Ask The Wizard #250

I have determined that this rule provides a slight advantage of 0.10% for each additional player present at the table, excluding yourself. Based on my calculations, the house edge stands at 0.48% before factoring in this rule, or the rule allowing for doubling down with three cards. Doubling down on three or more cards is valued at 0.23%. If we make an informed assumption that doubling down on exactly three cards is worth approximately 0.20%, this brings the house edge down to 0.28%. When we incorporate the 2-1 blackjack payout rule, the resulting house edge varies based on the total number of players, including you. blackjack house edge calculator This topic was raised and thoroughly debated in the forum associated with my partner site.

Imagine a scenario where a casino offers a video poker game with a payout exceeding 100%, but players are restricted to playing until they have achieved a single royal flush. Would it be wise to adjust strategies in this case? Wizard of Vegas .

In this scenario, one can maximize profits by employing a strategy aimed at a royal flush win of 450. This adjustment decreases the expected return to 1.007534 and alters the probability of obtaining a royal to 1 in 46,415, leading to an expected profit of 46,415 multiplied by (1.007534-1), resulting in 349.68. The additional 4.6 bet units might not justify the effort required to grasp an alternative strategy.

To pinpoint the optimal royal flush value, you can utilize my approach, gradually reducing the pay for a royal flush until the total return nears 1. At that juncture, it's akin to playing for free until you score the royal flush, at which point you receive a bonus for it. In the full pay deuces wild case, this bonus totals 800 minus 450, equating to 350.

This situation isn't entirely theoretical. There are reports of slot managers actively restricting advantage players from video poker games, and these players often receive a swift warning after achieving a royal flush. video poker calculator There’s a fascinating account involving a player who was able to influence betting odds during a dog race in Australia. Could you clarify how he managed to pull it off?

It’s quite an intriguing tale. The betting terms used in Australia differ somewhat from those in the U.S. To my knowledge, Australians don’t separate bets for place and show but rather use a single place bet. This type of bet pays off on the first two dogs in races featuring seven or fewer competitors, and the first three dogs when there are eight or more. In the specific race we’re discussing, eight dogs competed, with two favored to win. Below, I’ll outline how the winning odds are generally determined in a three-dog place pool in Australia, which contrasts with U.S. methods.

Then, distribute the winnings among the dogs based on the pro-rata amounts relative to the overall pool and the bets placed on each dog. If a dog’s total bets surpass its proportion of the pool, bettors on that dog will receive a refund.

1. Let’s examine a case: if $100,000 is wagered on place bets in an 8-dog race, and the winning dogs attract bets totaling $5,000 on dog A, $10,000 on dog B, and $15,000 on dog C, we would first deduct the 17% cut, leaving us with $83,000. This amount would then be equally distributed among three winning slots, equating to $27,667 for winners of each dog. Winning bets on dog A would receive $27,667 divided by $5,000, which equals 5.53 for 1, prior to any rounding. Similarly, winning bets on dog B would get $27,667 divided by $10,000, or 2.77 for 1, and dog C payouts would be $27,667 divided by $15,000, or 1.84 for 1.
2. Divide the rest into three pools.
3. In this situation, the bettor exploited the system by placing such massive bets that he essentially controlled the odds. For simplicity’s sake, let’s presume he was the sole bettor. The article mentions that he wagered $350,000 on the two favorites and $5,000 on each of the remaining dogs. This resulted in a total pool of $350,000 multiplied by 2 plus $5,000 multiplied by 6, yielding $730,000. After the take-out and split, each winning dog received $201,997. Since $350,000 exceeded $201,997, bets on the two favorites were refunded due to the applicable regulations. However, the share of the pool allocated to the third dog was significant compared to the amount wagered. Thus, the winning odds for the third dog were $201,997 divided by $5,000, equal to 40.4 to 1, resulting in a profit of $5,000 multiplied by 39.4 equating to $197,000. In the end, he took home approximately $170,000, seemingly due to additional wagers placed on the third dog.

It's worth noting that this tactic wouldn't be applicable in the U.S., as our system requires deducting the initial wagers placed on each winning dog from the total show pool, later adding them back after dividing by three. This deduction would result in negative pools for the two favorites, yielding minimal returns of just $0.10 for every $2 bet.

Some gambling literature suggests that the correct approach involves assessing advantage against variance. Nevertheless, you claim that this serves merely as an approximation and that the more precise method is to focus on maximizing the expected logarithm of the bankroll following a bet. My inquiry is regarding the level of error present in the variance approximation.

The advantage-to-variance ratio provides a respectable estimation. Let’s examine an instance, for example. The variance calculation would suggest making a bet of 0.000295 times your bankroll. However, using the precise Kelly criteria, the bet would be adjusted to 0.000345 times the bankroll.

Imagine a scenario where a casino offers a video poker game with a payout exceeding 100%, but players are restricted to playing until they have achieved a single royal flush. Would it be wise to adjust strategies in this case? Wizard of Vegas .

Uncover the Top Online Casinos Available in Your Region full pay deuces wild Calculator for Estimating Lottery Jackpot Ticket Sales