loonshot's avatar

loonshot

0 points

I'm a bit relieved that the mistake might have been with the book. I'm also surprised because I had previously looked up errata for MOP and found nothing about the calculation on page 132.

I'm glad I posted here because your responses helped to clarify other things too.

Cheers!

April 2, 2021 | 11:38 p.m.

I'm a tad confused. Indulge me a while longer please ...

The SB starts hand with a stack of 2000 (before he posts his 0.5). When SB shoves and BB folds, wouldn't SB finish the hand with a stack of 2001? i.e. his gain is simply the '+1' that was posted by the BB?

April 2, 2021 | 4:30 p.m.

Hi BigFiszh,

Thanks for chiming in. I was especially pleased that it was you because some of your other posts have helped me with earlier chapters of MOP :)

Following the steps you laid out, it was very useful to confirm that both formulas are indeed the same. I then went back to page 132 but my answer there (and annoyingly, only there!) continues to differ from that shown in the book. I'd like to show my working to see where I'm making a mistake:

p(BB folds) = 1222/1225 (rounding to 0.9976)
p(BB calls) = 3/1225 (rounding to 0.0024)
p(SB wins pot) = 0.13336
p(SB loses pot) = 0.86664

So EV of jamming ATs

= p(BB folds)(+1) + p(BB calls) * (p(SB wins pot)(+2000) + p(SB loses pot)(-2000))

= 0.9976 + 0.0024 * (266.72 - 1733.28)

= 0.9976 + 0.0024 * (- 1466.56)

= 0.9976 - 3.519744

= - 2.522144

The answer given in MOP is - 2.09525

If it's somehow not a mistake with my maths, could it be a rounding error?

Thanks again

April 2, 2021 | 3:47 p.m.

(MOP at page 132): Jam-or-Fold NLH with stacks of 2000. SB posts 0.5 and BB posts 1. Calling range for BB is AA only.

To calculate the "equity" (which I read as EV) of SB jamming ATs, their approach is:

EV of jamming ATs

= p(BB folds)(pot) + p(BB calls)p(SB wins pot)(pot size) - cost of jam
= p(BB folds)(1.5) + p(BB calls)p(SB wins pot)(4000) - 2000

Working through MOP, my approach to these calculations has generally been:

EV of jamming ATs

= p(BB folds)(+1) + p(BB calls) x (p(SB wins pot)(+2000) + p(SB loses pot)(-2000))

So my reference point is stacks at start of the hand and the numbers I plug in are gain/loss relative to that.

I've liked this approach because it reflects the narrative in my head as I'm thinking about the game tree: "When SB jams, the BB will fold sometimes and SB will win his blind. When the BB does call, SB will gain 2000 some % of the time and lose 2000 the remaining % of the time ..." This approach has led me to correct answer for MOP games until now. But I'm just not arriving at the book's answer for this calculation.

My assumption was that both calculations were the same thing but just laid out differently.

What's my blind spot here?

April 2, 2021 | 10:08 a.m.

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