Correct Sklansky hand answer, I think

[Howard Lederer]

I would first like to thank Tom for leading me back in the right direction
on this problem. The artifact left over when the "pot = near infinity"
meant almost no bluffing, and made it easy to discourage Player B from
betting. At Tom's urging, I started over and allowed each player to take
any action allowed in normal limit poker. I think I have solved for
Player A's optimal action in the 1 bet pot. I now think Player A should
bet using these frequencies:

Bluff .2
Check after missing .3
Check after improving .1
Bet after improving .4

This initial action should make Player B indifferent as to calling or
folding after Player A bets, and betting or checking after Player A
checks. It will also allow Player A to check raise in a frequency that
will make Player B indifferent to calling or folding after he has been
raised. Player B can never raise Player A without disastrous results,
either after Player A bets or check raises. Therefore, analysis only had
to go as deep as check-bet-raise followed by a call or fold. I have only
solved for pot size 1 so far. But, I am working on a general formula for
pot size N. To the three people that still care about this problem, I
will post a full analysis when I get home.

To Tom, I have disagreed and continue to disagree with much that you have
posted here. But, I do not dismiss your opinion as a matter of course.
You pointed out the inconsistency and broken logic of my earlier analysis
and I listened. Again thank you for that.

In the meantime, if any of you would like to show why my new betting
frequencies are correct, I would welcome that input. If, however, any of
you can refute my new strategy, I would welcome that even more. I could
then avoid posting further mistaken analysis. :-)

Howard Lederer

_________________________________________________________________
Posted using RecPoker.com - http://www.recpoker.com

[eleaticus]

What did I miss? How does a hand improve .50 of the time? That is part of
the setup assumptions/imaginings/

Eleaticus

"Howard Lederer" <how...@lvcm.com> wrote in message
news:woQO9.539417$%A6.61...@news.easynews.com...

[Howard Lederer]

If you go to google advanced search, and you put Howard Lederer in the
author line and "contains" Sklansky in the title, you should get all the
relevent threads to this problem. After doing that, you might want to
then look at a post titled "Analysis of Howard's example" by Tom Weideman
from the 24th.

Howard Lederer

[Tom Weideman]

Howard Lederer wrote:

> I would first like to thank Tom for leading me back in the right direction
> on this problem.

Don't thank me yet, heh (see below).

> The artifact left over when the "pot = near infinity"
> meant almost no bluffing, and made it easy to discourage Player B from
> betting. At Tom's urging, I started over and allowed each player to take
> any action allowed in normal limit poker. I think I have solved for
> Player A's optimal action in the 1 bet pot. I now think Player A should
> bet using these frequencies:
>
> Bluff .2
> Check after missing .3
> Check after improving .1
> Bet after improving .4

> This initial action should make Player B indifferent as to calling or
> folding after Player A bets, and betting or checking after Player A
> checks.

When checking after missing, if Player B bets, what is player A's action?
Are we to assume player A always calls but never folds or raises? If so,
then player B can bet whenever he improves and Player A checks to while
folding to a check-raise. He is not indifferent to betting when checked to,
because he loses 1 bet 1/4 of the time (when player A check-raises) and wins
1 bet 3/4 of the time (when Player A check-calls).

If Player A's "check after missing" is with the intention of folding to a
bet, then Player B can exploit by bluffing when he misses and checking when
he hits. Then 3 times out of 4 he will win the 1-bet pot, and 1 time out of
4 he will be check-raised and he can fold, losing 1 bet.

If Player A intends to check raise every single time Player B bets, then
player B can bet and call the raise whenever he improves, winning 2 bets 3
out of 4 times, and losing 2 bets 1 out of 4 times.

The problem is that you need to include all the strategies for Player A,
such as:

"check with the intention of folding after missing"
"check with the intention of raising after missing"
"check with the intention of calling after missing"

(even these only go one round, and don't take into account possible
additional raises - in such problems you really should to make a rule that
limits the number of raises in order to get around this)

and the optimal solution will mix some or all of these "pure strategies" (as
they are generally called). Some strategies are of course "dominated"
(meaning that they will have frequency zero in the optimal solution), such
as:

"check with the intention of folding after improving"

but it is not always easy to prove that a strategy is a dominated choice.
For example, you have stated but have not really proven that Player B can
never raise (or reraise when check-raised). It may come out to be true, and
instinctively it seems like a very bad play, but instincts can be deceiving.
(With the numbers you give above you can show that player B cannot raise a
bet by Player A - he risks 2 bets to win 2 bets but only has a 1/3 chance of
succeeding even if Player A always folds to a raise when he hasn't improved.
But of course you have to assume that the raise frequency is NOT zero, and
then have the solution show you that it is, not the other way around.)

