# Ask the Wizard #84

Anonymous

For this problem we can use the normal approximation to the binomial distribution. The variance of the number of heads is 1000*(1/2)*(1/2)=250. So the standard deviation is 250^{1/2}=15.8114. The probability of less than 548 heads is normdist((548+0.5-500)/15.8114) = 0.998920, where normsdist is the Excel function for the probability a random variable with a normal distribution of mean 0 and standard deviation 1 will fall under the given Z score. Next we subtract the probability of less than 452 heads. This is normdist((452-0.5-500)/15.8114) = 0.001080. So the answer is 0.99892-0.00108 = 0.997840. Again, this is an approximation. The actual answer is 0.997856, but is more tedious to derive.** On average, after establishing a point in craps how often will the player make the point?**

Given that a point was made 5/12 of the time it will be a 6 or 8, 4/12 a 5 or 9, and 3/12 a 4 or 10. The probability of making a 6 or 8 is 5/11, a 5 or 9 is 4/10, and a 4 or 10 is 3/9. So the probability of making a point, given that a point was established is, (5/12)*(5/11)+(4/12)*(4/10)+(3/12)*(3/9) = 40.61%.

WD

The math is quite easy. The probability of a royal flush is 1 in 649740. So the expense of reseeding the jackpot is $10,000*(1/649740) = 1.54%. For every dollar bet you keep 40% for profit and reseeding the jackpot. 40%-1.54% = 38.46% profit/house edge. It does not make any difference what you pay on the smaller jackpot or if there is a maximum win. Ultimately the 60% that goes to the meter goes to the players one way or another, it doesn’t matter to you how it gets divided up.

Anonymous

For those not familiar with the rules there are five up cards. So the question is asking what is the probability that in 5 cards dealt from a single deck, without replacement, that at least three will be of the same suit. There are combin(52,5)=2598960 ways to deal 5 cards out of 52. The number of ways to deal 4 of the same suit is 4*combin(13,5)=1144. The number of ways to deal 4 of a suit is 4*combin(13,4)*39=111540. The number of ways to deal 3 of a suit is 4*combin(13,3)*combin(39,2)=847704. So the total combinations is 960388 and the probability is 36.95%.

Anonymous

Good question. Let’s think of this in units as opposed to $100 bets. You will always have a bet on the pass or come. On any given roll the probability there is an old pass or come bet on the 4 is 3/9. This is the probability that by looking back at old rolls you will find a 4 before a 7. Likewise the probability of having a bet on 5 is 4/10 and on 6 is 5/11. So the average overall bet is 1+pr(4)+pr(5)+pr(6)+pr(8)+pr(9)+pr(10) = 1+3/9 + 4/10 + 5/11 + 5/11 + 4/10 + 3/9 = 3.3758 units. This average will not true at the beginning, while you are getting in to the game. It will only apply after all point numbers and the 7 have already been rolled at least once.

Bill K.

The probability of your number hitting exactly x times is combin(1000,x)*(1/38)^{x}*(37/38)^{1000-x}. The following table shows the probability of all number of hits from 0 to 6 and the total.

### Wins in 1000 Roulette Bets

Number | Probability |

0 | 0.00000000000262 |

1 | 0.00000000007078 |

2 | 0.00000000095556 |

3 | 0.00000000859146 |

4 | 0.00000005787627 |

5 | 0.00000031159330 |

6 | 0.00000139655555 |

Total | 0.00000177564555 |

So the answer is 0.00000177564555, or 1 in 563175. I hope this didn’t happen at an Internet casino.

You may wonder why I didn’t use the normal approximation as I did with the coin flipping problem above. That is because it doesn’t work well with very high and very low probabilities.