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# Gaming Guru  ### Double Bonus - Break Up Those Pairs?

2 October 1999

The Video Poker game known as Double Bonus is certainly popular and one can easily see why.  The payout for Quads is at least as good as that for a Straight Flush and in some cases a lot better; they are fun to hit.  There is a down side to the game, however, since the payout on Two Pair is 1 for 1 rather than 2 for 1 as in regular Jacks or Better.  Nevertheless, when played using optimum strategy, the 10 - 7 version of the game returns 100.17% to the player in the long run.  Those last three words are important; in the short run the game is very streaky and without a good sprinkling of the larger payouts, the 1 for 1 payout on Two Pair can eat into your bankroll rather quickly.  So can poor play.  Now I am not going to preach to you about learning a professional strategy; I'm just a casual player myself and I do make my share of mistakes.  Suffice it to say, however, that it is better to play well than to play poorly, and there is plenty of professional help available at this web site if you want to achieve the former; it will indeed pay off for you to work at improving your game.  Rather, my purpose here is to teach you some mathematics, show you how it relates to gambling games, and explain how an understanding of the mathematical principles underlying a game can prevent you from embracing fallacies.  All of which brings me to the subject mentioned in the title.

Last summer I was playing Blackjack at Binion's Horseshoe in downtown Las Vegas.  Binion's is a friendly joint and I was chatting with the dealer as I played.  Somewhere in the conversation I had occasion to mention that I do mathematical analysis for gaming developers.  The dealer immediately had a question for me.  He told me that he liked to play Double Bonus Video Poker.  Then he went on to say that whenever he gets Two Pair and one of them is high, he tosses the other pair and keeps the high pair.  His reasoning was something akin to this:  "If I keep only the high pair, I get exactly the same payout as I would if I kept both pairs and, in addition, I get a chance at Four of a Kind, which has a big payout in this game."  That is a reasonable sounding argument.  I might even buy it were it not for the fact that I know to be wary of my own intuition (and anyone else's).  The dealer went on to say that a friend of his told him that tossing the lower pair was the wrong play; that he should always keep both pairs.  "Who is right?"  Let's have a look.

We have already seen in these articles that it is easy to make a mistake when evaluating games.  Chuck from the Chuck-a-Luck game added probabilities using non disjoint events, the New York State Lottery was rather careless with their Quick Draw promotion, and the Birthday Problem taught us to be wary of our intuition.  As I pointed out in my July 1999 article, it is imperative when analyzing a gambling situation that we ask the right question.  In the case at hand, the question is: "Which course of action, holding only the high pair or holding both pairs, will produce the higher long run return to the player?"

Now, before I address the above question, let me confess to you that there are computer programs available that will instantly answer it.  Bob Dancer, one of the writers for Rolling Good TimesTM, markets a dandy program called WinPokerTM that will do the job.  Video Poker TutorTM, which was written by and is maintained by Gary Catlin (no relation), is another such program.  My purpose here, however, is not just to answer the question at hand but to expose you to mathematical ideas and techniques that we can use in situations where there are no computer programs available.  So we are going to address the question using only mathematical ideas and, hopefully, some common sense.  Besides, its challenging and fun.

We already know most of the ideas necessary to go to work.  We will need to refer to the Keno calculation from my August 1999 article.  Let me review this for you.  If a set S is partitioned into two sets R and T having r and t elements respectively (note that necessarily the set S must have r + t elements), then the probability p of randomly choosing a subset of S having u elements from R and v elements from T is given by

 p =   C(r, u) x C(t, v)          C(r + t, u + v) (1)

where C(r, u),  for example, represents the number of ways to choose u elements from a set of size r and can be calculated using

 C(r, u) =       r!                      u!(r - u)! (2)

The derivation of (2) can be found in my August 1999 article.  The numerator (top) of (1) represents the number of different ways of choosing u elements from R and v elements from T ; it is this idea that we will use over and over in what follows.

In >last month's article you were introduced to the idea of a conditional probability.  We wrote this as P(A | B).  This notation represents the probability that event A occurs given that we know that event B has occurred.  We are going to use this to introduce a new idea called conditional expected value.  This is a simple generalization of an idea with which we are already very familiar.  Given a partition A1, A2, ... , An of a sample space (think of these as being all of the different types of final Video Poker hands including the losers) and a random variable f that is constant on each event in the partition (in all of our examples so far, f has been a payoff function for the game under study and that will be the case here, as well), the expected value of  was calculated by multiplying the probability of each event in the partition by the value of  f on that event and then adding up all of these products.  In symbols

 exp( f ) = P(A1) f(A1) + P(A2) f(A2) +  ... + P(An) f(An) (3)

You will recall (see my July 1999 article at this site) that in a game where it costs one unit to play, exp( f ) represents the long run average return per unit risked.  The negative of this number expressed as a percentage is called the house edge.

