WAY tougher two round game theory problem

[David Sklansky]

Playing pot limit draw or lowball. Both players ante $100. Player A has a good pat hand. Player B has a 20% chance to draw one and beat it. Again there are two rounds of betting. But this time Player B gets to draw a second time if he misses! We will assumme independence on the draw for simplicities sake which means that Player B will nake his hand 36% total if he always draws.

So what is the best strategy and the expectations for both players?

Why I think this is especially hard is because Player B may be correct in sometimes standing pat as a bluff after the first draw.

PS Assume player A is first and checks. Almost certainly that is correct even on the first rouund

PPS Pink Bunny and BB were right about the other one.

[onlinechamp]

Ok you got me this time sklansky I am stumped maybe fishy can bail me out this time!

[David Sklansky]

Notice that no one can say that this question is theoretical only. There have been some pot limit triple draw loball games spread recently.

[onlinechamp]

Hi david do you know who I am?

[onlinechamp]

No use in real life huh David?

[Fishy]

LOL!

Draw poker and lowball is a well erm....drawing game, very automatic. It's no fun, Champ stay away from boring games, let's go play somewhere else :P

Time to get back into the internet page clown zone:)

[ThePinkBunny]

Does the betting take place after each drawing round or after both are complete?

Also, where's your answer to the 7 card stud 8 or better problem?

D

[jaytorr]

First time poster, long time lurker. Interesting problem so I'll take a shot at it.

When player B makes his hand on the 1st draw (20% of the time), he obviously stands pat on the 2nd draw and bets for value. Therefore on the 2nd draw he should also stand pat and bluff with 10% of his hands to keep the 33% bluffing frequency after standing pat.

Now, let's say that he bluffs after the 1st draw with frequency x. That means that he will draw a 2nd time with(x-0.10) of all hands. He should make his hand 20% of the times he draws a 2nd card, and should bluff an additional 10% of the times he draw a 2nd card:

Bets 2nd draw for value: 0.2*(x-0.10)
Bluffs after missing 2nd draw: 0.1*(x-0.10)

We also know that on the 1st round, he must bet 1.5 times the number of times he bets on the 2nd round (this was shown in the previous problem), therefore:

x + 0.2 = 1.5*(0.2+0.1)*(1+(x-0.10))

x = 37.3% (1st round bluffs)
Player A's EV = -$14.55
Player B will draw a 2nd card: 27.3%

[David Sklansky]

There is a pot size bet allowed after each round. So if there was no bet on the first round the second round bet would be $200 rather than $600.

I'd take the 763suited almost for sure.

[onlinechamp]

Talking to me?

I don't need quizes I prove it on the poker tables!

[BB King's]

Same solution as the simpler one, but in this one A should only call 35 % of the time on the first betting-round.
EV(B)=123 - EV(A)=77.

I think. I'm working on an explanation - if someone is interested ?

[ThePinkBunny]

Does player B draw one card, then draw a second one (if he decides to), then have two rounds of betting?

Or...

Does player B draw one card, round of betting, draws again (if he chooses to), and then the second round of betting?

Thanks,
D

[BB King's]

Does player B draw one card, then draw a second one (if he decides to), then have two rounds of betting? This would be trivial !!!

Does player B draw one card, round of betting, draws again (if he chooses to), and then the second round of betting? Yep - this is the one.

[jaytorr]

In my original post I forgot that when Player B decides to check on the first round, he gets a free card and can still win the pot by making a hand or bluffing after the 2nd draw. This doesn't change the 1st round bluffing frequency but it makes a huge difference in overall EV.

When checking on the 1st round, player B should still bluff 10% of the time on the 2nd round since he will be making his hand 20% of the time.

Correction to the solution in the previous post:

x = 37.3% (1st round bluffs)
Player B bluffs 1st round, then draws 2nd round = 27.3%
Player B checks 1st round: 1-0.373-0.2 = 42.7%
Player B checks 1st round, bluffs 2nd round: 0.1*0.427=4.3%

Probability that Player B bets in *at least one* round
= 0.2+0.373+0.3*0.427 = 70.1%

Since Player A does no better by calling any bets,
his EV = -100*(0.701)+100*(0.299)= -$40.18 Ouch!

