Ask The Wizard #180
A nearby casino initiated a special offer on their over/under 13 side bet in blackjack. If your initial two cards are of the same suit, you earn a $5 'action chip.' If both your first two cards and the dealer's up card match in suit, you are awarded a $10 action chip. These action chips can only be used for one bet; although players keep any winnings earned, they always forfeit the action chips. The minimum bet is set at $10, and the game involves six decks of cards. What is the house advantage associated with the over/under 13 bet?
When examining the bonuses, the house edge is lower for the over bet at 6.55%, as I demonstrate in my work. The odds of drawing three suited cards is calculated as 4×combin(78,3)/combin(312,3) = 4×776076/5013320 = 0.060699. The chances that the player's two cards are suited while the dealer's card is not comes from (4×combin(78,2)×234)/(combin(312,2)×310) = 2810808/15039960 = 0.186889. Assuming the action chips hold a value of 49.5% of their stated value, the bonuses come to 0.495×(0.060699×$10 + 0.186889×$5) = $0.76301. The anticipated loss on the over bet stands at $10×0.0655 = $0.655. Thus, each $10 wagered on over 13 yields $0.76301 - $0.655 = 10.8 cents in value. The players enjoy an overall advantage of 1.08% on a $10 over 13 wager. blackjack appendix 8 Hello, Wizard. Here at the casino, players have the option to take over another player’s hand if they decide to fold. However, the new player must place the requisite bet themselves. Should I take over if I can see that one of the dealer’s cards is a low rank, from 2 to J? What would my advantage be? I appreciate your guidance.
Absolutely, you should make that move. Noticing that the dealer is showing a card between 2 and Jack, the chances significantly favor you raising your bet regardless of the situation. This tactic indeed brings about a positive player edge, and I delve into these details further in my published guide. Three Card Poker The standard baseball season consists of 162 games. If a team secures 92 victories, it's likely to proceed to the playoffs. Given that the team has a 55% chance of winning any given match, what are the odds of achieving exactly 92 wins? Additionally, how likely are they to win at least 92 games?
The probability of winning exactly 92 games and losing 70 can be calculated as 162!/(92!×70!)×0.55^(92)×0.45^(70) = 0.056868. To determine the probability of winning at least 92 games, you would need to aggregate this calculation for all outcomes from 92 to 162 wins. The result for achieving 92 or more wins is 0.353239. Gambling 102 .
With numerous online platforms offering betting during halftime, I'm curious to know your thoughts on the concept of 'dutching.' While it can be apparent at times, in what scenarios would you consider placing a bet that contradicts your original game wager at halftime? Moreover, what stake would you consider if you identified a favorable opportunity?
I take it that by 'dutching' you're referring to the strategy of hedging. In the sixth chapter of my work, I discuss the principle 'Thou shalt not hedge thy bets.' The sole exception I would make is if the hedge bet itself presents a positive expected value, or if there is a substantial amount of money at stake.92₧0.4570I'm keen on trying online gaming, but I noticed that the Bodog casino you suggested does not permit Canadian players. Can you explain the reasoning behind this?
Thank you for your interest. Bodog is headquartered in Vancouver, and they have concerns regarding the legality of accepting bets from Canadian residents, prompting them to maintain a clean reputation by refraining from doing so.
How many five-card combinations can you create with cards from precisely two suits? ten commandments of gambling The suits can be distributed in a 4 and 1 manner, or a 3 and 2 split. Let's first examine the 4/1 division. You can choose from 4 suits to comprise the one with four cards, leaving you with 3 choices for the suit holding just 1 card. There are combin(13,4)=715 ways to select 4 ranks from the 13. There are also 13 options for choosing a singular rank. Hence, there are 4×3×715×13=111,540 possible ways to achieve a 4/1 distribution across the two suits. Employing similar reasoning, there are 4×3×combin(13,3)×combin(13,2)=267,696 ways to create a 3/2 split. Therefore, the overall probability comes out to (111540+267696)/combin(52,5) = 14.59%.
