Wild Cards in Profusion: Part 2

Brian Alspach

Poker Digest, Vol. 5, No. 9, April 19 - May 2, 2002

In today's article, let's suppose we have four wild cards all of the same rank (deuces wild is a commonly played variant). Consider five-card poker hands first. We have quints if a hand has four wild cards, three wild cards and a pair, two wild cards and trips, or one wild card and quads.

There are 48 hands with four wild cards. There are four ways of choosing three of the wild cards, 12 choices for the rank of a pair, and six ways of choosing a pair of the given rank. Multiplying yields 288 quints consisting of three wild cards and a pair.

In a similar way, we obtain 288 quints with two wild cards. Finally, with one wild card, there are 48 quints.

Summing the preceding numbers gives us 672 hands which are quints when we have four wild cards all of the same rank.

We perform the same kind of calculation for other combinations of wild cards to produce the following table. The assumptions on wild cards are two of the same rank, four of the same rank, four of one rank and two of a second rank for six wild cards, and four of two different ranks for eight wild cards. The columns correspond to the number of wild cards and the rows correspond to five-card hands and seven-card hands.

One interesting aspect of the numbers in the preceding table is the vast difference between five-card hands and seven-card hands. There are 2,598,960 five-card hands altogether. This means the probability of having quints with eight wild cards is about .0031. There are 133,784,560 seven-card hands. The probability of having quints with eight wild cards for seven-card hands is then approximately .0366, or more than ten times as likely.

Counting straight flushes is more complicated for the simple reason that their numbers vary according to the ranks of the wild cards. Since we are interested in determining whether or not a straight flush is ever a stronger hand than quints, when there are eight or fewer wild cards, we shall do the counting for a situation which minimizes straight flushes. If the number of straight flushes is larger than the number of quints in a minimal case, then the numbers are larger for all cases.

Let's first look at two wild cards. Letting the two wild cards have rank 5 minimizes straight flushes (5 is not the only rank minimizing straight flushes). For five-card hands, there are 30 straight flushes not involving any wild cards. There are 132 combinations of four cards that become straight flushes when any of the two wild cards is included. This produces 264 straight flushes with one wild card. There are 220 combinations of three cards that become a straight flush upon inclusion of the two wild cards.

Summing the preceding numbers gives us 514 straight flushes for five-card hands, when two of the cards of rank 5 are wild. From the table above, we see there are only 72 quints with two wild cards. Thus, we see that quints is a much better hand than a straight flush for five-card hands, when we have two wild cards.

Now we move to seven-card hands with two wild 5s. There are 41,584 straight flushes among seven-card hands. This number already is much larger than 27,600, the number of quints with two wild cards, and we have not included any of the many additional straight flushes arising from using one or two of the wild cards.

Upon making similar calculations for four wild cards, where we again let all four cards of rank 5 be wild, we find there are 2,180 five-card straight flushes. This compares with 672 quints, so it is easily seen that quints is a better hand for five-card hands with four wild cards.

For seven-card hands, there are 609,760 quints. Obtaining an exact count of the straight flushes is a bit of an ugly problem, but an estimate will serve perfectly well.

We must avoid duplicate counting in making an estimate. One way of avoiding duplication is to keep track of some kind of identifying feature which distinguishes between the straight flushes included in the count. I shall use the number of wild cards in the hand as an identifier.

There are 100 sets of four cards not containing a wild card that become a straight flush upon adding any wild card. There are four choices for the wild card. There are 44 cards left over that are not wild cards, but some of them from the same suit as the original four cards, may not be included or else we could end up with a seven-card hand that could have come from more than one set of four cards. Throwing away four of the cards from this suit avoids this problem. The point is that we have 40 cards from which we may choose any two to complete the four cards plus a wild card to a seven-card hand.

Multiplying 400 by gives us 312,000 distinct seven-card hands that contain a straight flush, and exactly one wild card that must be used to make the straight flush.

There are 152 sets of three cards not containing a wild card that become a straight flush upon adding any two wild cards. There are six choices for the two wild cards. In order to avoid any possible overlap with other seven-card hands, we choose the remaining two cards from the 36 not in the straight flush suit. Multiplying 912 by gives 574,560 seven-card hands containing straight flushes, exactly two wild cards and do not overlap with what was constructed earlier.

We already have more straight flushes than quints, and many of the straight flushes have not even been included.

For six wild cards, under the assumption four of the wild cards have the same rank and the other two have the same rank, and for eight wild cards, under the assumption the wild cards have one of two different ranks, the story is the same. Quints is a better hand than a straight flush.

Our final conclusion is that five-of-a-kind hands are definitely better hands than straight flushes in all the wild card scenarios we have examined.