# Ask the Wizard #61

Emi from Manila, Philippines

(8^{2}/combin(416,2))* (7^{2}/combin(414,2)) = 0.00000043, or 1 in 2308093

Emi from Manila, Philippines

There are lots of them, and they are all worthless.

And last, could you please help me understand why it’s a "fallacy" that a win becomes more probable after a series of losses? The way I see it, since the expected outcome is an approximately 99.5% return, then if after 1000 hands, you’re at 78%, then, by definition, it would necessitate that the next hand being a win must be more likely to occur. People say that cards "don’t have a memory", but isn’t the natural curve, in essence, its memory??? Please help me understand this point! Thanks a lot.

Steve from Canada

You’re right, if you used a robot player against an ordinary game you would only lose more. However some casinos do offer games with a player advantage if played properly. Unified Gaming had a blackjack game with an 0.5% player edge for several months, but no longer. Many Real Time Gaming casinos offer a joker poker game with an expected return of 100.18%. Other casinos have promotions in which the player who plays the most hands in a period of time wins a prize, in which a robot player would have a clear advantage. About your second question the bell curve is a forward looking estimate of the sum of many random variables. You can not mix together past and future events. Once an event has happened it is no longer a random variable but a cold hard outcome. If you played 1000 hands of blackjack with a return of 78% then you fell on the tail end of the bell curve during that play. Starting from hand 1001 your results could fall anywhere on a new bell curve. I hope this helps, but it really takes a course in statistics to truly understand.

Ruby from Tacoma, USA

Good question. In Stanley Ko’s booklet __Mastering Three Card Poker__ he says that if you had a concealed computer to take maximum advantage of the information then seeing the first hand would lower the house edge from 3.37% to 3.31% on the second hand. Even if you could see all seven hands at the table the house edge would still be 2.32%.

Anonymous

It depends on the number of decks. If the number of decks is n then the probability is 2*pr(ace)*pr(10) = 2*(1/13)*(16*n/(52*n-1)), which is conveniently about 1 in 21.
Here is the exact answer for various numbers of decks.

### Probability of Blackjack

Decks | Probability |
---|---|

1 | 4.827% |

2 | 4.780% |

3 | 4.764% |

4 | 4.757% |

5 | 4.752% |

6 | 4.749% |

7 | 4.747% |

8 | 4.745% |

Matt from Radford, USA

If the probability of a blackjack is p then the probability of not getting any blackjacks in 10 hands is 1-(1-p)^{10}. For example in a six deck game the answer would be 1- 0.952511^{10}= 0.385251.

Joe P from Parma Heights, USA

I’m going to assume there is never a shuffle between hands. The three other players don’t matter. The answer would be 2^{3}*(16/52)*(4/51)*(15/50)*(3/49)*(14/48)*(2/47)= 0.00004401, or about 1 in 22722. If there were a shuffle between hands the probability would increase substantially.

4 times without throwing a 7?

3 times without throwing a 7?

2 times without throwing a 7?

1 times without throwing a 7?

How does the math work for this? Thanks

Stan Abadie from Harahan, Louisiana

You’re welcome, thanks for the kind words. The probability of throwing the dice n times without a 7, and then throwing a 7, is (5/6)^{n}*(1/6). The probability of throwing n non-sevens, without specifying the next throw would be (5/6)^{n}. So the probability of throwing the dice at least four times without a seven would be (5/6)^{4}=625/1296=0.4823.

Stephanie from Evans, USA

I don’t know when they shuffle but I would speculate after every hand. From my blackjack appendix 10 you will learn that the player’s odds improve slightly if the dealer plays exactly n hands between shuffles (including one) rather than playing to cut card, finishing the hand, and then shuffling.