I think you will find that as the number of possible actions grows, it
becomes much more difficult to solve than you first expected. But at least
it should be clear by now why Sklansky chose to simplify it, heh. I hope I
saved you some time on your follow-up post for pot size equaling an
arbitrary number of bets.

Tom Weideman

[JonCooke]

I think there are 12 pure sensible (undominated) strategies for B & 8
for A.

A's pure strategies are:
1. Bet 100%
2. Bet improve - check & call not improve
3. Bet improve - check & fold not improve
4. Bet improve - check raise not improve
5. Check raise improve; Bet not improve
6. Check raise improve; check & call not improve
7. Check raise improve; check & fold not improve
8. Check raise 100%

B's pure strateges are:
Action behind a check if improved (Bet intending to call/check/bet
intending to fold) x Action behind a check if not improved (Bet/check)
x Call a bet or not f improved
All combinations of the above = 3 x 2 x 2 = 12 strategies.

Optimising this means to find a "mixed" optimal strategy for each
player.
The way to solve this formally (without any shortcuts).

First build the pure strategy payoff matrix.

P1 plays
1 2 3 4 5 6 7 8
P2 plays
1 # # # # # # # #
2
3
4
5
6
7
8
9
10
11
12

Where the numbers represent what player 1 earns over player 2 over the
full range of outcomes (here 4 possibilities)

Once you have that matrix (called the normal form of the game) you can
calculate the Nash equilibriums of the game using linear programming.

Fortunately you don't need to know anything about that because there
are engines on the web that do the work for you.

http://levine.sscnet.ucla.edu/Games/zerosum.htm

will do it for you.

[JonCooke]

Solution (I think)
------------------

P1 Pure strategies:
I n
1 C CR
2 C CF
3 C CC
4 C B
5 B CR
6 B CF
7 B CC
8 B B

Column I contains P1 action if he improves, column n if he doesn't.
If P1 improves he will always Check-raise (C) or Bet (B)
If he doesn't improve, he will Check-raise (CR), Check-fold (CF), Bet
out (B) or Check-call (CC)

His opponent has no viable strategy where he can raise a bet (which I
proved for myself but is intuitively obvious anyway)

-----------------------------------------------------------------------------
P2 pure strategies:
Ic Nc Ib Nb
1 Bf/ Bf/ C F
2 Bc/ Bf/ C F
3 C/ Bf/ C F
4 Bf/ C/ C F
5 Bc/ C/ C F
6 C/ C/ C F
7 Bf/ Bf/ F F
8 Bc/ Bf/ F F
9 C/ Bf/ F F
10 Bf/ C/ F F
11 Bc/ C/ F F
12 C/ C/ F F

Ic is his action when he improves behind a check
Nc not improve behind a check
Ib Improves behind a bet
Nb Not improves behind a bet.

Bf/ means bet with the intention of folding
Bc/ bet & call

----------------------------------------------------------------------
The payoff matrix (pot size 1; bet size 1) B result - A result
-----------------

1 2 3 4 5 6 7 8

1 -3 -1 -1.5 -1 -2.5 -0.5 -1 -0.5
2 -1.5 -1.5 -2 -1.5 -0.5 -0.5 -1 -0.5
3 -1.5 -0.5 -1.5 -0.5 -1.5 -0.5 -1.5 -0.5
4 -2 -1 -0.5 -0.5 -2 -1 -0.5 -0.5
5 -0.5 -1.5 -1 -1 0 -1 -0.5 -0.5
6 -0.5 -0.5 -0.5 0 -1 -1 -1 -0.5
7 -3 -1 -1.5 -2 -2 0 -0.5 -1
8 -1.5 -1.5 -2 -2.5 0 0 -0.5 -1
9 -1.5 -0.5 -1.5 -1.5 -1 0 -1 -1
10 -2 -1 -0.5 -1.5 -1.5 -0.5 0 -1
11 -0.5 -1.5 -1 -2 0.5 -0.5 0 -1
12 -0.5 -0.5 -0.5 -1 -0.5 -0.5 -0.5 -1

