Ask The Wizard #294

At the Spielbank Bad Homburg casino, near Frankfurt Germany, they have a special card in their blackjack game. This card is an ace and replaces one of the regular aces in a six-deck shoe. The card features the image of the casino founder Francois Blanc. If the player gets a blackjack including this card it pays 3 to 1 instead of the usual 3 to 2. It is also an automatic winner for the player, but an ordinary blackjack for the dealer. Six decks are used. How does this rule affect the house edge?


I show the combined value of these rules is worth 0.31%.

The blackjack rules they follow, which seem to be standard at all Speilbank casinos, are:

  • Six decks.
  • Dealer stands on soft 17.
  • No dealer hole card.
  • Double on 9 to 11 only.
  • Double after split allowed.
  • Re-split any pair, including aces, up to four hands.
  • No surrender.
  • Continuous shuffler.

Using my blackjack house edge calculator, I get a house edge before the Francois Blanc card of 0.55%. The Francois Blanc card rule lowers the house to 0.24%!

According to the Spielbank Bad Homburg rule book, the Francois Blanc is exclusive to their casino.

For how to play under these rules, please visit my European basic strategy.

What is the effect of removing a single particular card (2 to ace) from a six-deck shoe in blackjack?


The following table shows the effect on the player's expected return by removing one card from a six-deck shoe, according to whether the dealer hits or stands on a soft 17. For example, if the dealer stands on soft 17, and the burn card is a five, then the house edge drops by 0.146%.

Effect of Removal in Blackjack

Card Stand
Soft 17
Soft 17
2 0.069% 0.071%
2 0.070% 0.072%
3 0.084% 0.089%
4 0.114% 0.122%
5 0.146% 0.148%
6 0.079% 0.085%
7 0.041% 0.038%
8 -0.010% -0.012%
9 -0.041% -0.046%
10 -0.092% -0.098%
Ace -0.101% -0.091%

The table above assumes otherwise "liberal Strip rules," which allow double after split, late surrender, and re-splitting pairs (including aces) up to three times. The table was created using the Composition Dependent Combinatorial Analyzer at

Hello. I am writing because I am a big fan of video poker (although I am not a gambler and play strictly for fun in simulations). I wanted to play around with some simulations and found your page My Methodology for Video Poker Analysis immensely helpful. Although it took several tries to actually wrap my mind around it, I eventually managed to write a little simulator to auto-play games and can check my results against your Video Poker Hand Analyzer page and get the same answers.

But the methodology page only deals with a standard 52-card deck with no wild cards. I have tried to expand the algorithm on my own to allow adding one to four jokers to the deck, but so far the results have been slow and inaccurate. Part of the problem is that wild cards seem to add a lot of complexity as my algorithm substitutes 52 possibilities for the wild cards and furthermore I don't think having duplicate indexes is allowed by the algorithm. I was wondering if the methodology page could be expanded to give tips for dealing with wild card situations in a similarly efficient way that it already deals with the standard deck.

I love the website and want to thank you for the assistance it has provided to this point, as I don't think I could have completed my project without it. Have a great day!


Thank you for all the kind words!

Cycling through all 52 ways to use a wild card will definitely be slow, especially if there are multiple wild cards. Here is how I score a hand with wild cards:

  1. If you have a natural royal flush, then score the hand as such.
  2. Otherwise, count the number of wild cards in the hand.
  3. Then score the hand according to this number and the value of the other cards.

For example, suppose you know you have two wild cards. You could use this pseudo-code to score the hand based on the other three cards:

  • If the three cards are suited and the lowest one is at least a ten, then you have a wild royal.
  • Otherwise, if the three cards are suited and the difference between the highest and smallest is four or less, then you have a straight flush.
  • Otherwise, if the three cards are suited, the highest is an ace, and the second highest is a five or less, then you also have a straight flush (A2345).
  • Otherwise, if you have a pair, then you have a four of a kind.
  • Otherwise, if the three cards are suited, you have a flush.
  • Otherwise, if the difference between the highest and smallest is four or less, then you have a straight.
  • Otherwise, if the highest is an ace, and the second highest is a five or less, then you also have a straight (A2345).
  • Otherwise, you have a pair.

You already have code for no wilds, so you'll have to write separate code for one to either four or five wilds, depending on whether you may wish to analyze Deuces and Joker Wild.

Suppose you have a deck of ten red cards and ten black cards. After a random shuffle, you draw cards without replacement. What is the probability of drawing all ten black cards before five red cards?


That is the same as asking what is the probability that 14 random cards contain all 10 black cards. There are combin(10,4)=210 ways to choose 4 red cards out of the 10 in the deck. There is of course only one way to choose all ten black cards. There are combin(20,14)=38,760 ways to choose 14 cards out of 20. So the answer is 210/38,760=0.005418, or 1 in 184.57.