April 3, 2010 – 10:41 am | 7 Comments

I’ve compiled a short (just 7-pages) e-book, an introduction to the mathematics of poker. It’s basically covers how to calculate your expected value in a certain spot – starting with explaining what EV is, all …

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Home » Featured, Theoretical

Pre-Flop Shoves: Opponent’s Range

Submitted by on November 28, 2009 – 9:52 am2 Comments

In the last post, we took an initial look at pre-flop shoves, and came to the rough conclusion that if you’re short-stacked (with less than 10 big blinds), you should go all in with the following hands:

  • any Ace, any King, Any Queen, Any Jack other than J2 offsuit
  • T5 suited or higher, T7 offsuite or higher
  • 96 suited or higher, 97 offsuit or higher
  • 87 suited
  • Any pocket pair

We assumed, however, that we’d always get one (and only one) caller. Most of the time, however, this isn’t true. More often than not, your opponents will not call your all-in with any two cards.

We need to therefore put our opponent on a hand. Let’s assume that you’re playing against a relatively tight opponent, and you’ll only get called with the following hands:

  • Any pocket pair
  • A7 or higher
  • KT or higher
  • QT or higher
  • JT

For ease of calculation, we’ll still assume that you won’t get a call from more than one player (I’ll consider multiple players in a future post, I promise).

This calling range gets dealt about 21.6% of the time. So therefore, the other 78.4% of the time, your opponent will fold. When this happens, you recover your bet, and gain the blinds.

So, (using the same notation of s = your stack, b = one big blind and p = the probability of your hand winning), your new expected value is now

EV = 0.784 * (s + 1.5b) + 0.216 * p * (2s + 1.5b) – s
EV = 0.784s + 1.176b + 0.426 * p * s + 0.324 * p * b – s
EV = (0.784 + 0.426p – 1) s + (1.176 + 0.324p) b
EV = (0.426p – 0.216) s + (1.116 + 0.324p) b

Again, in order for the call to be correct, the expected value must be more than 0, which would mean

EV > 0
(0.426p – 0.216) s + (1.116 + 0.324p) b > 0
0.426*p*s – 0.216s + 1.116b + 0.324 * p * b > 0
1.116b – 0.216s > -0.324 * p * b – 0.426 * p * s
1.116b – 0.216s > p (-0.324b – 0.426s)
p > (1.116b -0.216s) / (0.324b + 0.426s)

Therefore, given a short stack of 10 big blinds,

p > (1.116b – 0.216 * 10b) / (0.324b + 0.426 * 10b)
p > (1.116b – 2.16b) / (0.324b + 4.26b)
p > -1.044 / 4.584
p > -0.228

This would always be true, so therefore, it seems like given an opponent of this tightness, it would make sense to all-in with literally any two cards whenever you are short-stacked.

Which leads to another question: just how short-stacked do you need to be for the “any two cards” strategy to work? Against the above calling range (A7+, Pocket pairs, Broadway), the worst performing hand is 7-2 offsuit, with a 27.487% chance of winning. So taking this probability

EV > 0
(0.426p – 0.216) s + (1.116 + 0.324p) b > 0
(0.426 * 0.27487 – 0.216) s + (1.116 + 0.324 * 0.27487) b > 0
-0.09891s + 1.205b > 0
0.09891s < 1.205b
s < 12.18 b

So therefore, against an opponent with this sort of calling range, you should shove with any two cards, as long as you’re below 12 big blinds.

To be honest, I’m personally having trouble coming to terms with this math. I don’t think I could bring myself to shove with a 6-4 (not even mentioning 7-2), even with a stack below 12 big blinds. I know I’m assuming a relatively tight opponent, and assuming only one opponent, but even then, mathematically, it seems like a reasonable model.

I’ll look into the other variations (multiple players, a different calling range, etc) in due time, but for now – what do you think? The math seems to support a shove with literally any two cards. I myself don’t really know if I agree, but did I miss anything in the math? Let me know.

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