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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+0.5500)/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((4520.5500)/15.8114) = 0.001080. So the answer is 0.998920.00108 = 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)^{1000x}. 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.