# Ask the Wizard #83

How does multiple doubling up in video poker cut the house edge?

Anonymous

This question refers to advice I gave in my newsletter, suggesting video poker players take the double up option when playing for a bonus at a Playtech casino. Doubling up does not lower the house edge of video poker, but does introduce another bet with zero house edge. For the same reason you should take or lay odds in craps it cuts the overall house edge playing video poker if you take the double up feature. Playtech does count double up bets towards the play requirement, so it gets the player to the bonus with less expected loss. So unless you are playing a machine with a return of over 100% and you don’t mind extra volatility then I recommend accepting the double up option.

There is an online sport book that is offering a bet in which you are assigned a random horse (to win) and you receive a guaranteed payout of 15.3 for 1. There are 17 total horses listed. How would you calculate the house advantage on this bet? Does it matter what the odds are on the horses? If so, they are below. Thank you, and I apologize if this is a stupid question.

Anonymous

Assuming you had an equal chance at getting each horse then the probability of winning would be 1 in 17, regardless of how the odds on each horse were distributed. The player’s expected return can be expressed as (1/17)*15.3 + (16/17)*-1 = -0.0412. In other words the house edge is 4.12%. However there is another simple formula you can use. If a = actual odds paid and f = fair odds for bet then the house edge is (f-a)/(f+1). In this case the fair odds are 16 to 1. So the house edge is (16-15.3)/(16+1) = 0.7/17 = 4.12%.

For those games which returns higher than 100%, have you thought of writing computer programs to play against them? I saw this idea mentioned somewhere else so I bet you must have known it. What is the problem with it then? No big return? Too many people have already done it? Casinos can easily find out and bar you?

Anonymous

I addressed robot players in the September 20, 2001 column. Assuming you could create a robot then it should do well against a game with over 100% return. However I would recommend programming it to play at a human speed and reasonable sitting times. Some online casinos have been known to blame players for using robots even when they weren’t, as an excuse not to pay in my opinion.

I know from your chart that PUT bets w/10x odds beats ALL place/buy bets. Lets talk about PUT vs. COME. A ’put’ on the 6/8 with 10x odds has a house edge of .83% while a come bet w/10x odds is .18%. You’re giving the house an additional 0.65% (.83-.18) edge by using a PUT vs. a COME bet. In return, you have the potential of getting more $ since you win on every 6/8 hit vs. winning every other time the # is made if you did a COME. That’s my theory at least. Now in the long run, who comes out ahead with more $? I say the guy w/the PUT bet comes out with more $ since you are hitting twice as often. With a come bet, you have make the #, then wait till it hits again, cutting your hits in half. If there’s something wrong w/my logic, please let me know!

Anonymous

I get challenged a lot on my support for come bets with odds. Those against come bets are always quick to say a number has to hit twice with a come bet and only once with a place or put bet. That is not a good way to look at it. First of all with come bets you have a 8/36=22.22% chance of winning on the first roll as opposed to only 11.11% of losing. You get no such advantage on the first roll of a place or bet bet. Even if you do roll a point number it can be any number. In other words there are six point numbers that can win on a come bet, and only one on a place or put bet. Ultimately the reason the come bet with odds beats place or put bets is the house edge is less.

Dear wiz, How do you calculate the probability of getting three sevens, three colored sevens, and three suited sevens in blackjack?

Geoff

Let’s assume six decks of cards and the player always takes a third card (whether by hitting or splitting). The number of ways to draw 3 suited sevens is the number of suits (4) times the number of ways to choose 3 out of 6 sevens of that suit in the shoe. In other words 4×combin(6,3)=4×20=80. The number of ways to draw 3 colored sevens, including 3 suited sevens, is the number of colors times the number of ways to pick 3 out of the 12 sevens in the shoe of that color, or 2×combin(12,3)=2×220=440. The number of ways to draw any 3 sevens, including 3 colored and suited sevens, is the number of ways to pick 3 cards out of the 24 sevens in the shoe, or combin(24,3)=2024. The total number of combinations for any 3 cards out of 312 is combin(312,3)=5013320. So the probability of 3 suited sevens is 80/5013320=0.000015957. The probability of three colored, but not suited, sevens is (440-80)/5013320=0.0000718. The probability of three sevens of mixed colors is (2024-440)/5013320= 0.00031596.

Good job and well done. The question: I notice from your May 5, 2003 Column that you actually CALCULATE your blackjack odds. I am a bit surprised that you were not using your computer to SIMULATE the results. Or is this a stupid question, i.e., the computer will take a million years to do the job?

Anonymous

Yes, I calculate blackjack odds using a combinatorial approach, analyzing every possible ways the player and dealer cards can come out, taking the greatest expected value at every decision point. This is harder to program than a simulation but I feel is more elegant and a nice challenge in recursive programming. However I still respect my peers to do simulations. With today’s computers it doesn’t take long to run off a billion bets, which gets very close to the optimal strategy return.