# Ask the Wizard #281

Given that a fair coin is tossed n times, what is the probability of seeing at least one streak of t tails?

Anonymous

The answer is 1-F^{(t)}_{n+2}/2^{n}, where F^{(t)}_{n} is the n-th number in a t-step Fibonacci sequence.

What is a Fibonacci sequence, you might ask? The first number is a one. In a t-step sequence every subsequent number is the sum of the previous t numbers. Assume any number before the first number is zero.

Let's look at a two-step sequence. The first number is 1. The second is the sum of the previous two numbers. Assume a zero before the one, so the second number is 0+1=1. The third number is 1+1=2, the fourth is 1+2=3, the fifth is 2+3=5.

The first twelve 2-step Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

Let's do an example. What is the probability of getting two consecutive tails at least once in ten flips?

We use the two-step Fibonacci sequence, because we need only two tails. The 12th number in the sequence (two more than the number of flips) is 144. So, the answer is 1-F^{(2)}_{10+2}/2^{10} = 1 - 144/2^{10} = 1 - 144/1024 = 85.94%.

How about the probability of getting five consecutive tails in the 20 flips?

The first 22 5-step Fibonacci numbers are 1, 1, 2, 4, 8, 16, 31, 61, 120, 236, 464, 912, 1793, 3525, 6930, 13624, 26784, 52656, 103519, 203513, 400096, 786568.

The answer is thus 1 - F^{(5)}_{20+2}/2^{20} = 1 - 786,568/1,048,576 = 1 - 75.01% = 24.99%.

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

I didn't realize this until I started hitting W2-Gs from casinos, but are there any ways to get paid without giving a Social Security number? I'm a bit paranoid about casinos getting more information than necessary. I hit a few royals and hand pays and do not like that the casino holds on to my license and I have to write in my social.

djatc

First, you'll have to produce photo identification, or the casino will hold onto the money until you do. If you show identification but decline to produce or declare a valid social security or other tax identification number, then 25% to 30% will be withheld depending on whether the jackpot is more or less than $5,000, and whether you are from the United States or a foreign country with a reciprocal tax treaty.

As of 2011, such countries were Armenia, Australia, Austria, Azerbaijan, Bangladesh, Barbados, Belarus, Belgium, Bulgaria, Canada, China, Cyprus, Denmark, Egypt, Estonia, Finland, France, Georgia, Germany, Greece, Hungary, Iceland, India, Indonesia, Ireland, Israel, Italy, Jamaica, Japan, Kazakhstan, Kyrgyzstan, Latvia, Lithuania, Luxembourg, Mexico, Moldova, Morocco, New Zealand, Norway, Pakistan, Philippines, Poland, Portugal, Romania, Russia, Russia, Slovakia, Slovenia, South Africa, South Korea, Spain, Sri Lanka, Sweden, Switzerland, Tajikistan, Thailand, The Czech Republic, The Netherlands, Trinidad and Tobago, Tunisia, Turkey, Turkmenistan, Ukraine, United Kingdom, Uzbekistan and Venezuela.

I've been trying to figure out the rules exactly, but it is giving me a headache. Please refer to IRS rules for issuing a W2G form for more information.

My thanks to Marissa Chien, co-author of Tax Help for Gamblers , and MathExtremist for their help with this question.

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

According to CardPlayer.com, Amir Lehavot, who is one of the nine players to make the 2013 final table in the World Series of Poker, is selling any winnings above the minimum $733,224 for ninth place at a price of $29,248 for each 1% share. Is that a fair price?

Anonymous

First, let's review the chip stacks.

### 2013 WSOP Final Table Chip Stacks

Player | Chips |
---|---|

JC Tran | 38,000,000 |

Amir Lehavot | 29,700,000 |

Marc McLaughlin | 26,525,000 |

Jay Farber | 25,975,000 |

Ryan Riess | 25,875,000 |

Sylvain Loosli | 19,600,000 |

Michiel Brummelhuis | 11,275,000 |

Mark Newhouse | 7,350,000 |

David Benefield | 6,375,000 |

The next table shows the win for each final outcome in the tournament.

### 2013 WSOP Final Table Prize Money

Place | Win |
---|---|

1st | $8,359,531 |

2nd | $5,173,170 |

3rd | $3,727,023 |

4th | $2,791,983 |

5th | $2,106,526 |

6th | $1,600,792 |

7th | $1,225,224 |

8th | $944,593 |

9th | $733,224 |

Assuming each player is of equal skill, the probability of winning could be estimated as the share of the total chip stack. However, it gets more complicated for every position after that. To help answer the question, I developed my poker tournament calculator.

After putting in the information above, you'll see that Amir has an expected win of $ 3,658,046. Then subtract out the minimum prize of $733,224 for 9th place and you get $2,924,822 in expected non-guaranteed winnings. Each 1% share has a value of $29,248.22. This is conveniently the price quoted in the cardplayer.com article.

By the way, Lehavot finished third, for $3,727,023 in prize money. Subtracting the $733,224 guaranteed money for 9th place and dividing by 100, each 1% share returned $29,938. The original cost per share was $29,248, so each share would have seen a 2.36% profit.

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

I think I have a winning betting system. However, I need more than the 3,000 baccarat shoes that you have on your baccarat page to test it. Can you make more?

gpl2112

I hope you're happy; I just created a quarter million of them. Better to find out your system will fail eventually, which they all do, for free than with real money in the casino.

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

What good is the math on your site when every casino has a secret room with a computer that controls the outcome of every bet. It can cause the ball in roulette to fall wherever players will lose the most. In card games, it controls the shuffling machine to give players bad cards. Are you just naive or in on the conspiracy?

mik

All I'm going to say is the "secret room" at the Venetian has cookies.

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