Progressive Blackjack à la Wildhorse

Many of you are familiar with the Midnight Skulker, who writes articles on this site as well as for other gambling publications. Well, a few weekends ago the ole Skulker was playing Blackjack at the Wildhorse Casino, a Native American casino located near Pendleton, Oregon. There he encountered a blackjack side bet that he had not seen before that involved Aces and had a progressive jackpot; the Wildhorse simply calls the bet Progressive Blackjack and I'll describe it in a moment. Anyway, the Skulker correctly observed that the bet did not look very good but wondered what the progressive should be before the bet turned positive for the player. He wrote to our editor John Robison wondering if any of the "RGT gurus" could help out on this question. John forwarded his letter to me. I don't think I qualify as "guru" but I can answer the Skulker's question.

The basic blackjack game is dealt from a six-deck shoe. Here is the side bet wager. The side bet costs $1 to play. The payoffs are as follows:

$25 if the player's first two cards are unsuited Aces
$100 if the player's first two cards are suited Aces
$250 if the player's first three cards are unsuited Aces
$2500 if the player's first three cards are suited Aces
Progressive jackpot if player's first four cards are Aces of the same color

As we will see, this is not a good bet at all. Why then spend the time and energy writing about a crummy wager in an out-of-the-way casino? Worse yet, why read about such? There are several reasons. First of all, of course, is to answer the Midnight Skulker's question. Second, there are some interesting calculations involved in the analysis of this bet. Thirdly, the analysis will show how to evaluate games having a progressive jackpot; the calculation for the player and the calculation for the house, though similar, are different. Finally, and this is really interesting, I think that this side bet was poorly planned and could easily be improved.

Let me discuss the last item first. Suppose that you are playing this side bet and are lucky enough to be dealt three unsuited Aces as your first three cards. Fine, you have a lock on $250. Now with incredible luck you draw another Ace giving you four Aces of different colors. Your additional prize? Zilch! There is no payout for such a hand; you win the $250 and that's it. How crummy is that? Worse, try this. You are lucky enough to be dealt three suited Aces as your first three cards. Great! You have a lock on $2500. Then with uncanny luck you manage to receive a fourth Ace but of a different color than the three suited Aces. This, as we will see, is a very unlikely scenario. Your additional prize? Again, zilch. How awful! Even if the pay tables were generous (they are not) I would not take this bet simply because of the two events just described. Even though these events are rare, knowing that they could happen to me really annoys me and I'll bet it will annoy others as well. In my opinion, someone should have thought of this when designing this game.

Well, let's get to the analysis. To begin with this bet is settled within the first four cards dealt. Moreover, assuming optimum play, the first two Aces will be split (and resplitting Aces is allowed in this game) so there will always be at least four cards dealt. So our analysis is based on dealing 4 cards from a 312-card shoe. (Note: unless some ploppie stands on the Ace pair, the following analysis is valid even if the pair is hit rather than split.) The ordering of the cards is essential here so the total number of possible 4 card hands is 312 x 311 x 310 x 309 = 9,294,695,280 distinct ordered hands.

There are 231,264 hands that contain four Aces of different colors. Of these, 5,760 are formed by three suited Aces in the first three cards and an Ace of a different color in the fourth card. This number is obtained as follows. There are four ways to pick the suit of the first three cards and then there are 6 x 5 x 4 ways to pick the three suited first three cards. There remain 12 Aces of an opposite color from the chosen suit so there are 4 x 6 x 5 x 4 x 12 = 5760 such hands. One could also obtain four Aces of a different color by starting with three unsuited Aces of the same color and then choose a fourth Ace of a different color. In such a hand, two of the first three cards must be suited. There are 4 ways to choose the suit of these two cards and then 6 x 5 ways of choosing them in order. There are 6 cards left for the off suit, same color card. These three cards can then be arranged in three ways (ssu, sus, uss). As above, there are 12 Aces of a different color to pick for the fourth card. Hence there are 4 x 5 x 6 x 6 x 3 x 12 or 25,960 such hands. Finally, and I'll leave the calculation to you, there are 199,584 hands that can be formed from 3 unsuited different colored Aces plus any Ace as the fourth card.

