# Ask the Wizard #94

Is it ever a good idea to "double-up" in video poker?

Anonymous

Sure. The double up is one of the few bets with no house edge. For the same reason I advocate the odds in craps I like the double-up in video poker. However if you are playing a game with over 100% return then I recommend against it. Also, if your stomach isn’t strong enough for the extra volatility that comes with the double up feature then you shouldn’t play it. It is also interesting to note that in land casinos cash back does not apply to double-up bets but at Playtech Internet casinos you get 0.1% cash on all bets, including the double-up.

Not that I plan to, but if I did want to card count, wouldn’t it be easier and just as effective to count only the high cards as opposed to tracking the hi’s vs. the lo’s? Assuming you are good at estimating how many cards were left in the shoe.

Anonymous

No, it would be both harder and less powerful to only count the large cards. You may misunderstand how the plus/minus counts work. The high cards count as -1 and the low cards as +1 and you keep a running total as you play. So most cards offset each other and the running count tends to stay close to zero. So you are only keeping track of one number. If you only tracked high cards the total would run high and then you would have to carefully divide it by the number of remaining cards. Even assuming you were good at estimating the number of remaining cards the division would be difficult to do well.

If I have any give number of dice what is the probability that if I roll them all of them at least one will land on a one?

Anonymous

The probability that all the dice will not be a one is (5/6)^{n}. So the probability of at least one 1 is 1-(5/6)^{n}. Let’s take an example of five dice. The answer would be 1-(5/6)^{5} = 59.81%.

Hello, wiz. Really great site. Thanks for all the valuable information that saves us readers countless money on sucker bets. The society needs more people like you to educate us common folks. I come from southern California, and instead of charging 5% on winning, the local casinos here charge commission by each hand you play ($1 for every $100 bet). My question is, what’s the house edge for both banker and player in this case?

Anonymous

Thanks for the compliment. As I state in my pai gow poker section the probability of a banker win is 29.98%, a player win is 28.55%, and a tie is 41.47%. So if you are charged 1% the expected return as banker in a head to head game would be .2998-.2885-0.01 = 0.0043, or a player advantage of 0.43%. As player the expected return is .2855-.2998-0.01 = -0.0243, or a house edge of 2.43%.

You mention the really low house edge in Spanish 21, but I have such a hard time playing the game because I get verbally abused playing your strategy. Not that I question anything you say about gambling, but man you are right about hitting a 17 vs an Ace. The worst slack I got was from a guy who was playing $400 split 8’s vs a 3 and got two 11’s and doubled down and got 19 on both. I hit my 14 vs a 3, busted with a ten. The dealer had 13 and pulled an 8. Now again I question nothing you say, but man when they had to almost call security it was really scary.

Anonymous

I get verbally abused too when I play Spanish 21. When I lived in Baltimore I played it a lot in Atlantic City because the house edge is lower than blackjack there. These idiots doing the abusing don’t understand that removing the tens from the decks makes hitting less dangerous because the probability of busting is less. Don’t bother to try to explain this, the logic won’t make it through their thick skulls. I used to just bite my tongue in these situations but the next time I may not be so nice.

Recently I played at one MG casino (Viper version) High Limit Baccarat and by betting on Banker only, I get an awful result as follows:

Player 44 (64.7%)

banker 19 (27.9%)

Tie 5 (7.4%)

Total 68

What’s the chance of this happening? I appreciate your reply if you can, and hopefully with the formula so that I can calculate it myself next time.

Anonymous

It is bad practice to look back at past play and ask about the odds. Rather, I prefer to state a hypothesis and then gather data to prove or disprove it. However, if we must, I would phrase your question this way: "I played the banker bet 68 times and lost 25.95 units (44-0.95*19). What is the probability of losing this much or more?"

To answer this question we must first find the variance of a single bet on the banker. Here are the possible outcomes and their probabilities, as found in my baccarat section, based on the Microgaming single-deck rules.

Win: 45.96%

Loss: 44.68%

Push: 9.36%

So the variance on a single wager is .4596*(.95)^{2} + .4468*(-1)^{2} +.0936*0^{2} - (-0.010117)^{2}= 0.861468877.

The variance on 68 of these bets is simply 68 times the variance of one bet, or 68*0.861468877= 58.57988361. The standard deviation of the 68 bets is simply the square root of the variance, or 58.57988361^{1/2} = 7.653749644.

The house edge on the banker bet in a single deck game is 1.01%. So over 68 bets you could expect to lose .67 units. You lost 25.95 units, which is 25.28 more than expectations. So your results were 25.28/7.653749644 = 3.30 standard deviations below expectations. You then use a normal distribution table to find the probability of this. Excel has a feature to do this calculation, simply put: =normsdist(-3.30) in any cell and the result is 0.000483424, or 1 in 2069. So this is the probability of losing as much as you did or more. I appreciate that you didn’t make any accusations about foul play. However, if you had, I don’t think this rises to the level to prove anything. It could easily be explained as simple bad luck.