[ThePinkBunny]

I probably could have figured that out for myself, had I spent a few minutes thinking about it.

D

[ThePinkBunny]

Player B bets the 20% of the time he makes his hand and bluff 27.65% of the time. He'll stand pat on half as many bluffs as his made hand (10% overall) and bluff again.

So now he's going to draw on that remaining 17.65% (overall) which were initial bluffs, betting the times he makes his hand, and half as many times as that that he doesn't.

So then on the remaining 52.35% of the hands where he didn't bet initially, he'll obviously draw the second time around betting his made hands and still bluffing once for every two made hands.

All of this makes player B's expectation $26.72. Player A's would obviously then be -$26.72.

Player A plays the same as in the other problem, flipping a coin as BB King put it.

-D

PS: If you're wondering where the numbers come from. Notice that 17.65% is exactly half of the percentage of the time he bets on both rounds.

PPS: If this is right, David, then I think I deserve a cookie or something.

[ML4L]

This might be a little ambitious for my first post, but I think I might have gotten the answer (or at least a viable method). I have yet to look at others' answers, so I'm very curious about how I did...

If A were ignorant, the best strategy would be for you, 30% of the time, to bet on the first round (every time that you made your draw, and 10% of the time bluffing that you made your draw), to stand pat after A calls, and then to bet the second round. For the other 70% of the hands (having never made your draw any of these times), you would check behind A in the first round, and then bet 30% of the time on the final round (again, 20% would be value betting after making your draw and 10% would be bluffing, offering A 2-1 odds on his call to match the 2-1 pot odds). This would yield (using net profit rather than gross win/loss) an EV for player B = .2(900) + .1(-900) + .14(300) + .07(-300) + .49(-100) = +62

However, if A knew that this was your strategy, he would never call when you bet on the first round because he will only have a break-even proposition on the second round, meaning that he loses the entire $200 on the first betting round. So, when you bet the first round, rather than always standing pat, you have to follow your bet up with a draw often enough for it to be correct for A to call your first round bet. If A were to just fold every time you bet, his EV would be -100. His EV every time that you bet and stand pat (assuming you have it 2/3 of the time and bluff the other 1/3) is -300; his EV every time that you bet and draw is +120 (this is .2(-900) + .1(900) + .7(300)). Taking these two scenarios, you can induce him to have an EV of –100 by drawing 10 times for every 11 times you stand pat ((10 * 120 + 11 * -300) / (10 + 11) = -100). Thus, it doesn’t matter if he folds or calls when you bet out on the flop.

So, to summarize... (assuming that A always calls, because he is indifferent)

20%: Make draw, bet, stand pat, bet, win 900
10%: Miss draw, bet, stand pat, bet, lose 900
5.45%: Miss draw, bet, make draw, bet, win 900
2.73%: Miss draw, bet, miss draw, bet, lose 900
19.09%: Miss draw, bet, miss draw, check, lose 300
8.55%: Miss draw, check, make draw, bet, win 300
4.27%: Miss draw, check, miss draw, bet, lose 300
29.91%: Miss draw, check, miss draw, check, lose 100

EV for player B = 40.18 (you also get this number by assuming that A always folds to a bet).

ML4L

[jaytorr]

In your strategy, player B bets 47.65% on the 1st round and he bets both rounds 35.3%.

Assuming that Player A folds to any 2nd round bet (EV=0), he would need almost 3:1 odds to call the first round bet (he wins pot 12.35%, loses 1st round bet 35.3%). Since player A is better off by never calling the 1st round bet, player B should bluff more often on the 1st round.

[ThePinkBunny]

You're right. I guess I really shouldn't have been trying to figure this out while playing 5 games online. Stupid mistake.

Let's try this again.

Player B bets the 20% of the time he makes his hand and bluff/semibluff 37.27% of the time. He'll stand pat on half as many bluffs as his made hand (10% overall) and bluff again.

So now he's going to draw on that remaining 27.27% (overall) which were initial semibluffs, betting the times he makes his hand, and half as many times as that that he doesn't.

So then on the remaining 42.73% of the hands where he didn't bet initially, he'll obviously draw the second time around betting his made hands and still bluffing once for every two made hands.