To begin with, I must emphasize that I fully understand and concur with your perspective on betting systems. It’s fundamentally clear: if you face a disadvantage in a single hand, that same disadvantage persists in multiple hands, irrespective of the betting amounts. That's a definitive statement. I realize that the longer I engage in gambling activities at a casino, the greater my likelihood of walking away empty-handed.
My inquiry doesn't revolve around securing long-term victories through systems, as we recognize that to be impossible. However, might these systems serve a purpose in 'tailoring' the losing experience? For instance, Player A might prefer that each visit to the casino results in either a small win or a moderate loss (acknowledging he’ll lose a little more often than he wins). In contrast, Player B may favor an opportunity to earn a small profit four out of five visits while experiencing significant losses on one of those trips.
Though both players will ultimately lose money over time, is there a betting strategy that could assist each in achieving their individual objectives?
Indeed, while no betting system can alter the house edge, they can still be leveraged to enhance the odds of meeting specific trip goals. Player A seeks to take minimal risks, meaning he should opt for flat betting. On the other hand, Player B desires a higher chance of winning during his visits, so he should consider increasing his bets following a loss, accepting that this method could lead to greater losses overall. Additionally, though you didn't ask, someone aiming to either minimize losses or hit a big win should ramp up their bets following a win. This approach generally results in losses, but it can also yield occasional substantial wins.
What is the likelihood that two bingo cards share no common numbers? Conversely, what are the odds that they contain all the same numbers?
The chance of two bingo cards having no overlapping numbers is (1-[(combin(11,4)/combin(15,4)] = 1 in 83,414. Conversely, the probability that both cards feature all 24 numbers identically is (1/combin(15,5)).
What are the chances of being dealt an unsuited combination of 2-3-4-5-7? Many thanks, your site is excellent!
Dear Mr. Wizard, I have recently been attempting to determine the probability of achieving a flush in Texas Hold 'Em if I am dealt two suited hole cards. However, my calculations consistently yield 5.8%, which appears to be inaccurate. Your assistance in clarifying this would be greatly appreciated.
The likelihood of forming a flush with three cards matching the suit of your hole cards is computed as (combin(11,3)×combin(39,2))/combin(50,5) = 122265/2598960 = 0.057706. The chance of completing a flush with four additional cards in the same suit is calculated via combin(11,4)×combin(39,1)/combin(50,5) = 2145/2118760 = 0.001012. The probability of achieving a flush with five additional cards of the same suit is given by combin(11,5)/combin(50,5) = 462/2118760 = 0.000218. The likelihood of making a flush with cards in a different suit is 3×combin(13,5)/combin(50,5) = 3861/2118760 = 0.001822. Summing these probabilities results in 0.057706 + 0.001012 + 0.000218 + 0.001822 = 0.060759.
I have been involved with a guy for three years. He claims he doesn't want a commitment but expresses a desire to always be with me. Currently, our relationship revolves solely around physical intimacy. Do you think he will ever want something more, or is it common for a guy to continue this arrangement while seeking other relationships? I feel sad about not receiving more from him. combin (10,5)/combin(15,5))4It appears he’s enjoying the benefits without any commitment, suggesting he may not feel inclined to pursue a more serious relationship.4×(1/combin(15,4)) = 1 in 111,007,923,832,371,000.
Accurate mathematical strategies and insights for various casino games such as blackjack, craps, roulette, and many others are available for players.
Thanks. (45-4)/combin(52,5) = 1020/2598960 = 1 in 2,548.
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A nearby casino is running a special offer on their over/under 13 side wager in blackjack. If your initial two cards are of the same suit, you earn a $5 'action chip.' If both your pair and the dealer's visible card match in suit, you receive a $10 action chip. These action chips can only be used for a single wager; any winnings can be retained, but you lose the chips. The minimum bet starts at $10, utilizing six decks of cards. What is the casino's edge on the over/under 13 bet?