The above breakdown shows that there are 199,584 + 25,960 or 225,504 hands that begin as an unsuited three Ace hand and end up as a four Ace hand of different colors. There are 5760 hands that begin as three suited Aces and end up as a four Ace hand of different colors. From the earlier discussion of these disappointing hands you can see why I made this breakdown.

Now we can begin in earnest. I am not going to do all of the calculations for you but here are a couple of them. How about the number of two-card unsuited Ace hands? Well, there are 4 packets of suited Aces each containing 6 cards. There are 4 ways to pick the suit of the first Ace and 6 ways to pick that Ace. This leaves 3 ways to pick the suit of the second Ace and 6 ways to pick it. The third card must be a non-Ace and there are 288 of them. Finally, the fourth card can be anything and there are 309 cards left in the shoe. The number of such hands is, therefore, 4 x 6 x 3 x 6 x 288 x 309 or 38,444,544.

How about the number of losers? These can occur in two ways. First, if a non-Ace occurs as the first card, the hand is a loser. There are 288 non-Aces and then the next three cards can be anything. The number of such hands is, therefore, 288 x 311 x 310 x 309 or 8,579,718,720. Next, a losing hand can occur with an Ace as the first card and a non-Ace as the second followed by any two cards. This number is 24 x 288 x 310 x 309 or 622,100,480 hands. Adding these two numbers give us a total of 9,241,819,200 losers.

Here is a table that breaks down the frequency of the hands relevant to this side bet.

Note that according to the frequencies involved one could easily include three suited Aces + an off-color Ace in the progressive and could also make a payout in excess of $250 for four different colored Aces arising from three unsuited Aces. At the very least, these two hands could be combined as simply a bonus for four different colored Aces.

Because the side bet is structured the way it is, the three suited Aces + off colored Ace hands will be included with the three-card suited Ace hands. In addition, the hands with four cards of different colors that originate with three unsuited Aces will be included with the three unsuited Ace hands. This done we can now answer the Midnight Skulker's question. To this end we construct the following table:

For a positive game the number in the box in the lower right hand corner of Table 2 must be positive. This will be true when

or equivalently

The number on the right of (2) is approximately $250,699.39. The four Aces of the same color occur on average once in every 388,965 hands. Assuming that the progressive is seeded at $25,000 and that $0.30 of every dollar wagered goes to fund the progressive jackpot (these are just guesses) the average progressive jackpot will be around $141,690. According to the Skulker the jackpot was hit two weeks prior to his visit and that payoff was around $131,000. I realize that one sample does not a statistic make, but this figure is at least consistent with my assumptions.

There is more to this. For a particular player playing at a particular time the house edge this particular player faces is a function of the progressive jackpot, specifically the X in Table 2. For the house, however, the edge is fixed. Sound contradictory? Let me explain. Using the assumptions in the previous paragraph, for each $1 wagered on this side bet, the house only collects $0.70; $0.30 is returned to the player. No, not each player but to some player, namely, the player who wins the jackpot. If the jackpot is seeded at $25,000 then that is the amount that is due from the house, the rest of the progressive jackpot comes from the 30 cent contributions already deducted from the house's take on losing hands. In other words if $9,294,695,280 was wagered, the house keeps 70% of this, or $6,469,273,440, and returns the remainder to the players. The situation, assuming everything goes according to probability theory, is given in Table 3.

Naturally the larger the progressive seed and the larger the percentage contribution to the progressive jackpot, the smaller the house edge will be. I seriously doubt that the actual numbers for this wager exceed the assumptions I made but there is certainly room to do so. The house edge for this wager is the net profit given in the lower right corner of Table 3 divided by the total amount wagered and is approximately 27.87%.

The probability of winning anything in this game is approximately 0.57%. Assuming 60 hands per hour (typical for a full table) this works out to about one win for every three hours of play. Although he obviously wasn't making this miserable side wager, the Skulker told me that he had three pairs and an unsuited triple in 12 hours of blackjack play so his experience matches that estimate.

One might be tempted to keep track of Aces for this wager but the above numbers are so overwhelming that I really don't think it is worth the effort to investigate such.

Many thanks to the Midnight Skulker for bringing this thought provoking wager to the attention of Rolling Good Times' readers. If any of you have questions I can be reached via email by clicking on the word Technigame in the "About the Author" box at the end of this article. See you next month.