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Ask the Slot Expert: Figuring out Split Card video poker hand more difficult than I thought

By John Robison

Last week I described a hand I had playing Split Card video poker. I wasn't sure if I had played the hand correctly and I said that I would eventually do the math to find the best play. Eventually was supposed to be this week. I thought I could do the math in an hour or so. I was wrong.

Figuring out the expected value of the combination of cards you hold is usually fairly easy. To pick an easy example, let's say you were dealt Kc Kd Kh 8d 9h and you wanted to find the expected value of holding the kings. The first step is to find the number of ways you can make each possible ending hand.

The best we can achieve is four-of-a-kind. There's one king to complete the quad and 46 cards left in the deck to complete our hand. There are 46 ways we can get four-of-a-kind.

The next best hand we can get is a full house. Calculating the ways is a little trickier now because we have 10 ranks that are still complete and two ranks missing one card. We have to consider them separately.

Looking at the complete ranks first, there are 10 of them and six ways to choose two of the four cards, so there are 60 (10x6) ways we can get a full house using the complete ranks.

There are two ranks missing a card and three ways we can draw two of the three cards, so there are 6 ways we can get a full house using the incomplete ranks.

That gives us a total of 66 ways to get a full house.

Those two cases are fairly small and we could use brute force to figure them out using a deck of cards and pencil and paper (and more paper and lots of erasers). We're sure the numbers are right.

The next possibility is that we don't improve our hand and end up with 3-of-a-kind. We could just say that there are 1081 to draw two of the remaining 47 cards, and we know that 46 of those combos give us a four-of-a-kind and 66 give us a full house, so there are 969 (1081-46-66) ways we can end with 3-of-a-kind.

But that would be too easy. I wanted to show how to get the result in the same way I got the results for the other hands -- counting the ways to achieve the hand.

After an hour, I realized what was wrong with my initial attempt and corrected it, sorta. After another hour, I finally realized what was wrong with my correction and finally got the right answer.

"Fasten your seat belts. It's going to be bumpy night."

There are three components to counting the ways to get 3-of-a-kind. We could use one card from two of the complete ranks, one card from each of the incomplete ranks, or one card from a complete rank and one card from an incomplete rank.

First, the complete ranks. There are 45 ways we can choose two ranks from the 10 complete ranks. Then there are four cards we can choose from the first rank and four cards from the second tank. That gives us 720 (45x4x4) ways to get 3-of-a-kind using the complete ranks left in the deck.

Using the incomplete ranks is easier. There are three cards left in each rank, so there are 9 (3x3) ways to get 3-of-a-kind using the incomplete ranks.

Now for the mixed case. There are 10 complete ranks with four cards in each rank, and there are two incomplete ranks with three cards left in each rank. That gives us 240 ways (10x4x2x3) ways to get 3-of-a-kind in this scenario.

The total number of ways to get 3-of-a-kind is 720+9+240=969. Tada!

I didn't expect it to take me so long to get right the example I wanted to use to show why it's taking me so long to get the Split Card calculations right.

To recap, Split Card sometimes gives you two cards in one position of your hand. In my case, I was playing Deuces Wild and was dealt Qd 2s/2s 2h Td 8h. I had two 2s in the second position.

Should I hold Qd 2s/2s 2h Td and be guaranteed a dirty royal and still have a chance at 4 Deuces?

Or should I just hold 2s/2s 2h and take a chance on getting 5 Deuces, which pays 2000?

Figuring out the first option is easy. The only better hand I could achieve is 4 Deuces. There are two deuces left in the deck. The other 45 cards leave me with the wild royal. Done with that option.

The second option's calculations started off easy. I'm holding 2s/2s 2h. To get 5 Deuces, I need the two deuces left in the deck and one of the remaining 45 cards. So there are 45 ways to get 5 Deuces.

Figuring out 4 Deuces is also easy. There are two ways we can choose one of the two deuces left in the deck and 990 ways to choose two of the other 45 cards, giving 1980 ways to get 4 Deuces.

We're on a roll, aren't we?

Let's do 5-of-a-kind. We need to draw three cards of the same rank. There are nine complete ranks, three ranks missing one card, and deuces. So we have nine choices of complete ranks and we have to choose three of the four cards left, and we have three incomplete ranks and we have to choose all of those cards. There are nine ways to choose a complete rank and three ways to choose three of the four cards in the rank and three ways to choose an incomplete rank and only one way to choose all of the cards remaining in that rank. So 9x4+3x1=39 ways to get 5-of-a-kind.

Next up is figuring out the ways to make a dirty royal. That's when I remembered that it is much easier to count the ways to make a set than to count the ways to make a sequence. Adding wild cards into the mix makes it harder still.

Maybe I'll have it figured out by next week.

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