# Ask the Wizard #226

Mick from Australia

Based on six decks, I show a player 21 pushing against a blackjack lowers the house edge by 0.37%. A blackjack tie paying 3 to 2 lowers the house edge by 0.32%. No strategy changes are required.

This may be balanced by the fact that going for a 2-pointer in final seconds you are more likely to be fouled and get 2 easy points, but even still, the best foul shooters run around 85%, meaning a 72% chance of making both, followed by 50% chance of winning in overtime, for a total of 36%. What is your take on these this?

Nick K. from Scarsdale, NY

I hope you’re happy. My knowledge of basketball rules and strategy is pretty weak, so I asked some friends stronger than me in that area, and never got the same answer twice. Some answers were direct opposites of each other. Two theories I got out of the discussion are (1) the overall field goal percentage for the NBA is more like 50% (source), and (2) there is a chance going for a 2-point shot that the shooter will get fouled, and make the shot anyway. Sorry I can’t do better than that.

Les from Fallbrook, CA

If you are implying the casino is changing the odds of the game while you're sitting there playing it, then I would say that is just a myth. To change the odds of a game, the slot maker would have to open up the game and change the EPROM chip. With server based game, where this can be done remotely, regulations require that the game be unplayed for a certain number of minutes before any changes can made.

If you are implying that the casino sets a slot machine loose for the first so many days, to draw new players, and then switches the EPROM to a stingier one, then I would disagree as well. That could easily be done, and legally, but I doubt it is. In my slot machine survey I found that any given casino was fairly consistent in how loose or tight they set their slots.

1) A will score more than B

2) B will score more than A

3) Game finishes as a tie.

Is the information provided enough to calculate the probabilities for each outcome?

Dimitar from Sophia, Bulgaria

That does not take into account that the individual scores should be somewhat negatively correlated, and that the average points each team gives up is just as important as the average points scored. If we can assume that 1.5 and 1.2 are the expected number of points scored in the game, considering both offense and defense, and we ignore the correlation factor, then we can get a decent estimate on your three probabilities. There are lots of Super Bowl props like this, but based on who will score more touchdowns, field goals, interceptions, etc..

The first step is to use the Poisson distribution to estimate the probability of each number of goals for each team. The general formula is the probability that a team has g goals, with a mean of m, is e^{-m} × m^{g}/g!. In Excel, you can use the formula poisson(g,m,0). The following table shows the probability for 0 to 10 goals of both teams, using this formula.

### Probabilities for 0 to 8 Goals for each Team

Goals | Team A | Team B |

0 | 0.223130 | 0.301194 |

1 | 0.334695 | 0.361433 |

2 | 0.251021 | 0.216860 |

3 | 0.125511 | 0.086744 |

4 | 0.047067 | 0.026023 |

5 | 0.014120 | 0.006246 |

6 | 0.003530 | 0.001249 |

7 | 0.000756 | 0.000214 |

8 | 0.000142 | 0.000032 |

The next step is rather mundane, but you have to make a matrix of all the 81 possible combinations of 0 to 8 scores for each team. This is done by multiplying the probability of x scores for team A and y scores for team B, from the table above. The following table shows the probability of every score combination from 0-0 to 8-8.

The next table shows the winner according to each combination of goals, where T represents a tie.

### Winner Combinations for Both Teams

Goals Team A | Goals Team B | ||||||||

0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |

0 | T | B | B | B | B | B | B | B | B |

1 | A | T | B | B | B | B | B | B | B |

2 | A | A | T | B | B | B | B | B | B |

3 | A | A | A | T | B | B | B | B | B |

4 | A | A | A | A | T | B | B | B | B |

5 | A | A | A | A | A | T | B | B | B |

6 | A | A | A | A | A | A | T | B | B |

7 | A | A | A | A | A | A | A | T | B |

8 | A | A | A | A | A | A | A | A | T |

Finally, you can use the sumif function in Excel to add the corresponding cells for all three possible outcomes of the bet. In this case the probabilities are:

A wins = 44.14%

B wins = 30.37%

Tie = 25.48%

Appendix C in Sharp Sports Betting by Stanford Wong gives the win/lose/tie probabilities for bets like this. For this case he lists 44%, 30%, and 25%. If anyone knows a simple formula for this kind of problem, I’m all ears.

**Follow Up**: I received an e-mail from Bob P., who always keeps me on my toes when it comes to math. Here is what he wrote.

Looked up the distrib of the difference between 2 uncorrelated Poissons. It’s a Skellam (new to me).Anyway, the question can then be posed as P(Z=0), P(Z>0), and P(Z<0) where Z is a Skellam with parameters 1.5 and 1.2.

If you haven’t already done it, you’ll be pleased to know

P(Tie) = P(Z=0) = .254817

P(A beats B) = P(Z>0) = .441465

P(B beats A) = P(Z<0) = 1 - .254817 - .441465 = .303718

almost exactly your answers.

The Wikipedia entry for a Skellam mentioned Bessel functions, which is about the point in calculus where I get scared to go further. So, I’m going to take Bob’s word on this one.

Anonymous

The answer and solution can be found on my companion site, mathproblems.info, problem 201.