# Ask the Wizard #191

Jim from Las Vegas

In ten years of running this site I steadfastly denied the myth that bad players cause other players to lose in blackjack. However, you are the lucky 1000th person to ask, so I took the trouble to prove it by random simulation. The rules I put in are the standard liberal Vegas Strip rules as follows.

6 decks

Dealer stands on soft 17

Double on any first two cards allowed

Double after split allowed

Late surrender allowed

Player may re-split to four hands, including aces

Cut card used

First, I had both players follow correct total-dependent basic strategy. Over almost 1.6 billion rounds, the loss of the first player to act was 0.289%, and the second player to act of 0.288%.

Second, I had the first player follow the same correct strategy, and the second player follow the same correct strategy except:

Always hit 12 to 16

Always double 9 to 11

Split any pair

Never surrender

Never soft double

In a simulation of 1.05 billion hands the loss of the first player was 0.282%, and the second player was 11.260%. So the house edge of the basic strategy playing first player was almost the same, regardless of whether the second player played correctly or wildly incorrectly. I hope this puts and end the third baseman myth, but I doubt it. As I have said many times, the more ridiculous a belief is, the more tenaciously it tends to be held.

Scott F. from Philadelphia, PA

In a six-deck shoe, the probability of such a blackjack is 2 × (6/312) × (12/311) = 0.001484. I’m going to assume that if the dealer has a blackjack also, then the hand is a push. That said, the probability of the dealer not getting blackjack, given that the player did, is 1 - 2 × (23/310) × (95/309) = 0.954379. So the probability of winning such a blackjack is 0.001484 × 0.954379 = 0.001416. The value of an additional 8.5 units whenever that happened is worth 8.5* 0.001416 = 1.2039%. Assuming otherwise liberal Vegas Strip rules, with a house edge of 0.28%, the __player__ edge with the 10 to 1 rule would be 0.92%.

Dave S. from New Haven

I assume that you assume the probability of a royal is 1 in 40,000. Playing 4,000 hands per session the expected number of royals per session is 0.1. A very close appoximation for the probbility of zero royals per session is e^{-0.1} = 90.48%. The reason it is not 90% is because sometimes you will get more than one royal per session. The expected number of royals in 50 sessions is 0.1 × 50 = 5. The probability of zero royals in 50 sessions can be closely approximated at e^{-5} = 0.67%. The exact probability is (39,999/40,000)^(200,000) = 0.67%, as well.

Second, I hear that pai gow players are rated much lower than other table games per bet due to the slower pace of the game, true? Do you have an educated guess what the average bet in pai gow would have to be to equal the comp privileges of a $25/bet Blackjack player?

Uncle Mo from Parker

Your timing is good with this question. The summer of 2007 shall be referred to as my "summer of pai gow," because I am devoting a great deal of analysis to the game. About banking, yes, you can change your mind about banking, if a player makes a bet larger than you are comfortable with. It is a rule that you must have enough chips in front of you to cover all the action. My May 5, 2007 column shows that pai gow is rated at 30 hands per hour, at least by the casino that gave me the figures. If you multiply the assumed house edge and hands per hour in that table, you can see the blackjack is rated at 0.525 bet units per hour, and pai gow at 0.495, so blackjack is only slightly better at the same bet size. Although the expected loss is greater in pai gow, the standard deviation is much less. This makes pai gow a great game if you are playing for a rating, and wish to minimize risk.

Ray from Egg Harbor Township

You’re welcome, thanks for the compliment. Without knowing anything about the probabilities, if those were the payoffs, then there would exist a player advantage on at least one bet. The way you can tell is to take the sum of 1/(1+x), where x is what the bet pays on a "to one" basis, over all the bets. If this sum is less than 1, then at least one bet has a player edge. In this case, according to your odds, this sum would be 1/2.5 + 1/3 + 1/4 = 0.9833. This trick may come in handy, for example, if you see an amateur putting up sports betting futures.

What is probably the case here is that six cards pays 2 to 1. Based on that assumption, and six decks, the house edge is 5.27% on four cards, 8.94% on five cards, and 4.74% on six cards. For more information see my baccarat appendix 5.