Gambling Guide
Ask the Wizard #116
"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.
"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.
"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.
"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.
"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.
"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.
"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.
"Anonymous" .
Not being allowed to split aces increases the house edge by 0.18%. You should only double against a six, otherwise hit.
"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%.
"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.
"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%.