For example: P1 employs strategy 4, P2 employs strategy 9.
4 situations:
ab (neither improves): P1 bets, P2 folds: result 0 -1 = -1
Ab (1 improves): P1 checks, P2 bets & folds: -1 -2 = -3
aB (2 improves): P1 bets, P2 folds: -1
AB (both improve): P1 checks, P2 checks: -1

Average result for 4 v 9 = -1.5 (see table)

--------------------------------------------------------------------------
Nash Equilibrium:
So, what is player 1's optimal strategy:

-3,-1,-1.5,-1,-2.5,-0.5,-1,-0.5
-1.5,-1.5,-2,-1.5,-0.5,-0.5,-1,-0.5
-1.5,-0.5,-1.5,-0.5,-1.5,-0.5,-1.5,-0.5
-2,-1,-0.5,-0.5,-2,-1,-0.5,-0.5
-0.5,-1.5,-1,-1,0,-1,-0.5,-0.5
-0.5,-0.5,-0.5,0,-1,-1,-1,-0.5
-3,-1,-1.5,-2,-2,0,-0.5,-1
-1.5,-1.5,-2,-2.5,0,0,-0.5,-1
-1.5,-0.5,-1.5,-1.5,-1,0,-1,-1
-2,-1,-0.5,-1.5,-1.5,-0.5,0,-1
-0.5,-1.5,-1,-2,0.5,-0.5,0,-1
-0.5,-0.5,-0.5,-1,-0.5,-0.5,-0.5,-1

Feed into: http://levine.sscnet.ucla.edu/Games/zerosum.htm

1 2 3 4 5 6 7 8

0 0 0 1/8 1/32 11/32 3/16 5/16

This is one optimal strategy for P1.

Value of the game is 0.71875 to P1

[Howard Lederer]

I posted the frequencies of Player A's initial actions only. Of course,
Player A would use game thoery on his check raise frequency. Based on the
frequency of .2 bluff, .3 check, .1 check after inproving, and .4 value
bet, I would have Player A check with the intention of raising 13.33% of
the time. This would include a frequency of 10% "check after improving"
and an "intend to check raise bluff" rate of 3.33%. Which would mean that
Player B would be a 3-1 underdog to win if he called the check raise,
while getting 3-1 pot odds. Of course, he would have to fold if he was
caught bluffing, and then got check raised. I guess I would have to prove
this, but it seems obvious. This would mean that I would have Player A
check with the intention of calling or folding 26.67% of the time. This
would be the 30% check frequency less the 3.33% bluff check raise
frequency. The more I have thought about this 2-1 ratio of "check and
call or fold" to "check with the intention of raising", I think it might
not be optimal for A. I need to do some more work on it. I may not have
the neccesary math skills at this time to fully solve this problem. I
know I am close though.

Howard Lederer

_________________________________________________________________

[Howard Lederer]

Thank you for the link, and the insight. I hope this is the right answer.
I guess I'll spend the next couple of years checking it. ;-) If such a
simple, totally defined situation in poker can be so complicated, I wonder
how an entire poker game could be solved by computer.

Howard Lederer

_________________________________________________________________

[Tom Weideman]

JonCooke wrote:

> Solution (I think)
> ------------------
> P1 Pure strategies:
> I n
> 1 C CR
> 2 C CF
> 3 C CC
> 4 C B
> 5 B CR
> 6 B CF
> 7 B CC
> 8 B B
>

> 1 2 3 4 5 6 7 8
> 0 0 0 1/8 1/32 11/32 3/16 5/16
>
> This is one optimal strategy for P1.
>
> Value of the game is 0.71875 to P1

Now why would you want to go and spoil Howard's fun?

Anyway, to translate here...

For every 32 hands played, there will be 8 where Player A improves while
Player B does not, 8 where Player B improves while Player A does not, 8
where both improve, and 8 where neither improves.

Player A will employ the "check-raise if I improve, bluff if I don't"
strategy 1/8 of the time overall. Since he improves half the time, and this
is the only strategy that includes check-raising, it means he check-raises
for value a total of 1/16 of the time overall.

Player A employs a "bet if I improve, <other> if I do not", 7/8 of the time.
He improves 1/2 the time, so he bets out for value 7/16 of the time. Thus
Player A check raises for value only 1/7 as often as he bets out for value,
which is quite a departure from the "infinite pot" ratio of 1/2. Obviously
the smaller the pot is the less one has to "discourage" the opponent by
throwing in some check-raises. Obviously this leads to greater EV for
Player A than when he needs to check-raise more often.

