Issue 345
April 23 - April 29, 2007
Volume 7
page 3

Those pesky low pairs in 9/6 Jacks
By Don Catlin

Whenever I play 9/6 Jacks Video Poker I am always troubled by three things:
  1. Missing them seems to be my most frequent error.
  2. They seem to occur with a greater frequency than I would have guessed.
  3. Even though they are listed quite high on the strategy table, they seldom seem to lead to anything useful.

I'm going to try to address these three issues and, hopefully, bring some light to bear.

The first point is simply carelessness. When you miss a low pair it generally means you are playing too fast. Jean Scott's advice, in both her book and her software, is to pause before you hit that draw button and make sure you are not missing an outside straight, a three-card straight flush, or especially a low pair. In a hand like 4S, 9H, JH, QH, 9D that nice three-card inside straight flush with two high cards sort of jumps right out at you. The right play, however, is to hold the two 9s. So remember that pause; it is darn good advice.

Now let's address the question of how frequently a low pair hand occurs. By this I mean a hand that when correctly played results in holding the low pairs and discarding the rest. During this discussion I will use the notation C(n, k) to represent the number of combinations of n things taken k at a time; this has been discussed many times in previous articles.

In a fresh deck there are 9 packets of low cards in the range 2 to 10, each containing four cards. There are C(9,1) ways of choosing a packet and then C(4,2) ways of selecting two cards from it. The product of these two numbers (9 and 6) is 54 so there are 54 ways of selecting exactly two low cards.

What about the remaining three cards? Well, one could obtain 3 of a kind. There are 2 cards that match the low pair, 12 packets containing cards that don't match, so the number of 3 of a kind that one can obtain is C(2,1)C(12,2)C(4,1)2 which is 2 x 66 x 16 or 2,112.

There are 48 way to obtain quads; that one is easy. For 2 pair we have to select one of the 12 packets, choose two cards from it, chose one of the remaining 11 packets and choose 1 card from it.

Altogether this gives us C(12,1)C(4,2)C(11,1)C(4,1) or 12 x 6 x 11 x 4 or 3168 two pair hands. A Full house can occur in two ways. We can choose three matching cards from the 12 packets, C(12,1)C(4,3), or we can match the original pair (2 ways) and then pick 2 matching cards from the remaining 12 packets, that is, C(12,1)C(4,2).

Hence the number of Full Houses is just 12 x 4 + 2 x 12 x 6 or 192. If we want to have a hand that doesn't have a match to the pair for the remaining three cards we simply select 3 of the 12 packets and choose one card from each. This is just C(12,3)C(4,1)3 or 14,080. Let me tabulate what we have so far:

Type Hand


3 of a Kind


4 of a Kind


2 Pair


Full House


Pair + 3




I should note that C(50,3) = 19,600 so that is an independent check that our calculations are correct.

Now I labeled the last category Pair + 3 because not all of these hands are hands in which we would keep the pair. Specifically there are some three-card Royals, some four-card Royals, some four-card Straight Flushes, and some four-card flushes. The three-card Royals arise as follows. There are two suits that don't match the pair. Choose one of these and then pick 3 cards out of the 5 card set of 10 through Ace assuming that the pair does not consist of tens. There are clearly 2C(5,3) or 20 ways to do this. If the pair is 10s then there are only 2C(4,3) or 8 ways to do this. The four-card Flushes occur as follows. There are two suits that match one of the pairs. Choose this suit and pick 3 of the 12 remaining cards in that suit. There are 2C(12,3) or 440 ways to do this. Finally we can now determine how many of the 14,080 hands really are low pair hands. If the low pair is not tens then there are 14,080 - 20 - 440 = 13,620 low pair hands. If the pair consists of tens then there are 14,080 - 8 - 440 = 13,632 low pair hands.

Very well, here at last is our calculation. There are C(8,1)C(4,2) x 13,620 low pair hands where the low pair does not consist of tens and C(1,1)C(4,2) x 13,632 paired ten hands that are really low pair hands. These numbers are, respectively, 653,760 and 81,792 and the total is 735,552. There are 2,598,960 possible five-card hands that can be dealt from a 52-card deck. Dividing this by our total we get a ratio 3.5333. In other words, on average, a true low pair hand (meaning that you only hold the pair) will occur about two times in every seven hands. So they really do occur quite frequently.

Now let us address the third point. If you only hold a pair and draw three from the remaining 47 cards, what can you expect? If you would like to see how to calculate the following numbers you can write to me and I'll be happy to show you how to do it - it is a bit tricky. I am not doing the calculations here because you can get these figures from a Video Poker simulator such as Win Poker. Here are the results:

Type Hand


3 of a Kind


4 of a Kind


2 Pair


Full House






From this table one can easily deduce that when drawing to a low pair you will get a losing hand 71.286% of the time. So why are they so high on the strategy table? Because when they do produce a winning hand, it pays well -- at least 2 to 1 and as high as 125 to 1. See you next month.

Don Catlin can be reached at

About the Author

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 the Author

Purchase Don's must-have book, available online here.


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