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Ask the Wizard #176

What is the house edge of gold pontoon on the new Microgaming platform?

Steve from KC

There are two significant rules changes to the Microgaming pontoon game. First, the dealer stands on soft 17, which is worth 0.47% to the player. Second, there is no drawing to or resplitting of aces, which is worth 0.49% to the dealer. So these two rules almost offset each other. Overall the house edge is 0.39%, with correct strategy, which is indicated in my Microgaming section.

In regards to your Wizard Ace Five Count, what effects does real casino play versus your simulation have on the bank role needed to have a reasonable risk of ruin. More specifically, is your simulation played head to head with a dealer, where when the count is high you are betting the higher bets more times per shoe than at a real table where 6 players are using up cards when the count is high, resulting in fewer chances to bet when the count is high. Thank you.

Rich from Greenwood, IN

Measured in terms of the house edge, the number of other players doesn’t matter. Yes, they will eat up cards when the count is good, but they will also do so when the count is bad. In the long run it doesn’t matter. However, in terms of expected wins per hour, when you have an advantage, it helps to have fewer players, because you will play more hands in the same amount of time.

First, great site. Second, sorry if this is in the wrong category, I did my best. Finally, Appendix 3B mentions a program used to come to all of the composition dependant strategy. I was wondering if this program was available for purchase or if you could just give me the formula used to determine what any given hand’s optimal play would be given the removal of any number of other cards (lise the perfect play calculator at Thanks.

Chuck from Spring, TX

Thanks. My program is not very user friendly. I would recommend you use the blackjack calculator at, which will give you perfect advice for any situation and deck composition.

I have looked all over the net about the probability of at least getting a pair by the river card in hold’em if you have two different cards dealt to you. I have tried to work it out using a probability tree but my answer seems too high. Also on the net I have read different answers some suggesting it is around 1/3 or 2/5 or 1/2. What is the probability of at least pairing and is it possible to work this out using a probability tree? Your help will be much appreciated thanks.

Nathan S. from New Plymouth

For those not familiar with the hold 'em terminology, you are asking for the probability of at least a pair in six cards, given that the first two are (the hole cards) of different ranks. I hope you'll forgive me if I just do the probability of getting exactly a pair, including hands that also form a straight or flush.

The number of ways to pair one of your hole cards is six (2 hole cards * 3 suits remaining). The other three cards must all be of different ranks from the 11 left. There are combin(11,3)=165 ways to choose 3 ranks out of 11. For each of these there are four suits to choose from. So the number of ways to pair one of your hole cards is 6*165*43=63,360.

Now let's look at the number of ways to get a pair outside of the two hole cards. There are 11 ranks to choose from for the pair. Once the pair is chosen there are combin(4,2)=6 ways to choose 2 suits out of 4. For the other two cards there are combin(10,2)=45 ways to choose 2 ranks out of the 10 fully intact ranks left. For both of those ranks there are 4 possible suits. So the total combinations for a pair, not including the hole cards, is 11*6*45*42=47,520.

The total number of ways to choose 4 cards out of the 50 left in the deck is combin(50,4)=230,300. So the probability of getting exactly a pair in six cards is (63,360+47,520)/230,300 = 48.15%.

I have been watching High Stakes Poker on the Game Show Network and there are two terms that have not been explained by the commentators. One is "Straddle" and the other is "Props". Could you please explain what these terms mean in the context of the Poker Game that is being played? Thanks very Much. By the way, Gambling 101 is a great book. Very nice Job!

Edward from Baltimore, MD

A straddle, often called a “live straddle”, is when the player after the big blind makes a raise before looking at his cards. For example in a $3/$6 game the large blind would be $3 so the straddle would be $6. I asked my friend Jason about the reason for this. He said, “The reason some people do this is to stimulate action in a "tight" game. The person who straddles also has the option to raise after the big blind acts. Card rooms like this and allow it because is almost assures a larger pot and therefore more rake.”

There are two uses of the term "props" in poker. First, a Prop Player is one who is paid by the poker room an hourly wage to play. The reason for this is to keep a certain minimum number of players at each table. For more information this questions is answered in much more depth on Second, a Prop Bet is a side bet made among the players, often on the flop. There is an article about it

We are in a disagreement between workers. there is a bar down the street that has a shake a day. which is you must throw five dice at once and all five must end up being the same "like yahtzee" but he gives you three chance at it. but you must pick up all the dice all three times. so the questions is what’s the odds to do it in one shake and what’s the odds to do it in the three shakes allowed. Thanks , if you already answered this before i am sorry but i couldt find it.

Dan and co workers at maple island from Forest Lake

The probability of a five-of-a-kind on one throw is 6*(1/6)5 = 1/1,296. This is because there are six different five-of-a-kinds (one to six) and the probability each die will be that number is (1/6). The probability of not getting a five-of-a-kind is 1-(1/1,296)=1,295/1,296. The probability of going three attempts without a three of a kind is (1,295/1,296)3=99.77%. So the probability of getting at least one five-of-a-kind in three tries is 100%-99.77% = 0.23%.

What advantage (%) would a player have if the 10’s were used in a stand on all 17, 8-deck spanish 21 game?

Kevin from Toronto

Keeping all the tens in the deck is worth 1.89% to the player. The house edge under those rules is normally 0.40%. So with all the tens in the shoe, the player edge would be 1.89%-0.40% = 1.49%.


Follow up: In the February 1, 2006 column, a reader was mad because of the short expiration dates on slot machine tickets. I took her side, saying they shouldn’t expire at all. Many readers took me to task, saying that casinos routinly honor expired tickets. So I then did an experiment in which I collected $2 tickets up and down the Strip. After they expired I went to cash them in all of them were honored. So I have amended my answer and I offer my apologies to the casinos for my earlier harsh words.