No Limit Theory and Practice (Terms of Expectation)

[MikeyG80]

Just purchased Sklansky and Miller No Limit Theory and Practice and I had a question from page 21 "Thinking in terms of expectation - Playing the nuts on the river"

It gives three examples on the river bet, holding the nuts, of a small bet at $50. A medium sized bet of $150. A large sized bet of $450- All in push. (Take into consideration the example given is from a proposed $5/$10 no limit cash game table where you have $450 left in front of you making this river bet)

It says your expectation for the $50 bet using this formula is $40. $40 = (.80)($50)
expectation for the $150 bet
is $60. $60 = (.40)($150)
expectation for the $450 bet
is $90. $90 = (.20)(450)

What do the amounts of $40, $60 and $90 mean in this example? I'm just a little unclear of these figures and hope i can get a supplement to this example from someone who has this book or knows this example. Is this the amount on average you'll make, in this situation with this much money going into the pot as the bet? It says clearly shown by these amounts, your best move is the all-in push for $450 as it maximizes your expectation.

[PocketRuler]

Expectation, or more appropriately expected value, is from probability theory. It is the average amount you are expected to win over numerous trials.

In the book, Sklansky’s first example is betting $50 on the river. Say you could play the hand 100 times. Of that 100 times, you think that your opponent will call you 80 (from the book) times out of that 100 and the other 20 times he/she will fold. For every time that he/she calls, you make $50 and for every time your opponent folds, you make $0. So for the 80 times out of 100 that your opponent called, you would make $4000 and for the 20 times out of 100 he/she doesn’t call, you make 0$. The average of this example results in an expected value of $40 ($4000/100). Therefore, on average, you’d expected to make $40 on a $50 bet with the nuts on the river. With the other two examples, you are expected to make $60 on a $150 bet and $90 on a $450 bet. Expectation is more about making plays that will benefit you in the long run.

I hope this helps.

[Pirana]

Either there are errors or I don't understand some of the expectation equations. For example, on the top of page 54, the book states you "win nothing when a non-pairing diamond comes (7/44)." Actually, you LOSE the $100 pot when your opponent hits his 7-outer flush.

Your expectation if you check goes down to $86.38, NOT $102.28.
(35/44)($100) + (7/44)(-$100) + (2/44)($500)
= $86.36.
The book's conclusions become incorrect for this example. You would NOT prefer that your opponent draw for free.

[agoldenbear]

Pirana, the example says you check, which has no immediate risk, so it is in fact (7/44)($0). You are not risking $100 dollars in this spot because what's the in the pot does not belong to you. It is my understanding that expectation is always considered in terms of the current risk/reward scenario.

Look back at the first example in this same chapter. Instead of having QQ and top set, you have AA here with the same board. Now any diamond gives your opponent a winning hand. The math for checking would look like this:

(35/44)($100) + (9/44)($0) = $79.54.

Compared to your $100 expectation if you had just bet more than he would call, you now lose $20.46 by allowing him infinite odds to draw.

Incidentally, this issue confused me when I read a different book subsequent to T&P. Angel Largay's NL cash game book, in his section on expectation, treats EV as something you can quantify for the entire hand.

He assumes in one example that with the nut straight on the flop, the $10 pot already belongs to you. If you move in with your remaining $98 here, and never get called, this play has +$10 EV. If you just bet the pot, and someone with top two calls, the math works out to +$7.45 EV, which he then adds to the original $10 pot, saying "Our expectation for the entire hand has increased from $10 to $17.45. Note that, while both plays have positive expectation, the play with the higher expectation is the one you want" (pp.60-61).

Is this rationale flawed somehow? I will always trust Sklansky's figures, but is there ever any reason to consider the pot "yours for the taking" as Largay puts it?

[Pirana]

[ QUOTE ]
the example says you check, which has no immediate risk, so it is in fact (7/44)($0). You are not risking $100 dollars in this spot because what's the in the pot does not belong to you. It is my understanding that expectation is always considered in terms of the current risk/reward scenario.

[/ QUOTE ]
That does not make sense to me. By checking, you can win the $100 pot if the river card is not a diamond, but you risk LOSING the $100 pot if the river card is one of the seven diamond outs. You can't have it both ways: getting the $100 or $500 pot if you win, but not losing the pot if you lose?!

[ QUOTE ]
Look back at the first example in this same chapter. Instead of having QQ and top set, you have AA here with the same board. Now any diamond gives your opponent a winning hand. The math for checking would look like this:

(35/44)($100) + (9/44)($0) = $79.54.

Compared to your $100 expectation if you had just bet more than he would call, you now lose $20.46 by allowing him infinite odds to draw.

[/ QUOTE ]
For this first example, you would expect to lose the $100 pot if a diamond appears. The correct expectation is
(35/44)($100) + (9/44)(-$100) = $59.09
You lose more than $20.46 by giving your opponent a free card, i.e., $40.91 ($100 - $59.09).

[agoldenbear]

[ QUOTE ]
Either there are errors or I don't understand some of the expectation equations.

[/ QUOTE ]

Are you sure you're willing to consider the possibility that maybe you don't fully understand? Because your most recent post doesn't seem to allow for it. Despite your insistence that you're correct, your thinking is still flawed. Here's why:

When you give a free card that beats you, you have not LOST any money in the immediate sense, you just expect to WIN LESS MONEY, on average, because you didn't bet. Your odds of winning haven't changed, but you expect to win less money than if you had bet when your hand does hold up. Therefore, your expectation, while still positive, has decreased.

So the equation for checking allows for three different scenarios to play out, all with different likelihoods and different dollar figures attached to them. You add all three possibilities together to determine the total EV of initially checking. When you check:

1.) 79.5% (35/44) of the time, you see the river risking nothing, no diamond comes, and you WIN $100.
(.795)($100) = $79.54

2.) 15.9% (7/44) of the time, you see the river risking $0, a non-pairing diamond comes, and you DO NOT WIN $100.
(.159)($0) = $0

3.) 4.5% (2/44) of the time, you see the river risking nothing, a diamond comes and pairs the board, and you WIN the $100 in the pot, plus your opponent's $400 stack, totaling $500.
(.0455)($500) = $22.73

Total EV of checking: $79.54 + $0 + $22.73 = $102.27

Notice how in scenario 2 I used the phrase DO NOT WIN instead of LOSE. There is a difference. In the example your hand is way out in front, so any choice you make (besides folding) gives you a positive expectation. Even in the example with AA, you expect to make money on average if you check, because you're favorite at that point in the hand. However, checking here has the lowest possible positive expectation (other than folding), because you are giving your opponent a shot to beat you and not charging him for it. The play still makes you money, but less money than if you bet and forced a fold or got him to call more that he profitably should. So, you always want to choose the play with the highest positive EV.

In order for the equation to have a (-$100) anywhere in it, we would have to bet $100 on the turn, which would give us a much higher EV than checking. If our opponent always folds to the $100 bet, the EV of that play is, as was stated in the book, $100. If we can anticipate a call of our $100 bet, the equation now looks like this:

EV = (35/44)($200) + (7/44)(-$100) + (2/44)($500)
($159.09) + (-$15.91) + ($22.73) = $165.90 EV

In order to LOSE $100, we must bet $100 AND have our opponent fill his flush with a non-pairing diamond. When our opponent calls our $100 bet and then misses his draw, which will still happen 79.54% of the time, we win $200.

The lesson in all of this is a one of poker semantics. The pot never belongs to you, so in terms of EV, you never stand to LOSE it. It's all about what you DO NOT WIN. You can lose money on bets, as I demonstrated above, but the pot is never factored in, except when thinking of your potential win.