Ask The Wizard #204
Greetings Wizard, I appreciate the excellent website. Do you have any information regarding how the Jumbo Jackpot functions at Station Casinos? Are the chances of winning increased when the jackpot amount rises due to more draws, or are there additional factors to consider?
Thank you! While I cannot confirm this definitively, here's my understanding of the situation. The occurrence of the jackpot is determined randomly within a range of $50,000 to $100,000. I believe that all values within this range have an equal chance of selection.
When the jackpot meter reaches the predetermined threshold, every player with an active slot card will receive $50 in free play. To qualify as 'playing,' a player must have their card inserted and have placed a bet in the last ten seconds. At this point, a machine is randomly selected from all currently active machines to win the Jumbo Jackpot. The size of the bet does not seem to influence selection, so each qualifying machine shares the same odds. Therefore, if a player actively engages with multiple machines, they can increase their chances of winning the jackpot based on the number of machines they are using, while also qualifying for free play on all of those machines.
I would like to thank Bob Dancer for his help with this question.
Note: This information was revised in June 2008 following a change in the Jumbo Jackpot rules.
Numerous casinos in Oklahoma now provide a variant of card craps resembling the game from California (like California, Oklahoma has its share of unusual gambling regulations). In the version I've experienced, a 54-card deck is employed, featuring nine cards each from ace to six, and the 'thrower' calls for 1 to 3 burn cards during throws. Suits are irrelevant in this game. Since cards are not returned to the deck, the odds differ from those of traditional craps. For instance, if a 5 and 4 is rolled initially, it reduces the likelihood of hitting those numbers again since they aren’t available to be rolled again. Thus, opting for the 'don’t pass' line may be a wiser choice at these tables. Moreover, players can partially track the cards (e.g., increasing odds on the 4 if only a few small cards have been dealt). How does the non-return of cards to the deck alter the odds for placing bets on the pass or don’t pass line?
This seems very promising! If that's accurate, there will be ample chances to count cards. I’m not sure if the casinos permit this practice, but opportunities are likely to be most significant on proposition bets. For example, the 'yo' bet, which offers 15 to 1 odds on an 11, starts off with a house edge of 9.43%. Yet, if neither a 5 nor a 6 is rolled in the first two turns, the odds shift to give the player a 5.80% advantage. This logic extends similarly to any bet involving two-number hops.
At the Betfair 'Zero Lounge,' they provide a payout of 976 instead of 800 for a royal flush on a 9/6 machine, which improves the expected return to 100%. This adjustment could influence video poker strategies, slightly favoring plays with potential royal flush outcomes over those that lack them. Could you consider releasing an updated strategy reflecting these odds? Thanks in advance.
Utilizing the optimal 9/6 strategy in this game yields a return of 0.999796, indicating a mere 0.02% error rate. I believe this is too insignificant to warrant learning an entirely new strategy.
I'm interested in knowing the break-even point for the Carribean Stud lottery in Sweden, as their jackpot structure differs slightly. The ticket cost is 5 Kronor, and it pays out 200 for a flush, 400 for a full house, 2,000 for four of a kind, 20,000 for a straight flush, and covers 100% for a royal flush. Thank you for your assistance!
The return rate stands at 34.53%, plus an additional 3.08% for every 100,000 Kronor in the jackpot. The break-even threshold sits at 2,126,825 Kronor.
Choose two random numbers ranging between 0 and 1, ensuring that they are evenly distributed. Now, pick the smaller of the two. What is the average of this selection? How does this apply in the general case of selecting n numbers?
For a pair of numbers, the resulting average is 1/3, and for n numbers, it equals 1/(n+1). I've posted the solutions on my page dedicated to math problems , questions 194 and 195.