# Ask the Wizard #256

I've been having a good laugh about Paul the Octopus and the "predictions" he’s been making. I tend to be a bit more analytical, and rely on math for my bets instead of an Octopus, but thought it was kind of cute and amusing anyhow.

I don’t suppose you have any thoughts on this? The reason I found it interesting was because the Octopus does seem to favor the German flag, perhaps because there’s other German flags in its aquarium setting. He’s also correctly picked the Serbia and Spain matches that Germany played. Are there any interesting mathematical odds or personal input you’d like to share in your next column or an article perhaps?

Michael

Paul’s record is 12 correct picks and 2 incorrect picks. The probability of getting exactly 12 out of 14 correct picking randomly is combin(14,12)×(1/2)^{14} = 0.56%. The probability of getting 12 or more correct out of 14 is (1+14+combin(14,2))×(1/2)^{14} = 0.65%. He was not given the choice of picking a tie, and there never was one in the games he handicapped. I’m not sure how his record would have been depicted if there were any ties, but I suspect they wouldn't have been included.

This is obviously dumb luck, possibly combined with some form of chicanery. While this may be fun, I don't consider it legitimate news. I think this story got more news coverage here than some civil wars in Africa.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

A player in Pai Gow Poker is dealt the following hand: Jh Qh Kh Ah Qs Ks Joker. How should the hand be set?

- Royal flush & A-K
- Two pair (KKQQJ) & AA

rdw4potus

My pai gow poker appendix 1 is useful for questions like this. To answer the question, add the low and high hand power ratings for all viable ways to play the hand. The following table shows the sum of the power ratings (for not banking) for both viable ways to play the hand. It shows that breaking up the royal, while painful, is the much better play.

### Pai Gow Poker — Power Rating Table

Low Hand | High Hand | Low Power Rating | High Power Rating | Total Power Rating | Expected Value |
---|---|---|---|---|---|

KQ | Royal flush | 0.452967 | 0.999507 | 1.452474 | 0.416162 |

AA | KKQQJ | 0.989071 | 0.821870 | 1.810941 | 0.765667 |

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Life expectancy for people of various ages has been calculated and summarized with data at the Social Security web site. However, I want to know the life expectancy of two people. Say I have two people: a thirty-year-old male (me) and a twenty-eight-year-old female (my gf). According to the chart, I will live another 46.89 years and she will live another 53.22 years. But, how long is it expected until we both are dead? How do I calculate this?

zachisbest

First, it would be appropriate to use cohort life tables, as opposed to the period life table you linked to. I tried to find cohort life tables online but was unsuccessful. However, we can still use the table provided. It may underestimate how long you will live slightly, because it won’t take into account future increases in life expectancy.

Answering your question involved creating a large matrix of the probability of each combination of year of death for you and the 28-year-old female. Forgive me if I don’t get into the details. The bottom line is that I show that first one of you will die in 41.8 years, and the latter death will be in 57.3 years. Both figures round down; in other words, you don’t get credit for partial years.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

I was wondering if I could get your help on computing the probability distribution table for Jacks or Better. I know that 52 choose 5 = combin(52,5) = 2,598,960, yet in every table that I have looked at for video poker, there are 19,933,230,517,200 total combinations. I was wondering why there are so many more than 52 choose 5, and how to compute them.

Mic

There are combin(52,5)=2,598,960 possible combinations on the deal. The reason my video poker return tables have almost 20 trillion combinations is you also have to consider what could happen on the draw. Here are the number of combinations according to how many cards the player discards.

### Combinations on theDraw in Video Poker

Discards | Combinations |

0 | 1 |

1 | 47 |

2 | 1,081 |

3 | 16,215 |

4 | 178,365 |

5 | 1,533,939 |

The least common multiple of all those combinations is 5×combin(47,5)= 7,669,695. Regardless of how many cards the player discards, the return combinations should be weighted so that the total comes to 7,669,695. For example, if the player discards 3, there are 16,215 possible combinations on the draw, and each one of them should be weighted by 7,669,695/16,215 = 473.

So the total number of combinations in video poker is 2,598,960 × 7,669,695 = 19,933,230,517,200 . For more on how to program video poker returns yourself, please see my page on Methodology for Video Poker analysis.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

On average, how many trials would you need in 38-number roulette before any number is repeated?

inversehelix

Counting the first trial, I show the mean is 8.408797, the median is 8, and the mode is 7.

The probability of two numbers without a repeat is 37/38 = 97.37%.

The probability of three numbers without a repeat is (37/38)×(36/38) = 92.24%.

The probability of four numbers without a repeat is (37/38)×(36/38)×(35/38) = 84.96%.

Following this pattern, the probability of no repeats in 8 numbers is (37/38)×(36/38)×(35/38)×...×(31/38) = 45.35%.

So the probability of a repeat within 8 numbers is 100% - 45.35% = 54.65%.

I suspect most people would estimate that that probability of a repeat within 8 numbers would be less than that. If you’re not above taking advantage of your math-challenged friends, propose a bet that it will take 8 or fewer numbers for at least one to repeat. So you would be betting on 8 or fewer, and your friend 9 or more. If he/she balks, then offer to take 7 or over, which would have a 55.59% chance of winning. Basically, whichever side covers the median of 8 is likely to win.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.