The conditional expected value of f given B, written exp( f | B), is defined by

 exp( f | B) = P(A1| B) f(A1) + P(A2| B) f(A2) + ... + P(An| B) f(An) (4)

You see what I did.  In (3) I simply replaced all of the probabilities with conditional probabilities.  Here is the idea.  Suppose that B1 represents the event of receiving two pair and discarding the single card and B2 represents the event of receiving two pair and discarding all but the highest pair.  Then whenever one receives a Video Poker hand containing two pairs, one of them high, exp( f | B1) represents the long run return one can expect by always keeping both pairs and exp( f | B2) represents the long run return one can expect by only keeping the high pair.  I think it is obvious that whichever of these numbers is the larger, that is the way the hand should be played.

Very well, we know what we want to do, now let's do it.  To begin with, let us make sure that we know the structure of the game in question.  The 10 - 7 Double Bonus game is a Jacks or better, 5 card draw, Video Poker game. The payout structure, which is always displayed on the front of the machine, is given by the following table.

 Hand One Coin Pay Five Coin Pay Royal Flush 250 4000 Straight Flush 50 250 Four Aces 160 800 Four 2, 3, 4 80 400 Four 5 - King 50 250 Full House 10 50 Flush 7 35 Straight 5 25 3 of a Kind 3 15 Two Pair 1 5 Jacks or Better 1 5

The 10 - 7 refers to the single coin payout on the Full House and Flush, respectively.  You can see by looking at the Royal Flush payout why most people play the full five coins.  Although the Royal Flush will be of no concern to us since there is no way to draw a Royal if you keep a pair, I'll use the per coin payout for 5 coins played since this seems to be common practice.

Let us suppose that we have been dealt JC, JD, 3S, 3H, 7D.  There are now 47 cards left in the deck and we can partition them into 13 packets as follows.  The Aces, Kings, Queens, Tens, Nines, Eights, Sixes, Fives, Fours, and Twos each have four cards in them for a total of 40.  The Jacks and Threes each have two cards in them and the Seven packet contains three cards giving us 7 more cards for a total of 47.  We're all set.

Suppose that we toss only the 7D.  What are the possibilities?  Not many.  We can draw a Jack or Three in 4 ways and have a Full House.  In other words

 P(Full House | Toss 7D) = 4/47 (5)

The other 43 cards leave us with our original Two Pair so

 P(Two Pair | Toss 7D) = 43/47 (6)

The conditional expected value of the payout given that we toss a 7D can be done in tabular form just as we did when we calculated house edges.  To keep our table neater, I will just list numerators in the probability column and divide by 47 at the end.  Here it is.

 Final Hand P(FH | Toss 7) Payout Product Royal Flush 0 800 0 Straight Flush 0 50 0 Four Aces 0 160 0 Four 2 - 4 0 80 0 Four 5 - King 0 50 0 Full House 4 10 40 Flush 0 7 0 Straight 0 5 0 3 of a Kind 0 3 0 Two Pair 43 1 43 Jacks or Better 0 1 0 Total 47 -- 83

Dividing 83 by 47, the conditional expected value is 1.76596.  Note that this is not an expected profit since it cost us one unit to play; the actual expected profit is 0.76596.

Well, that was easy; when three cards are tossed it is a bit trickier.  Specifically, suppose we toss the 3H, 3S, and 7D.  We must now pick three of the remaining 47 cards to fill our final hand.  As we know, there are C(47, 3) ways to do this, this number being 16,215 according to (2).  Let's dig in and do it.

