333 |

This question is discussed on my forum at Wizard of Vegas.

pokerbum

Let's first ask what is the probability that in a four-player game all four players have any ace-king?

The answer to that question would be (4*4/combin(52,2)) * (3*3/combin(50,2)) * (2*2/combin(48,2)) * (1/combin(46,2)) = 1 in 3,292,354,406.

However, it is possible that some of these ace/king hands will be suited. To be exact, the probability that none of them are suited is 9/24. So lower the probability to 1 in 8,779,611,750.

However, it is a ten-player game, and any of the combin(10,4)=210 sets of four players could be the four with non-suited ace-king. So, multiply that probability by 210 and the answer is 1 in 41,807,675.

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