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Ask The Wizard #73

I wanted to confirm that the optimal move in the Microgaming single deck game when holding A,A against an Ace is indeed to split, as you've indicated on your card. However, I have a feeling that hitting might be the better choice, especially when playing with four decks. With so many tens still in play, it seems like I could end up losing double my stake quite often. It's not that I doubt your expertise, but I’d appreciate your confirmation. Do you happen to have the expected value (EV) for hitting compared to splitting with this hand? basic strategy I've received inquiries about this matter multiple times, and I stand firm in my belief that even though committing the entire bet may seem risky, the opportunity to draw to split aces compensates for the European rule of not peeking at the dealer's cards. Therefore, I maintain that splitting is the superior strategy. When considering a single deck, the expected value of hitting is approximately -0.532849, while for splitting it's around -0.223277. This means that splitting is better by roughly 31% of a unit. The case is similar for the four-deck game, which is actually not advisable for Microgaming players since a one-deck alternative with identical rules is accessible.

William from Pittsburgh, USA

Hey there. You mentioned that all betting systems are destined to fail. If I place a bet of one unit on the numbers 1 to 12 and two units on the numbers 13 to 24 in roulette, wouldn't that give me a 66.66% opportunity to either break even or win?

Not exactly. Your chances would be 12 out of 38 to win 3 units, 12 out of 38 to break even, and 14 out of 38 to lose 3 units. The expected value can be calculated as follows: [(12/38)*3 + (12/38)*0 + (14/38)*-3]/3 = (-6/38)/3 = -2/38 = -5.26%. This calculation is valid for any combination of bets, as long as you steer clear of the troublesome five-number combo (0, 00, 1, 2, 3). If you're placing a single bet and want to enhance your chances of winning, it’s best to wager evenly across 35 of the numbers, which gives you approximately a 92.11% chance of earning 1 unit while facing a 7.89% chance of losing 35 units.

Atle from Porsgrunn, Norway

What fundamental strategy should one follow for the new Blackjack Switch game? Additionally, what is the house edge associated with it?

Waurkelter B. from Mashpee, Massachusetts

Everyone keeps asking me about this. Currently, I haven't been able to calculate the odds myself. Until I do, you’re on your own regarding this inquiry. Since this website isn't generating much income, I have to prioritize my paid consulting work.

In a scenario of 500 hands of baccarat where I'm only wagering on the banker, what percentage of the time can the player expect to win more than 46% of their decisions? Thank you.

First, I’ll assume you want me to disregard ties for this calculation. From my research, the likelihood of a player victory stands at 49.32%, excluding tied outcomes. For this scenario, we’ll apply the normal approximation to the binomial distribution. The expected number of player victories would thus be 500 multiplied by 0.4932, resulting in approximately 246.58. If we consider the 46% threshold for decisions, that translates to 230. The standard deviation is computed using the formula (500*(0.4932)*(1-0.4932)).

Haig

When playing Texas Hold'em poker, what is the likelihood of being dealt pocket aces? Also, what are the odds of receiving pocket aces in consecutive deals? baccarat The total number of ways to choose 2 cards from a standard deck of 52 is calculated as 52*51/2, giving a total of 1326 combinations. For aces, there are 4 available, and the ways to arrange 2 aces from these is 4*3/2, which equals 6 combinations. Therefore, the odds of being dealt pocket aces is 6 out of 1326, simplifying to 1/221. The chance of this happening two times in succession would be (1/221) squared.1/2= 11.18. So..

pr(player wins > 230) =
pr(player wins-246.58 > 230-246.58) =
1-pr(player wins-246.58 <= 230-246.58) =
1-pr(player wins-246.58+0.5 <= 230-246.58+0.5) =
1-pr((player wins-246.58+0.5)/11.18) <= (230-246.58+0.5)/11.18) =
1-Z(-1.44) =
1-0.075145503 =
0.924854497

So the answer is 92.49%.

I frequently enjoy playing Double-Down Stud video games when I'm in Las Vegas. I'm curious about how the overall payout of these machines compares to the different types of video poker available. The machines located at the Riviera (one of the few places I have encountered such machines) offer the following payouts:

Adam from Redding, USA

What is the expected return for optimal play on these machines?2= 1 in 48,841.

As a side note, these machines accrue slot points at the same rate as video poker (which is half the rate of standard slot machines), but I feel that I accumulate points more quickly due to the apparent double-down opportunities. Does this enhancement indirectly boost the expected return?

One-coin pay table
--------------------
Pair, 6-10’s 1
Pair, J-A’s 2
2 Pair 3
3 of a Kind 4
Straight 6
Flush 9
Full House 12
4 of a Kind 50
Straight-Flush 200
Royal Flush 1,000

2-5 coins, multiply 1-coin payout

5-coin Royal pays 20,000 coins

I really appreciate your website! Thank you!

With this payout table, the house edge is 2.10%, while the expected return is 1.68%. The strategy remains consistent with what I've indicated in my previous section. Factoring the increase for cash back feels like receiving an extra 25% on top of the original bet.

Providing mathematically sound strategies and insights for casino games like blackjack, craps, roulette, and many others available for play.

Tim from Newburgh, New York

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