# Ask the Wizard #120

Anonymous

The following table shows the probability of 0 to 4 aces in a totally random hand.

### Ace Probabilities — Random Hand

Aces | Formula | Combinations | Probability |
---|---|---|---|

0 | combin(48,5) | 1712304 | 0.658842 |

1 | combin(4,1)×combin(48,4) | 778320 | 0.299474 |

2 | combin(4,2)×combin(48,3) | 103776 | 0.03993 |

3 | combin(4,3)×combin(48,2) | 4512 | 0.001736 |

4 | combin(4,4)×combin(48,1) | 48 | 0.000018 |

Total | 2598960 | 1 |

Take the sum for 1 to 4 aces we see the probability of at least one ace is 0.341158. The probability of two or more aces is 0.041684.

The probability of there being at least one more ace, given there is at least one, can be restated per Bayes' theorem as probability(two more aces given at least one ace) = probability (two or more aces)/probability(at least one ace) = 0.041684/ 0.341158 = 0.122185.

For those rusty on Bayes' Theorem, it states that the probability of A given B equals the probability of A and B divided by the probability of B, or Pr(A given B) = Pr(A and B)/Pr(B).

The next table shows the combinations and probability for each number of other aces given that the ace of spades was removed from the deck.

### Ace Probabilities — Ace Removed Hand

Aces | Formula | Combinations | Probability |
---|---|---|---|

0 | combin(3,0)×combin(48,4) | 194580 | 0.778631 |

1 | combin(3,1)×combin(48,3) | 51888 | 0.207635 |

2 | combin(3,2)×combin(48,2) | 3384 | 0.013541 |

3 | combin(3,3)×combin(48,1) | 48 | 0.000192 |

Total | 249900 | 1 |

This shows the probability of at least one more ace is 0.221369.

For fun, let's solve it the same question using Bayes' Theorem. Assume that random hands are dealt until one is found containing the ace of spades. The probability of at least one additional ace, given the hand contains the ace of spades, could be rewritten as probability(at least two aces given ace of spades is in hand). According to Bayes' theorem, this equals Probability(hand contains ace of spades and at least one more ace) / Pr(hand contains ace of spades). We can break up the numerator as Probability(2 aces including ace of spades) + Probability(3 aces including ace of spades) + Probability(4 aces). Using the first table this equals 0.039930×(2/4) + 0.001736×(3/4) + 0.000018 = 0.021285. The probability of the ace of spades is 5/52 = 0.096154. So the probability of at least two aces given the aces of spades is 0.021285/0.096154 = 0.221369.

So the probability of two or more aces given at least one ace is 12.22% and given the ace of spades is 22.14%.

Anonymous

Let’s look at another simpler situation. Suppose woman A says "I have two kids and at least one is a boy." Woman B says "I have two kids and the older one is named John." We can assume nobody named John is a girl and no woman gives the same name to more than one kid. Using conditional probability the probability of both kids being a boy of woman A is pr(both boys)/pr(at least one boy) = pr(both boys)/(1-pr(both girls)) = (1/4)/(1-(1/4)) = (1/4)/(3/4) = 1/3. However the probability woman B’s younger kid is a boy, or both kids are boys, is ?, because saying the older child is named John tells us nothing about the younger child.

To make another example suppose you go to Jiffy Lube and they offer two deals for the same price. Deal A is they will check four parts and replace only the first defective one found. Deal B is they will check only one problem and fix it if found. Wouldn’t you rather take deal A? Your car came in with the same number of expected bad parts but the probability of finding a problem is greater under deal A, and thus you will leave with a small number of expected defective parts under that plan. Likewise a test for any ace will probably turn up the only ace, while a test for the ace of spades does not check for the other three suits, leaving them more likely to be aces.

Anonymous

Face down. Not being allowed to see other player cards until the end of the hand gives the player less information, which works against card counters.

Anonymous

The casinos like to corral their bettors according to how much they bet. One reason for this is the higher limit tables have fewer players so the big bettors get in more hands per hour. Another reason is that it is said players like to be around other players of similar bet size. If a player wanted to bet $1000 at a $5 table it might make other $5 players at that table feel nervous or uncomfortable. A third reason is it is a preventative measure against cheating.

Anonymous

As I’ve said hundreds of times there is no magic number of when you enter the "long run." However the more impressive your results the fewer the hands your need to make a case that they are not just random. In your case the probability of getting 54.5% or better out of 2391 games is just about 1 in 200,000. So I would say that record is worth to be taken very seriously. Here is how I arrived at that number:

Expected wins = 2391/2 = 1195.5

Actual wins above expectations = 107.5

Standard deviation = sqrt(2391*(1/2)*(1/2)) = 24.45

Standard deviations away from expectations = (107.5 + 0.5)/24.45 = 4.4174

Probability of 4.4174 standard deviations or more = normsdist(-4.4174) = 0.000005 = 1 in 200,000

Anonymous

Yes. I have never heard of a player being refused to use one.

Anonymous

The probability of getting 2 more of the same suit is 39*combin(11,2)/combin(50,3) = 0.109439. The probability of getting 3 more of the same suit is combin(11,3)/combin(50,3) = 0.008418. So the probability of getting at least 2 more of the same suit is 0.117857.