Player A bluffs in pure strategies 4 and 8, which sum to 7/16. He only
fails to improve 1/2 the time, so the total frequency that he bets as a
bluff is 7/32. We have already ascertained that he bets out for value with
a total frequency of 7/16, so Warren's partial answer that Player 1 bluffs
half as often as he bets for value is confirmed. This is as it should be,
since raising has been eliminated as an option for Player B, so when facing
a bet Player B is made indifferent to his only remaining options, calling
and folding.

Now suppose Player B has been check-raised after betting for value. This
happens 1/32 of the time overall when both players have improveds improved
(Player A's pure strategy #4 happens 1/8 of the time overall, and both
players improve 1/4 of the time), and 1/128 of the time when Player A has
not improved while Player B has improved. This means that it is 4 times
more likely that Player A has him beat than the reverse, and of course at
that point in the hand Player B is faced with a bet with a pot of 4 bets, so
he is indifferent to calling the check-raise.

Now think about when Player B is checked to and is considering betting as a
bluff. The total fraction of times that Player A checks is: 1/16 + 1/64 +
11/64 + 3/32 = 22/64. Player A will check-fold a total fraction of 11/64,
and when he does Player B will win the 1 bet that is in the pot. When
Player A does not fold (either raises or calls), Player B loses 1 bet.
Player A check-folds 11 out of 22 times that he checks, so Player B wins 1
bet half the time and loses 1 bet half the time, making him indifferent to
bluffing when checked to.

Okay, what about Player B betting for value? This one is a bit tricky.
Player B has the option of checking it down here, so what he gains or loses
must be compared to what he would have done if he had checked it down.
[Actually this was also true for the case above where Player B was
considering bluffing, but checking down an unimproved hand had an EV of
exactly zero, so it was no big deal.] The four possible scenarios that can
occur with Player B betting for value include Player A pure strategies 4
through 7. The probabilities that each occurs is 1/4 times the frequencies,
since Player A has to either improve (in the case of #4) or not (5 through
7), and Player B must improve, and these all have probabilities of 1/2. We
have already ascertained that Player B is indifferent to calling when his
value bet is check-raised, so we will assume he folds to all check raises
for simplicity.

case 1: Player 1 improves and check raises Player B's value bet
probability of occurring = 1/32, Player B result = -1 bet
(Player B would have won 0 bets if he had checked, but lost 1 when he bet.)

case 2: Player 1 does not improve and check raises Player B's value bet
probability of occurring = 1/128, Player B result = -2 bets
(Player B would have won 1 bet if he had checked, but lost 1 when he bet.)

case 3: Player 1 does not improve and check-folds to Player B's value bet
probability of occurring = 11/128, Player B result = 0 bets
(Player B would have won 1 bet if he had checked, won 1 when he bet.)

case 4: Player 1 does not improve and check-calls to Player B's value bet
probability of occurring = 3/64, Player B result = +1 bet
(Player B would have won 1 bet if he had checked, won 2 when he bet.)

The total extra EV that Player B achieves by betting for value instead of
checking is:

(1/32)*(-1) + (1/128)*(-2) + (3/64)(+1) = 0 bets

Therefore he is indifferent to betting for value compared to checking it
down.

I hope everyone who has read to here was able to follow all that, heh. I
have demonstrated that player B is indifferent to both bluffing and betting
for value (vs. checking) when checked to, so NOW we can just assume that he
always checks when he is checked to. I have also shown that Player B is
indifferent to calling or folding when bet into, so we can assume he folds
every time he faces a bet. Doing these things allows us to determine Player
A's EV easily:

Since Player B folds to any bet and checks behind all checks, Player A wins
1 bet every time he has the best hand (3/4 of the time), AND he wins 1 bet
whenever he bluffs with the worst hand (which he does 7/64 of the time - he
bluffs 7/32 of the time, and has the worst hand half those times), so his
pot equity is:

3/4 + 7/64 = 55/64 = 0.859375

Why doesn't this match Jon's number given above? I assume it's because
Jon's number takes into account that the two players anted equal amounts
(1/2 bet each), so whatever amount they win above 1/2 bet is profit. Player
A's expected profit is 0.859375 - 0.5 = 0.359375, which is exactly 0.71875
of the 1/2 bet that Player B anted to the pot.

One other way to look at this EV to compare it to what the "fair" EV is,
based on just letting the cards decide. Player 1 expects to get 3/4 of the
pot in this case, so the strategy portion of the game accounts for 7/64 of a
bet of his EV. In the infinite pot case, this marginal amount was 8/64.

