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Ask the Wizard #113
"Anonymous" .
You should try to sit as far to the left as possible, assuming you play the side bet. If you don’t make the side bet it doesn’t matter. The reason is if two more players get a straight flush or higher the player furthest to the left will get the benefit of the full meter, because the dealer pays players from her right to left. Subsequent players will get less after paying the first player. In the event of two royal flushes the first player would get the full meter and the second only $10,000, which is what the meter is generally reset to. However the odds of this are extremely unlikely. I would just play where you have the most elbowroom and are furthest away from any smoking players. Finally, yes, there are generally six spots.
"Anonymous" .
The house edge is 0.93%. More details can be found in my baccarat appendix 6.
 The examination shall consist of 7 subjects.
 For each subject, 60 multiple choice questions shall be asked.
 Each multiplechoice question shall have four possible answers, but only one correct answer.
 In order to pass, an examinee must obtain a general average of at least 75% and must not have a grade lower than 65% in any subject.
My question is, if an examinee merely guesses all his answers, what is his chance of passing the exam? Stated differently, what is the probability of passing the exam by sheer luck?
"Anonymous" .
To satisfy the 75% requirement the student must get at least 315 out of the 420 questions right. The expected number of correct answers from guessing is 420*0.25=105. The standard deviation is (420*0.25*0.75)^0.5 = 8.87412. So the candidate must exceed expectations by 210 questions, or 210/8.87412=23.66432 standard deviations. The probability of doing this is way off the charts. If every living thing on earth took this test, answering randomly, I doubt anyone or anything would pass. I won’t even get into the other requirement.
"Anonymous" .
I doubt the casino would cheat, why would they? The bigger concern is the other players. It would be very easy for players to collude over the phone or instant messenger. Whether they actually do or not I don’t know. There is probably a greater risk for that at the higher limit tables.
"Anonymous" .
But it is so much fun.
HEADS  bet that eleven to twenty numbers in the top half appear  even money
TAILS  bet that zero to nine numbers in the top half appear  even money
EVENS  bet that exactly ten numbers in the top half appear  pays 3 to 1
"Anonymous" .
The probability of the tie bet winning is combin(40,10)*combin(40,10)/combin(80,20) = 0.203243. Paying 3 to 1 the house edge is 18.703%. The probability of the heads (or tails) bet winning is (10.20343)/2 = 0.398378. Paying even money the house edge is 20.324%.
"Anonymous" .
This is a textbook Bayesian conditional probability question. In general the probability of A given B is the probability of A and B divided by the probability of B. In this case A is flipping 10 heads in a row and B is picking the twoheaded coin. The probability of A and B is 1/100. This is because there is a 1 in 100 chance of picking the twoheaded coin, and if you do the probability is 100% of flipping 10 heads in a row. The probability of flipping 10 heads in a row, assuming a randomly picked coin, is (1/100)*1 + (99/100)*(1/2)^{10}. That is because there is a 1% chance of picking the twoheaded coin, which has a 100% of getting 10 heads, and a 99% of picking a fair coin, which has a (1/2)^{10} chance of flipping 10 heads in a row. So, the probability that you picked the 2headed coin, given that you flipped 10 heads in a row, is 0.01/(0.01*1 + 0.99* 0.000977) = 0.911843.
"Anonymous" .
That is true only of the superbig jackpots like Megabucks and Wheel of Fortune. When somebody wins a representative of IGT (the slotmaker) verifies the win is legitimate and then pays the winner. A portion of each bet made goes to a fund to pay the progressive.
"Anonymous" .
The probability of any three of a kind or full house, based on "9/6" jacks or better is 0.085961. To make things easy I’ll divide by 13 to get the probability that the rank of the three of a kind is threes. This is obviously overstating the probability because you will see more in jacks through aces because correct strategy is to hold those cards more often. 0.085961/13 = 0.006612. Tripling the wins for 9 games is like getting 18 free games. 18* 0.006612= 0.119023. To this I would apply some kind of fudge factor to account for the disproportionately fewer three of a kinds in threes, perhaps 75%. 0.119023*0.75 = 0.089267. So whatever your normal return is multiply it by 1.089.
"Anonymous" .
I’ve played in Berlin, Hamburg, and Monte Carlo and the etiquette is more or less the same as in the United States. The main difference that I can think of is I didn’t see much tipping the dealer in any of these locales. Now that I think about the German players seemed to take their gambling very seriously and the casinos, especially in Berlin, were unusually quiet. In Monte Carlo the famed Grand Casino is very stuffy and formal but the Paris Casino and Sun Casino are much more fun and lively, not unlike an American casino. Have a good time!
"Anonymous" .
I never tested the machines in Ontario but did test a machine in Montreal. Quebec casinos are also government owned so the concern should be the same there. The 5cent machine (equivalent to 3 U.S. cents) I played was set to 89.975%. For a small coinage this isn’t too bad and comparable to the Las Vegas Strip. I have played blackjack at the casino in Niagara Falls, as well as Montreal, and the rules were the same as in Atlantic City, resulting in a house edge of 0.41%. I think this goes to show that the government there is not abusing their monopoly but giving the players a decent bet. See my slot machine appendix 3F for more information.