Ask the Wizard #204
Dan from Las Vegas, NV
You’re welcome. I don’t guarantee this as fact, but here is how I believe it works. First, the point at which the jackpot hits is randomly chosen between $50,000 and $100,000. I think each hit point is equally likely.
When the meter crosses the predestined hit point, everybody with a slot card in and playing will win $50 in fee play. To be considered “playing” the player must have his player card inserted, and have made a bet within the last ten seconds. Then, somehow, a machine is chosen at random from all those being actively played to win the Jumbo Jackpot. It does not appear that the bet amount matters, so all qualifying machines have the same probability of being chosen. As long as you can actively play multiple machines, that would muliply your chances of the jackpot by number of machines played, and you would qualify for the free play on all of them.
I would like to thank Bob Dancer for his help with this question.
Note: This answer has been updated in June, 2008, after a rule change to the Jumbo Jackpot.
KRISTIN from NORMAN, OK
This sounds very promising! If this is true, there would be lots of opportunities to count cards. I don’t know if they even allow them, but I think the best opportunities would be on the proposition bets. For example, the “yo” bet, which pays 15 to 1 on an 11, would have a 9.43% house advantage off the top of the deck. However, if no 5 or 6 appears in the first two rolls, the odds swing to a player edge of 5.80%. This same principle would apply to any two-number hop bet.
Brett from Dublin, Ireland
Playing the optimal 9/6 strategy in this game will result in a return of 0.999796. That is an error rate of only 0.02%, which is not worth learning a new strategy over, in my opinion.
Pelle from Malmoe, Sweden
The rate of return is 34.53%, plus 3.08% for every 100,000 Kronor in the jackpot. The breakeven meter is 2,126,825 Kronor.
Hagay
For two numbers, the answer is 1/3, and for n numbers it is 1/(n+1). I posted the solutions to my page of math problems, questions 194 and 195.