There are no Royals and no Straight Flushes.  By holding two Jacks, there is no way we can get four Aces or four low cards.  We can, however, end up with four Jacks.  This happens when we draw the remaining two Jacks and any one of the remaining 45 cards.  In other words

 P(Four Jacks | Toss Three) = 45/16,215 (7)

What about Full Houses?  There are many ways that these can occur.  For example, the pair of Jacks held can be matched with a triple of non-Jacks.  The triple cannot consist of Threes since there are only two of them left.  There are 10 packets that contain 4 cards each, so there are 10 x C(4, 3) = 40 ways to select three non-Sevens.  There is only 1 way to pick the 3 Sevens, so altogether there are 41 ways to match the pair of Jacks with a triple.  The other way a Full House can occur is if we pick one of the two Jacks and a pair of the remaining cards. There are two ways to pick the Jack.  First let's match the three Jacks with a pair of cards from one of the 4-packets.  There are 10 ways to pick one of the 4-packets and once this choice is made there are C(4, 2), or 6,  ways to pick a pair from the packet.  So, there are 2 x 10 x 6 = 120 ways to pick a Jack and a pair from one of the 4-packets.  We could also pick a Jack and a pair of the 3 Sevens.  There are 2 x C(3, 2) = 2 x 3 = 6 ways to do this.  Finally, we could pick a Jack and the two remaining Threes; there are 2 x 1 = 2 ways to do this.  Altogether, then, there are 120 + 6 + 2 = 128 ways to pick a Jack and a pair.  The total number of Full Houses is thus 41 + 128 = 169.  We have shown that

 P(Full House | Toss three) = 169/16,215 (8)

Now, I am not going to drag you through all of these, but I would like to do one more for you and then you can try your hand at counting the remaining hands.  There are no Flushes or Straights that occur, but there are Three of a Kinds.  Let me count these for you.  The only way to obtain Three of a Kind is to pick a Jack (2 ways) and then pick two unequal cards from the remaining 45.  Let's see how many ways there are to do the latter.  Well, we could pick two of the ten 4-packs and then pick one card from each; there are C(10, 2) x C(4, 1) x C(4, 1) = 45 x 4 x 4 = 720 ways to do this.  Or we could pick one of the ten 4-packets, pick one card from it and match it with one of the 3 Sevens; there are C(10, 1) x C(4, 1) x C(3, 1) = 10 x 4 x 3 = 120 ways this can happen.  Or we could pick one of the ten 4-packets, pick one card from it, and match it with one of the two remaining Threes;  there are C(10, 1) x C(4, 1) x C2, 1) = 10 x 4 x 2 = 80 ways to do this.  Done?  Not quite.  We could pick a Seven and a Three and there are 3 x 2 = 6 ways to do that.  Now we have it.  There are 720 + 120 + 80 + 6 = 926 ways to pick two unequal non-Jacks after we pick one of the two Jacks.  Since there are 2 ways to pick the Jack, there are 2 x 926 = 1852 Three of a Kinds.  We thus have

 P(Three of a Kind | Toss three) = 1,852/15,215 (9)

By now I think you see the similarity of these calculations to the Keno calculations.  You can try the Two Pair and High Pair hands and see if you can get the right numbers.  If you need some help, let me know and I'll show you how to do them.  Give it a good try though; it's a fun challenge and you'll feel good if you get them.

Very well, we're ready to fill in our table.

 Final Hand P(FH | Toss three) Payout Product Royal Flush 0 800 0 Straight Flush 0 50 0 Four Aces 0 160 0 Four 2 - 4 0 80 0 Four 5 - King 45 50 2,250 Full House 169 10 1,690 Flush 0 7 0 Straight 0 5 0 Three of a Kind 1,852 3 5,556 Two Pair 2,629 1 2,629 Jacks or Better 11,250 1 11,250 Total 16,215 -- 23,645

Dividing 23,645 by l6,215 we get the conditional expected value of 1.45822.  Notice that this is considerably less than the previous value of 1.76596.  Tossing the lower pair will, on average, cost you in excess of 30 cents per dollar played on Two Pair hands.  Expensive!

I told the Binion's dealer the facts and he accepted them.  Then he said, "What about Aces?  With such a big payout on 4 Aces, isn't holding a single pair of Aces better?"  Now that sounds reasonable.  In such a case the payout is 160 for 1 so in the above table we would have 7,200 (160 x 45) rather than 2,250 in the product column.  The new total in this column would be 28,595.  Dividing this by 16,215 we obtain 1.76349.  Close, but keeping the Two Pair is still better by around a quarter of a cent per dollar played.

Intuition is a wonderful thing, but we have to constantly be critical of it.  The dealer's argument for tossing the lower pair sounded reasonable but, as we have seen, accepting it is expensive.  In fact, if you reflect on it, his observations were entirely correct, but they just didn't tell the whole story; those Full Houses were an important consideration.  Now here's one for you.  What if the game were 9 - 7 Double Bonus, that is, the Full House only paid 9 for 1?  There are a lot of these games around.  In a Two Pair hand with Aces would you only hold the Aces in this game?  See you next month.

Recent Articles
Best of Donald Catlin
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

#### Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

#### Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers