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In an 8 deck baccarat, what is the probability of getting an Ace and an 8 of Diamonds for both the player and the banker in a same deal?

Emi from Manila, Philippines

(82/combin(416,2))* (72/combin(414,2)) = 0.00000043, or 1 in 2308093

Is there a progressive wagering system for baccarat? Is there a specific site for this?

Emi from Manila, Philippines

There are lots of them, and they are all worthless.

Love your site! It’s amazing. My question is regarding on of your answers about "robot players" for online casinos. You said: "These robots take a lot of expertise and time to use but if done properly can turn a computer into a money making machine," and that essentially, this is why casinos sometimes don’t pay out. What I don’t understand is, you’ve insisted, as any statistician will, that no matter what you do, you will end up losing in the long run. So, my question is, how can using a robot make any difference? Who cares, and why would the casinos see this as a problem? Even if they play perfect BS, the house still has the advantage, right? Regarding expected outcomes in BJ, I’ve seen your tables in appendix 4 about standard deviation and found it very helpful. I’m curious to know though, what are the chances of going down as soon as you start playing, and not coming back up to an average of 100% (after factoring in the loses due to the house edge)? What would the chances be over 100, 200...1000...10000...100000, etc., hands be?

And last, could you please help me understand why it’s a "fallacy" that a win becomes more probable after a series of losses? The way I see it, since the expected outcome is an approximately 99.5% return, then if after 1000 hands, you’re at 78%, then, by definition, it would necessitate that the next hand being a win must be more likely to occur. People say that cards "don’t have a memory", but isn’t the natural curve, in essence, its memory??? Please help me understand this point! Thanks a lot.

You’re right, if you used a robot player against an ordinary game you would only lose more. However some casinos do offer games with a player advantage if played properly. Unified Gaming had a blackjack game with an 0.5% player edge for several months, but no longer. Many Real Time Gaming casinos offer a joker poker game with an expected return of 100.18%. Other casinos have promotions in which the player who plays the most hands in a period of time wins a prize, in which a robot player would have a clear advantage. About your second question the bell curve is a forward looking estimate of the sum of many random variables. You can not mix together past and future events. Once an event has happened it is no longer a random variable but a cold hard outcome. If you played 1000 hands of blackjack with a return of 78% then you fell on the tail end of the bell curve during that play. Starting from hand 1001 your results could fall anywhere on a new bell curve. I hope this helps, but it really takes a course in statistics to truly understand.

Wizard -What is the advantage in Three Card Poker in playing two hands versus one, if there is one? Some casinos allow you to play two, playing the first hand before looking at and making a decision on the second hand. Some casinos will only let you play the second hand blind, which I’m sure is no advantage to the player. Thank you

Ruby from Tacoma, USA

Good question. In Stanley Ko’s booklet Mastering Three Card Poker he says that if you had a concealed computer to take maximum advantage of the information then seeing the first hand would lower the house edge from 3.37% to 3.31% on the second hand. Even if you could see all seven hands at the table the house edge would still be 2.32%.

In blackjack, what is the probability of a blackjack?

"Anonymous" .

It depends on the number of decks. If the number of decks is n then the probability is 2*pr(ace)*pr(10) = 2*(1/13)*(16*n/(52*n-1)), which is conveniently about 1 in 21. Here is the exact answer for various numbers of decks.

### Probability of Blackjack

Decks Probability
1 4.827%
2 4.780%
3 4.764%
4 4.757%
5 4.752%
6 4.749%
7 4.747%
8 4.745%

What is the probability that you play ten hands and never obtain a (two-card) 21? Assume the cards are reshuffled after each play?

If the probability of a blackjack is p then the probability of not getting any blackjacks in 10 hands is 1-(1-p)10. For example in a six deck game the answer would be 1- 0.95251110= 0.385251.

What are the odds of getting 3 blackjacks in a row with 1 deck 4 players and one dealer.

Joe P from Parma Heights, USA

I’m going to assume there is never a shuffle between hands. The three other players don’t matter. The answer would be 23*(16/52)*(4/51)*(15/50)*(3/49)*(14/48)*(2/47)= 0.00004401, or about 1 in 22722. If there were a shuffle between hands the probability would increase substantially.

First let me say I think your web site is absolutely outstanding. Thanks. I watched a new craps game being played at Grand Casino, Biloxi, MS. called "Four The Money". To win the shooter must throw the dice 4 times without a 7 coming up. What are the odds of throwing the dice:
4 times without throwing a 7?
3 times without throwing a 7?
2 times without throwing a 7?
1 times without throwing a 7?
How does the math work for this? Thanks