Part 13: Basic Semi-Bluff Shove Calculation
Now it's time for some poker. Suppose we're heads-up on some
street and our opponent has us covered. The pot size is P and we're considering making a
shove of size B. Our opponent's chance of folding is F, and when he calls our
equity is E. Let's find the EV of this
shove.
First we have to figure out all of the possible outcomes. The first decision that happens after we
shove is that either
Villain folds or
Villain calls. If
Villain folds the hand is over.
Outcome #1:
Villain folds
If
Villain calls, either we win the hand or we lose. (Note again we're ignoring ties since that's taken care of in our
equity %). Both of these outcomes end the hand.
Outcome #2:
Villain calls, we win
Outcome #3:
Villain calls, we lose
Now we have to find the chance of each outcome happening. The chance of outcome #1 is just the chance of
Villain folding, and we know that's F. The chance of outcome #2 is the chance of
Villain calling times the chance we win the hand after he calls (our
equity). The chance
Villain calls is (1-F) which just means 100% minus the % of the time he folds, and our
equity is E, so the chance of outcome #2 is E(1-F), and remember that means E TIMES (1-F). The chance of outcome #3 is the chance
Villain calls times the chance we lose after he calls. The chance
Villain calls is (1-F) like before, but the chance we lose after
Villain calls is (1-E), or 100% minus our
equity. So the chance of outcome #3 is (1-F)(1-E). In summary, the chances of each:
Outcome #1,
Villain folds: F
Outcome #2,
Villain calls, we win: E(1-F)
Outcome #3,
Villain calls, we lose: (1-F)(1-E)
(Side note: For this next part, it might be useful for those not super familiar with Algebra to know that (a+b)(c+d) = ac + ad + bc + bd.)
So we need to first make sure that all of these chances added together equal 1. So we have:
F + E(1-F) + (1-F)(1-E)
=
F + E - EF + 1 - E - F + EF
= 1
Originally Posted by spoonitnow
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