Ask the Slot Expert: Split Card calculations: Light at the end of the tunnel

In probably the least anticipated series finale ever, we're finally ready to see which option has the higher EV for our hand in Split Card poker.

Let's recap where we stand:

A few weeks ago I played Deuces Wild with the Split Card gimmick. Split Card sometimes gives you two cards in one position of your hand. Although it's possible to get 6 of a Kind and other six-card hands, the only bonus hand is 5 Deuces, which pays 2000.

I was dealt this hand: Q♦ 2♠/2♠ 2♥ T♦ 8♥.

I wasn't sure which combination was better to hold:

1. Q♦ 2♠/2♠ 2♥ T♦ and be guaranteed the Wild Royal and maybe get 4 Deuces
2. 2♠/2♠ 2♥ and take a chance on 5 Deuces

Here is the EV of option 1.

Here is what I have so far for the EV of option 2.

Now we have to do 4 of a kind. I thought this might be easy. I started with 4 of a Kind with an ace. Remove the remaining aces in the deck and the cards that could lead to a Straight Flush or Wild Royal. Find the combination of the number of cards left taken two at a time and "Bob's your uncle".

I've never had an Uncle Bob and this was completely wrong. The number was too large.

I didn't trust that I could figure out the number of ways to make the hands without some help. I created a SQL table with all of the three-card hands I could be dealt. I wrote queries against that table to check -- and correct -- the number of ways to make the scenarios I described in prior articles. After I found the combinations that would evaluate to the hand I was analyzing at the time, I stored the value of a combination in its record. The number I calculated was larger than the number of combinations I had left to evaluate -- all of which should have evaluated to 4 of a Kind.

What did I forget? I got rid of the cards that could give me 5 of a Kind, a Straight Flush, or a Wild with an ace, but I didn't do anything to prevent, say, getting a Straight Flush by being dealt a suited 6 and 7 along with the ace.

Looks like I'll have to go back to the procedure I used before: Deal two cards from the deck and find the number of ways to fill the fifth position without getting a higher-paying hand.

Starting with an ace, I don't need to look at a suited 3, 4, or 5, which would give me a Straight Flush, or a suited ten, jack, queen, or king, which would give me a Royal Flush. I have to look at a suited 6, 7, 8, and 9. It's always nice when I can eliminate some possibilities right off the top.

But I have to do some suits separately. Clubs and spades are the same, but diamonds is different because I discarded a ten and hearts is different because I discarded an 8.

Next I'd have to look at unsuited combinations and I'd have to look at A3, A4, A5, A6, A7, A8, AT, AJ, AQ, and AK. This is getting out of hand.

There's a very easy was to find the number of ways to get a 4 of a Kind. I've already figured out the number of ways to get the higher-paying hands. All I have to do is subtract the total number of ways to get a higher-paying hand from 16,215 (the combination of 47 things taken three at a time) and now Bob really is your uncle. As long as my calculations are correct, the result of the subtraction will be correct.

I was hoping that I could calculate the number of ways to get 4 of a Kind and therefore verify all of my calculations, but I didn't realize that there would be so many scenarios that would have to be individually analyzed.

Here's the complete table for option 2:

The EV of option 1 is 162.23 and the EV of option 2 is 169.61. Option 2 has the higher EV. I was hoping that option 2 would win, but I was surprised at how close the EVs of the two options are.

Which option did you think would win? Were you also surprised at how close the two options are?

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Send your slot and video poker questions to John Robison, Slot Expert™, at slotexpert@slotexpert.com. Because of the volume of mail I receive, I regret that I can't reply to every question.

Copyright © John Robison. Slot Expert and Ask the Slot Expert are trademarks of John Robison.