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Ask the Wizard #116

I received a promotion from a casino that offers to return half my wager if the dealer gets a blackjack. How would it affect the house edge and would there be any strategy change to play it optimally?

"Anonymous" .

That is a great offer. Assuming six decks the dealer will have a winning blackjack with probability 2*(4/13)*(24/311)*(1-2*(95/310)*(23/309)) = 4.53%. So a half bet every time that happens is worth 2.27%. Assuming a house edge of 0.5% the player advantage would be 1.77%. The strategy is the same as regular blackjack. Too bad I missed that one.

How did you decide to simulate 5,197,920,000 games of 10-hands each for your Texas Hold’em page? Is that number significant?

"Anonymous" .

I set my simulator to run a certain amount of time. Every 10,000 hands the program checks the time and when it has passed the ending time it stops wherever it is.

I know that in video poker, the cards are selected at random -- but are they selected at the instant the button is pushed for both deal and draw?

"Anonymous" .

I think different video poker makers do it different ways. On all at least the draw cards are determined when the player presses the button. I think some also determine the draw cards at this time. Others keep shuffling the remaining 47 cards until the player presses the button to draw the replacement cards.

Many Oklahoma Indian casinos can only use "class 2" poker machines, where "skill" is not allowed to be a factor -- does this mean the hands are somehow predetermined? And would the next hand be the same no matter who played it?

"Anonymous" .

Let me explain what a class 2 machine for the benefit of others. It is a slot machine in which the outcome is determined by the draw of bingo balls. If done well (and it often isn’t) the game will play just like a regular slot machine. I have been to two casinos in Tulsa and the closest thing I found to video poker were not class 2 slots but rather "pull tabs." With pull tabs the player makes his bet, presses a button, 5 cards appear on the screen, and a voucher drops if you won anything. You may take that to the cashier. Although there is a pay table for the 5-card stud hand I do not think the cards are dealt randomly. Rather it is just a visual aid to show you how much you won.

I’m getting ready to go to Vegas and it has been a few years since I have been there, I know Binion’s used to be the best place to play craps, however I understand since the Harrah’s takeover that it no longer holds true. Could you please tell me the best places to play craps are?

"Anonymous" .

I don’t think the takeover had any effect on the craps at Binion’s Horseshoe in Vegas. Although they used to offer 100x odds they ended that long before the federal marshals shut them down earlier this year. The best odds in Vegas can now be found at the Casino Royale (between the Venetian and Harrah’s), which offers 100x odds.

My question is based more from observations on your part and rumors that I’ve heard on my part. If it’s true that Las Vegas dealers, when they are taught at dealer school, learn how to spin and roll the ball the same way then is it true after observing how a dealer spins, one can determine the quadrant of the wheel the ball might land on?

"Anonymous" .

No. Dealers are taught only the basics and nothing nearly that skillful. In fact if a dealer had that control he could simply get an accomplice to bet wherever he planned for the ball to land and they could easily make millions.

Hi Michael. Why is it better to make Come bets with odds than to make Place bets? My math indicates that you make more on a place bet when betting equal units. On 4, to instance, if I were to place $10 I’d win $18. On the come bet I would only win $15 ($5 on the initial bet and $10 on the odds). The other advantage to place bets is that I get to choose which numbers I want to bet and that I win the first time that it is rolled. Am I missing something?

"Anonymous" .

Yes, you are. You are forgetting that a come bet wins on the first roll 22.22% of the time and loses 11.11% of the time. So you are missing the extra value of the first roll of a come bet. However if you had a crystal ball that told you that the first roll would result in a point number then you would be right.

At the Privilege Casino you can’t split aces, but you can double. How would it change the strategy assuming Cryptologic rules 6 decks and how does it increase the house edge?

"Anonymous" .

Not being allowed to split aces increases the house edge by 0.18%. You should only double against a six, otherwise hit.

If two people throw a pair of dice, What is the probability that it is the same number? Is there a formula to figure this out?

"Anonymous" .

Yes. You simply run through all total from 2 to 12 and determine the probability of rolling each twice. So the answer would be (1/36)2+(2/36)2+(3/36)2+(4/36)2+(5/36)2+(6/36)2+(5/36)2+(4/36)2+(3/36)2+(2/36)2+(1/36)2 = 11.27%.

Hi I’ve been playing the Java Let It Ride Game on your website and it’s really fun. Thanks for putting it out on your website. I was wondering, is it fairly accurate as to how a game in a live casino would be played? And does it use a virtual new shuffled deck for each new hand that is dealt?

"Anonymous" .

It plays just like the real thing. The casinos use a shuffling machine, which I understand to be very good. My program shuffles the deck after every hand too.

If a university’s football team has a 10% chance of winning game 1 and a 30% chance of winning game 2, and a 65% chance of losing both games, what are their chances of winning exactly once?

"Anonymous" .

If we assumed the games were independent then the probability of losing both would be 90%*70%=63%. But since you say the probability of losing both is actually 65% (which is more than the 63%), that means the two events are correlated. If the probability of losing both is 65% and just losing game 2 is 70%, then the probability of winning game 1 and losing game 2 must be 5%. Using the same logic the probability of losing game 1 and winning game 2 must be 25%. That only leaves 5% for winning both games. So the probability of winning exactly once is 25%+5% = 30%.