Okay, I'm done being so forthcoming for awhile - it's just too much work.
Happy Holidays.

Tom Weideman

[Howard Lederer]

"Tom Weideman" <zwi...@attbi.com> wrote in message
news:BA31FCBB.22885%zwi...@attbi.com...

> Now why would you want to go and spoil Howard's fun?

Not too far off. I was definitely enjoying the journey here. I felt like I
was learning quite a bit while I splashed about on this problem. You will
notice that I was headed in the right direction.

> Anyway, to translate here...
>
> For every 32 hands played, there will be 8 where Player A improves while
> Player B does not, 8 where Player B improves while Player A does not, 8
> where both improve, and 8 where neither improves.
>
> Player A will employ the "check-raise if I improve, bluff if I don't"
> strategy 1/8 of the time overall. Since he improves half the time, and
this
> is the only strategy that includes check-raising, it means he check-raises
> for value a total of 1/16 of the time overall.

So he checks after improving 2/32 of the time.

> Player A employs a "bet if I improve, <other> if I do not", 7/8 of the
time.
> He improves 1/2 the time, so he bets out for value 7/16 of the time. Thus
> Player A check raises for value only 1/7 as often as he bets out for
value,
> which is quite a departure from the "infinite pot" ratio of 1/2.
Obviously
> the smaller the pot is the less one has to "discourage" the opponent by
> throwing in some check-raises. Obviously this leads to greater EV for
> Player A than when he needs to check-raise more often.

> Player A bluffs in pure strategies 4 and 8, which sum to 7/16. He only
> fails to improve 1/2 the time, so the total frequency that he bets as a
> bluff is 7/32. We have already ascertained that he bets out for value
with
> a total frequency of 7/16, so Warren's partial answer that Player 1 bluffs
> half as often as he bets for value is confirmed. This is as it should be,
> since raising has been eliminated as an option for Player B, so when
facing
> a bet Player B is made indifferent to his only remaining options, calling
> and folding.

OK, he bets for value 14/32 and bluffs after missing 7/32 of the time. This
leaves checking after missing 9/32 of the time.

> Now suppose Player B has been check-raised after betting for value. This

> happens 1/32 of the time overall when both players have improved improved

> (Player A's pure strategy #4 happens 1/8 of the time overall, and both
> players improve 1/4 of the time), and 1/128 of the time when Player A has
> not improved while Player B has improved. This means that it is 4 times
> more likely that Player A has him beat than the reverse, and of course at
> that point in the hand Player B is faced with a bet with a pot of 4 bets,
so
> he is indifferent to calling the check-raise.

Now I am completely lost. I would have assumed that Player A would raise
with an improved hand three times as often as he bluffs. If I am not
mistaken, there are three bets in the pot after a check raise, not four.

<snip the rest> while I wait for a reply

Howard Lederer

[Tom Weideman]

Howard Lederer wrote:

> Now I am completely lost. I would have assumed that Player A would raise
> with an improved hand three times as often as he bluffs. If I am not
> mistaken, there are three bets in the pot after a check raise, not four.
>
> <snip the rest> while I wait for a reply

ante + Player B's bet + Player A's 2 bets = 4 bets

Tom Weideman

[Howard Lederer]

"Tom Weideman" <zwi...@attbi.com> wrote in message

news:BA330C2D.22963%zwi...@attbi.com...

Wow, what a brain lock. I spent hours thinking about this problem with
three bets in the pot after a check raise. Thanks.

Howard Lederer

[JonCooke]

> Now why would you want to go and spoil Howard's fun?

Well, I enjoyed tackling this, too, and I thought Howard might be
interested in the formal method. I get a lot out of reading his posts.

> 3/4 + 7/64 = 55/64 = 0.859375
>
> Why doesn't this match Jon's number given above? I assume it's because
> Jon's number takes into account that the two players anted equal amounts
> (1/2 bet each), so whatever amount they win above 1/2 bet is profit. Player
> A's expected profit is 0.859375 - 0.5 = 0.359375, which is exactly 0.71875
> of the 1/2 bet that Player B anted to the pot.

Right.
I quoted P1's supremacy over P2 per hand.

[Mark B [Diputsur]]

HOly shit, there was a part 2! And my 'bet' response was apparently the right answer 60% of the time according to Howard! And I didn't even have to do all that math to figure it out! WHo's retarded now? woot woot! :) Dip