# Ask the Wizard #223

Chris from Chicago

As I show in my page in the NBA, when buying a half point the probability of a win is 51.01%, a loss is 47.01%, and a push is 1.98%, assuming the bettor never buys the half point off of a spread of 0 or -1, which he shouldn’t do. If you only had to lay 110 for the extra half point, the expected return would be (0.5101 - 1.1×.4701)/1.1 = -0.64%. So, a free half point would not be enough to overcome the house edge.

Not that you asked, but if you lay 120, you can buy a half point at most sports books. If you were prepared to bet a game against the spread anyway, is the extra half point a good value? Laying 110, the house edge for a random picker is 4.45%, including ties. Laying 120, the house edge with the half point is 4.50%. So, buying a half point is marginally not worth the price.

The value of buying a half point in football depends a great deal on the point spread, because some margins of victory are much more likely than others. The only time it is worth buying the half point in the NFL is off of a point spread of 3. Unfortunately, the sports books know this too, and won’t let you do it off of 3, most of the time.

Alex from Greenwich, CT

I’ve been wondering this myself for years. In 2004 somebody accepted my betting system challenge, claiming he could beat blackjack without counting. The details are in my page on the Daniel Rainsong challenge. After I posted it, I received a message from a blackjack genius, who goes by the handle "Cacarulo." He challenged me under the same conditions and blackjack rules set forth in the Rainsong challenge.

Knowing how knowledgeable he is about blackjack, I felt that he was probably right, so I declined the challenge. I asked anyway how he would have gone about his strategy, but he wouldn’t tell me. I tend to think that he would have bet the minimum most of the time, except if it was late in the shoe, and the ratio of losses to wins was very high since the last shuffle, he would have bet the maximum. The reason is that losing is positively correlated to small cards being played, and winning to large cards. In other words, a benefit of losing is that it tends to make the count better. However, this is a weak correlation. My challenge allowed the player a bet range of 1 to 1,000, which is probably enough to overcome the house edge, but it will be hard to find a real casino okay with a jump in bet size by a factor of 1,000.

The short answer to your question is, no, tracking wins and losses will not help enough to warrant the bother of doing it.

Jon from Philadelphia

You're right. The probability of the same sequence of numbers chosen two nights in a row is 1 in 1000. The question that the writer was answering is what is the probability that 1-9-6 is drawn twice is a row, which is indeed one in a million. However, as you note, the pertinent question is what are the odds that any sequence repeats. The answer to that question is (1/10)^{3} = 1 in 1000.

Joe from Colorado

The probability of getting exactly 50 of each is combin(100,50)*(1/2)^{100} = 7.96%. Fair odds would be 11.56 to 1. So, at 3 to 1, it is a terrible bet, with a house edge of 68.2%. That is some friend you have. 50/50 is the most likely exact split between heads and tails. An interesting bet is whether the number of heads/tails will fall between 47 and 53, or not. The probability of falling inside that range is 51.59%. If you can find someone to bet that the total will fall outside that range, then at even money you would have a 3.18% advantage.

The following table shows the probability for each of 30 to 70 heads/tails.

### Probability of Total Heads/Tails in 100 Flips

Heads/Tails | Probability |
---|---|

30, 70 | 0.000023 |

31, 69 | 0.000052 |

32, 68 | 0.000113 |

33, 67 | 0.000232 |

34, 66 | 0.000458 |

35, 65 | 0.000864 |

36, 64 | 0.001560 |

37, 63 | 0.002698 |

38, 62 | 0.004473 |

39, 61 | 0.007111 |

40, 60 | 0.010844 |

41, 59 | 0.015869 |

42, 58 | 0.022292 |

43, 57 | 0.030069 |

44, 56 | 0.038953 |

45, 55 | 0.048474 |

46, 54 | 0.057958 |

47, 53 | 0.066590 |

48, 52 | 0.073527 |

49, 51 | 0.078029 |

50 | 0.079589 |

The general formula for the probability of w wins out of n trials, where the probability of each win is p, is combin(n,w) × p^{w} × (1-p)^{(n-w)} = [n!/(w! × (n-w)!] × p^{w} × (1-p)^{(n-w)}.