Ask The Wizard #62
I decided to try my luck at deuces wild video poker. I noticed a significant difference in payout odds across various machines. Interestingly, none of the odds aligned with those you had suggested as favorable for players. Could it be that the casinos adjusted the odds after observing your successful strategy to reduce players' advantages? If that's the case, I truly appreciate your insights, Wizard!
While I don't think I'm wholly to blame, it's possible to argue that gaming specialists like myself, particularly Bob Dancer, contributed to the decline of video poker's appeal. Nevertheless, without the guidance of these experts providing valuable strategies, only a select few would understand how to play effectively.
Hello wiz, amazing website. I placed a bet on all contenders for the Kentucky Derby futures market. My odds ended up at 5/2, which I recognize corresponds to approximately $7.00 for a $2.00 wager. However, while tracking the odds, I sensed it hovered closer to 3/1. Could you assist me in calculating the true payout for my $2.00 bet? The overall pool amounted to $577,889, with $125,353 staked on my selection. Thank you for your assistance!
I appreciate the kind words. Let's denote 'c' as the portion taken by the track. If the final odds were 5-2, then:
(577889*(1-c)-125353)/125353 = 2.5
577889*(1-c)-125353=313382.5
577889*(1-c)=438735.5
1-c=0.7592
c=0.2408
Therefore, the track's cut was 24%, which is quite typical for futures wagers. This exemplifies why betting on futures can often be unwise.
In Texas Hold'em, can you shed light on the odds of completing a one-gap or two-gap inside straight by the fifth street, starting from the flop?
For the benefit of my audience, this inquiry revolves around the likelihood of completing a one or two gap inside straight with two more cards remaining, considering there are 47 cards left in the deck. The probability for a one-gap scenario is calculated as 1-combin(43,2)/combin(47,2) = 0.164662. For a two-gap situation, the probability is 4.2/combin(47,2)=0.0148.
What is the increased house edge associated with allowing certain hands to 'ride' in the game Let it Ride?
1) Three unsuited high cards such as A-K-Q and K-Q-J, for instance.
2) Low consecutively connected straight flush cards, like 3-4-5.
3) A hand resembling J-10-7 of diamonds, with a gap of 5.
Thanks Mike, your site continues to impress (I’ll keep saying it every time).
Thank you for your kind remark. To clarify, you should 'let it ride' when you have suited cards like 3-4-5 (three sequential suited cards) and suited 7-10-J (three to a straight flush, featuring two high cards and two gaps). My personal approach aligns with this strategy. Here’s how it impacts your expected return for the other hands, detailed in units. For example, betting three units of $1 on unsuited A-K-Q would result in a cost of 18.62 cents.
Unsuited A-K-Q: -0.186224
Unsuited K-Q-J: -0.104592
In a two-player game of blackjack using a single deck, what are the odds of the dealer obtaining a blackjack?
The number of hands dealt does not affect this calculation. The probability is determined as 2*(4/13)*(8/103) = 0.0478.