Hole card odds - connectors - check my math

Poker Discussion
1

I need math help. I’m trying to calculate the percentage of the possible starting hands that are connectors (sequentially ranked, suit not important).

https://www.cardplayer.com/poker-tools/odds-and-outs gives me the odds, but doesn’t tell me how to calculate them. According to this resource 15.7% of two card combos are connected.

Here’s how I calculate my expected frequency of being dealt connectors (two cards of sequential rank).

There are 12 combinations of connectors, ignoring suits:

  1. AK,
  2. KQ,
  3. QJ,
  4. JT,
  5. T9,
  6. 98,
  7. 87,
  8. 76,
  9. 65,
  10. 54,
  11. 43,
  12. 32

Incorporating suits, there are 16 combinations of each connector hand, (4 Aces + 4 Kings, etc.). We can generalize these combinations by calling them R1R2.

There’s 52 cards in the deck. So you have a 4/52 chance of drawing an Ace, and a 4/51 chance of drawing a King. And so on for each of the twelve connectors.

Therefore, I calculate that we can expect to be dealt connectors:

12 * (4/52) * (4/51) = 0.07239819005 = 7.24% of the time.

I don’t see my error.

One thought I had in trying to figure out where I went wrong is that, for the non-edge cases (all the connectors that aren’t AK or 32), you can make a connector hand by drawing R1R2 or R1R0. In other words, if your first hole card is a Ten, then you can make connectors by drawing a J or a 9. This gives you 8 “outs” to hit connectors on the second card. So, if we work this out, we get a 44/52 chance of drawing any card that isn’t an Ace or a Two, and an 8/51 chance of drawing a card that is within one rank of whatever the rank was of the first card.

44/52 * 8/51 = 0.13273001508295625942684766214178 = 13.27%

This is closer to the 15.7% figure that cardplayer.com provides, so I think I’m on to something with this reasoning. The above calculation doesn’t incorporate first-card Ace, and first-card Two, as both of those situations only have 4 “outs” to make connectors, not 8, so I add those back in:

4/52 * 8/51 + 4/52 * 4/51 + 4/52 * 4/51 = 14.48%

Still far enough off from 15.7% that I am still not there.

So what am I missing?

2

Let’s think of the two cards as being dealt one after the other. The first card is irrelevant. Given the first card, there are 8 out of 51 cards that make the hand connected (even in the “edge cases”). Hence, the probability is 8/51 = 15.69%.

Edit: I think you just missed A2.

3 Likes
3

Ah, thank you! I did miss A2, and that of course takes care of the edge case because the Ace “wraps”. Makes sense now!