Ask the Wizard! (No. 204)
March 4, 2008 column
Hello Wizard, thanks for the great site. Do you have any insight or
knowledge of how the Jumbo Jackpot works at Station Casinos? Is it more likely to hit as the jackpot increases simply due to more chances being drawn, or are there other considerations? — 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 $100,000 and $150,000. I think each hit point is equally likely, but I could be wrong. It has been known to hit under $102,000 and over $147,000.
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 30 seconds. It is not clear if using free play counts as "playing." Then, somehow, a machine is chosen at random from all those being played to win the Jumbo Jackpot. It does not appear that the amount bet matters, so all qualifying machines have the same probability of being chosen. Players have been known to monopolize entire banks of penny machines, betting a penny at least once every 30 seconds, when the jackpot gets close to $150,000. This, indeed, would be a good strategy. However, I’m told once a player was doing this at the Palace Station, and was limited to playing only two machines.
I would like to thank Bob Dancer for his help with this question.
Many Oklahoma casinos are now offering a version of card craps similar to the California game (like CA, OK has some very silly gambling laws). The version I've played uses a 54-card deck, with nine each of ace to six, where the "thrower" calls 1 to 3 burn cards between throws. Suits don't matter. The cards are not returned to the deck, so the odds of the game are not equivalent to a true craps dice game. Obviously, if a 5&4 is the come out roll, this hurts the odds in hitting it again since the same 5 and 4 are not returned to the deck. So, it might make even more sense to play "don't pass" at these tables. In addition, it does allow the player to partially count cards (e.g., laying additional odds on the 4 if very few small cards have been seen). How does not returning the cards to the deck change the odds of playing the pass or the don't pass line?
— 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.
At the Betfair "Zero Lounge" they pay 976 instead of 800 for the royal flush on a 9/6
to bring the expected return up to 100%. This will have some effect on video poker
strategy (slightly favoring plays that have a chance at the royal over ones that
don't). Any chance of publishing an updated strategy for these odds? Cheers.
— 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.
I'm curious about the break even point on the Carribean Stud in
Sweden, since the jackpot is a bit different. The jackpot costs 5 Kronor and pays 200 for a flush, 400 for a full house, 2,000 for 4 of a kind, 20,000 for a straight flush and 100% for a royal flush.
Thank you in advance.
— 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.
Select two random numbers between 0 and 1 (evenly distributed).
Now select the smaller of the two. What is the average of the selection? What about the general case of n numbers?
— 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.
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