# Ask the Wizard #132

Do you have any advice for picking the terminal digit in Super Bowl pools?

Anonymous

The office pools I have seen randomized the tables by assigning a random digit to each row and column. However if you can choose the actual terminal digits the following table shows the frequency of each terminal digit for the final score of either team, based on every NFL game from 1983 to 2003.

### NFL Terminal Digits per Side

Digit |
Frequency |
Probability |

0 |
1887 |
17.75% |

1 |
1097 |
10.32% |

2 |
348 |
3.27% |

3 |
1382 |
13.00% |

4 |
1608 |
15.13% |

5 |
396 |
3.73% |

6 |
848 |
7.98% |

7 |
1945 |
18.30% |

8 |
631 |
5.94% |

9 |
488 |
4.59% |

Total |
10630 |
100% |

So this table shows 7 is the best choice, followed by 0, 4, and 3.

I am just learning how to play Baccarat and since every player can bet on either player and banker and are not really playing each other, I was wondering what game is played in the James Bond movies? For example, In Dr. No it seems as if Bond is against a woman and he is winning her money? Is there something I am missing or is it a different game? Thank you for your time.

Anonymous

Fortunately I am a big James Bond fan and have all the Bond movies on DVD. I checked Dr. No and it seems he is playing Chemin De Fer. The scene was spoken in French, which doesn’t help me. There is a similar scene in For Your Eyes Only. In that movie it looks like Bond is playing baccarat, acting as the banker, but after the player acts he pauses and another character tells Bond, "The odds favor standing pat". This would imply that Bond had free will in whether to take a third card, an option you don’t have in baccarat. As I understand my gambling history, the American version of baccarat is a simplified version of Chemin De Fer, in which the drawing rules are predetermined. Incidentally, according to www.casino-info.com American baccarat originated at the Capri Casino in Havana, Cuba.

In a 10-handed game of Texas Hold 'em, and the flop is three different ranks, what is the probability that three players have a set?

Anonymous

For those unfamiliar with the terminology, each player gets two cards to himself and the three flop cards are shared among all players. So this is the same as asking if you dealt three community cards, all of different ranks, and ten 2-card hands, what is the probability three of the 2-card hands would be pairs that match one of the three community cards.

The probability player 1 has a set is 3*combin(3,2)/combin(49,2). Then the probability player 2 has a set is 2*combin(3,2)/combin(47,2). Then the probability player 3 has a set is combin(3,2)/combin(45,2). However, any three players can the three sets, not necessarily the first three. There are combin(10,3) ways to choose the 3 players out of 10 that have sets. So the answer is combin(10,3)*(3*combin(3,2)/combin(49,2))*(2*combin(3,2)/combin(47,2))*(combin(3,2)/combin(45,2)) = 0.00000154464 = 1 in 64,740.

What is the ’statistical’ dollar value of a phantom bonus? Say I deposit $100 and get another $100 in phantom bonus. If my goal is to win $100 (total balance of $300), how much approximate value is the phantom bonus worth to me?

Anonymous

Ignoring the house edge, the probability of reaching your goal is 2/3 and the expected value of the phantom bonus is $33.33. For a phantom bonus of b, cashable chips of c, and a winning goal of g the probability of reaching your goal is (c+b)/g and the expected value of the phantom bonus is ((c+b)/g)*(g-b)-c. In general, the higher the winning goal the greater the expected value of the phantom bonus.

A Hold 'em tournament starts by high-carding for the button. Highest card wins, and spades beats hearts beats diamonds beats clubs. What is the average card that will win in a 10 person table? I've tried simulating it by assigning a number value to each card, but I can't figure it out for the life of me! Thanks and keep it up!

Stephen K. from Atlanta, GA

To simplify the question, let's say the cards were numbered 1 to 52. The following table shows the probability that the 10th to 52nd card is the highest card. There are combin(x-1,9) ways to choose 9 numbers under x and combin(52,10) ways to choose any number numbers out of 52. So the probability that x is the highest number can be expressed as combin(x-1,9)/combin(52,10). The expected column is the product of the probability and the number of balls. The sum of the expected column shows us that on average the highest ball will be 48.18. Rounding to the nearest card, the highest expected card is the king of spades.

### Highest of 10 Cards

Highest Card | Probability | Expected |
---|---|---|

10 | 0.000000000063 | 0.000000000632 |

11 | 0.000000000632 | 0.000000006953 |

12 | 0.000000003477 | 0.000000041719 |

13 | 0.000000013906 | 0.000000180784 |

14 | 0.000000045196 | 0.000000632742 |

15 | 0.000000126548 | 0.000001898227 |

16 | 0.000000316371 | 0.000005061939 |

17 | 0.000000723134 | 0.000012293281 |

18 | 0.00000153666 | 0.000027659882 |

19 | 0.00000307332 | 0.000058393084 |

20 | 0.000005839308 | 0.000116786168 |

21 | 0.000010616924 | 0.000222955411 |

22 | 0.000018579618 | 0.000408751587 |

23 | 0.00003144243 | 0.000723175884 |

24 | 0.00005165542 | 0.001239730087 |

25 | 0.000082648672 | 0.002066216811 |

26 | 0.000129138551 | 0.003357602319 |

27 | 0.000197506019 | 0.005332662506 |

28 | 0.000296259028 | 0.008295252787 |

29 | 0.000436592252 | 0.012661175306 |

30 | 0.000633058765 | 0.01899176296 |

31 | 0.000904369665 | 0.028035459607 |

32 | 0.001274339073 | 0.040778850337 |

33 | 0.001772993493 | 0.058508785267 |

34 | 0.002437866053 | 0.082887445794 |

35 | 0.003315497832 | 0.116042424112 |

36 | 0.004463170158 | 0.160674125694 |

37 | 0.005950893544 | 0.220183061136 |

38 | 0.007863680755 | 0.298819868684 |

39 | 0.010304133403 | 0.401861202713 |

40 | 0.013395373424 | 0.535814936951 |

41 | 0.017284352805 | 0.708658464999 |

42 | 0.022145577031 | 0.930114235312 |

43 | 0.028185279858 | 1.211967033891 |

44 | 0.035646089232 | 1.568427926212 |

45 | 0.044812226463 | 2.016550190844 |

46 | 0.056015283079 | 2.576703021634 |

47 | 0.069640622206 | 3.273109243697 |

48 | 0.086134453782 | 4.134453781513 |

49 | 0.106011635423 | 5.194570135747 |

50 | 0.129864253394 | 6.493212669683 |

51 | 0.158371040724 | 8.076923076923 |

52 | 0.192307692308 | 10 |

Total | 1 | 48.181818181818 |

Although you didn't ask, the median card is the ace of clubs. The probability of the highest card falling under the ace of clubs is 41.34%, exactly on the ace of clubs is 10.60%, and higher than the ace of clubs is 48.05%.