# Ask the Wizard #102

Hi Wizard - thanks for a great site! The basic Three Card Poker strategy is to play Q-6-4 or better to minimize the house edge. But what is the best strategy to minimize the element of risk (given ’5-4-1’ rules)? Your site reads that playing Q-6-3 loses 1.00255 units; but then betting that hand will dilute the element of risk, since the bet is suffering a house advantage of only 0.255% (which is less than the 2.01% element of risk).

Anonymous

You’re welcome. To minimize the element of risk you should raise until the point where the expected value of raising is less than the element of risk for the entire game. The expected loss by raising on Q-6-2 is 1.24% and on Q-5-4 is 2.10% to 2.15% (depending on how the cards are suited). So to minimize the element of risk you should raise on Q-6-2 or higher.

In practical terms, does the casino make more money on a bet with a high standard deviation, than a bet with a low standard deviation, even if the two bets have the same edge? Basically, I figure the probability of player bankruptcy is higher with a higher standard deviation. If the house already has all my money, I can’t play the next hand that might be a winner.

Anonymous

Assuming the player plays the same number of hands regardless of results then the casino would make the same amount of money either way, over the long run. Yet if the player will quit early if he reaches a certain loss point then he will play less on average and consequently the casino will make less money. It sounds paradoxical but if you quit playing when you go broke then you will lose more at a low volatility game, because there is a smaller chance of ruin and thus the house edge will grind you down longer. So by increasing your probability of ruin your expected loss actually goes down. For example if player A bets his entire $100 bankroll by betting on red in one spin of roulette then his expected loss is only $5.26. If player B bets $1 at a time for 8 hours on red in roulette his expected loss is 60*8*5.26% = $25.26 (assuming 60 bets per hour). So although player A has a much higher probability of ruin his expected loss is much less. This lesson is especially applicable to Internet bonus playing. If you get the bonus up front I recommend betting everything in one hand to start. By sometimes going broke before completing the play requirement you expose yourself to the house edge less and thus save time and lose less playing over the long run.

Dear Wizard, I was wondering if it reduces the expected loss to a player to play multiple lines in multi-play video poker as opposed to one line at a time (where all lines share the first 5 cards and independently draw the non-held cards). It seems that on multi lines, when you get lucky and the first five cards are a winner (such as a natural flush draw), it guarantees that EVERY hand of multi-play will also be a winner. However, no matter how bad the first five cards are, it does not guarantee that EVERY hand will be a loser on the draw because every hand will have an independent draw. It seems like there must be some difference in odds playing 10 lines sharing first five cards vs playing 10 independent hands?? I enjoy your website.

Kevin

The odds are exactly the same on a one line, 10 line, and n-line video poker machine. When you get a trash hand in 100-play you can expect to get about 36% of your original bet back. In 10-play it is still 36% but there is more volatility. In 1 play it is still 36% but you can get lucky and get a high paying hand on the draw. In other words you are more likely to hit it big on the draw in single play, but at the expense of lots more non-paying hands.

First, I’d like to say I’ve thoroughly enjoyed your site and the information found therein. My question regards multi-play video poker games I’ve found in most of the casinos in Tunica, Mississippi. These games allow you to play 3, 5, 10, 50, or 100 hands of video poker at a time keeping the cards you receive on your initial deal and receiving random cards for each hand on the draw. Should my betting strategy vary from the strategies you’ve set forth in your video poker section? How (if at all) does this affect the expected return per play? Thanks for your time.

Anonymous

Thanks for the kind words. Assuming the pay table is the same the strategy and expected return are exactly the same. Be warned that multi-play games tend to have worse pay tables than single line games.

Why is there so much written about card counting strategies in BJ, if they are not as rewarding "as television and the movies make it out to be"? This seems highly illogical to me and is really puzzling.

Anonymous

There is a lot written about card counting simply because the books sell. I suspect many people watch a movie like Rain Man and get interested in card counting. Then they buy a book and realize it is too hard or give it a try and get discouraged because they lose. Only the most patient, devoted, and well financed players stick with it.

I read somewhere that online casinos may shuffle the deck after every turn and that card counting then would be of no value. How can I find out for sure if a casino is doing this?

Anonymous

You could test for how often the same card appears in two consecutive hands. Considering only the four initial cards dealt and assuming all four are different the expected number of those cards seen in the next hand is 16/52 = 0.307692, assuming the cards are shuffled after every hand. If you see fewer repeats than would argue that the cards are not shuffled after every hand. If the cards are not shuffled between hands the expected number of repeats seen would be 4*(4d-1)/(52d-4), where d is the number of decks. It is easier, but less reliable, to just ask customer support.

Do Internet casinos deliberately let you win in fun mode?

Anonymous

I have heard this allegation a lot of times but have yet to see any proof either way.

Where are the single deck games in Vegas that still pay 3 to 2?

Anonymous

With the demise of Binion’s Horseshoe the number of true single deck games in Vegas has fallen by about 75%. Although it isn’t a priority of mine to keep up to date on this some that I know of are the Fiesta Rancho, Golden Gate, El Cortez, and the Western. Beware of single deck games that only pay even money or 6 to 5 on a blackjack, you are much better off at a shoe game that does pay 3 to 2.

If I throw 36 dice what is the probability of getting at least one six?

Anonymous

1-(5/6)^{36} = 99.86%

If I kept throwing and removed all the sixes each time, how would I predict the theoretical number of dice remaining after a particular number of throws?

Anonymous

Each roll the expectation is that 5/6 of the dice will remain. So the expected number of dice remaining after n throws would be 36*(5/6)^{n}. For example after 10 throws you would have 5.81 dice left, on average.