All of this makes player B's expectation $40.18. Player A's would obviously then be -$40.18.

Player A plays the same as in the other problem, flipping a coin as BB King put it.

-D

[ML4L]

Out of curiosity, did ThePinkBunny and I get the correct answer (EV of player B = 40.18)?

ML4L

[ThePinkBunny]

well at least I'm pretty sure of it. But obviously I do make mistakes. jaytorr also got the same answer.

D

[BB King's]

Forget about the above - here is a better try !?!

---

When B misses his draw he sometimes decieds to bet anyway as a semibluff - call that number b. Sometimes he will draw - sometimes not - so b=d+n.

If he is drawing: EV(d)= +180

If he is not drawing he is risking 600 :

q2*1200-600=180

Solving for q2 (the folding-frequency): q2=65 %

This is UGLY: On the second round of betting (when there were a bet in the first round) A should only call 35 % of the time !??!?

That's why - just has to: n=10

When B misses his first draw he could simply take a free card: EV= +60

Compare that to the semi-bluff:

q1*400+(1-q1)*600*0.2*1.5-200=60

Solving for q1 (the folding-frequency): q2=36.36 %

!??!?

We need to calculate d:

0.7*d*600-200*(20+n+d)=0 -- d= 27.27 %

---

It somewhat looks like what other posters had come up with but not quite !??!?

[BB King's]

... mention how A should play ??? Nice job though - so far !

[ML4L]

Well, I haven't really played with the problem to see if it would ever be correct for A to bet out, but my instincts say that it can't be. If I try to use specific theory to back that claim, I'll butcher it, so I'll just give my general line of reasoning. On the first round, a bet by player A would be similar to increasing the ante, which would merely multiply B's edge. The situation that player A is about to enter into is negative EV; why would he want more money in there?

On the second round, what does player A hope to accomplish by betting? He doesn't have to worry about being "bluffed out" on the end, because player B only bluffs enough to make it correct for player A to call when player B has made his hand. And, betting out does not prevent player B from bluffing on the end; it merely makes it more expensive for player A to call (or fold, since player B's strategy makes player A indifferent). Thus, the situation that player A faces on the second betting round is analogous that which he faces on the first betting round: a bet by player A would merely be multiplying his expected loss.

That explanation might not be perfect, but I'm nearly positive that player A should never bet (someone please correct me if I'm mistaken).

ML4L

[Utah]

I havent read anyones responses yet.

My answer is that the value of the game is $18.08 to player B.

There are three player B strategies

(Bet, Draw (bet,fold)
(Bet, Bet)
(Bet, Draw (Bet, Bet)

The first part relates to the first draw and the second part relates to the second draw. For example, The first strategy is bet if you hit, redraw if you miss, bet if you hit, fold if you miss).

All other strategies are dominated (e.g., you wouldn't redraw if you hit)

Player A has four Strategies

Call First, Call Second
Fold First, Fold Second
Call First, Fold Second
Fold First, Call Second

The Matrix Looks Like:

_______All Fold___All Call____Call First/Fold Second___Fold First/Call Second
BDBF______-28________260____________132_____________100
BB________100________-540___________-540____________100
BDBB______100________-252___________260____________-412

Equalize all strategies for B using p1,p2,p3
(i.e. set -28p1+100p2+100p3 = 260p1-540p2-252p3 = 132p1-540p2-260p3 = 100p1+100p2-412p3)

P1 = 64%
p2 = 20%
p3 = 16%

Value of the game for B = 18.08

Equalize all strategies for Player A using Q1,Q2,Q3,Q4

Same method

Q1 = 69%
Q2 = 1.5%
Q3 = 11.3%
Q4 = 18.5%

Value of the game for A is -$18.08

I didn't include all the math here as it got complicated nd I am sure I made mistakes. However, everything tied out perfectly.

[BB King's]

... how often should he call ???

[Utah]

I have read the other posts and I think I misunderstood the betting. Here is my answer under the assumption that cards can be received after 1 round of betting - i.e. draw, bet, draw, bet. This scenario is much more complicated

Player B logical options:

BDBF Bet if hit, Draw if miss (bet if hit, fold if miss)
BDBB Bet if hit, Draw if miss (bet if hit, Bet if miss)
BBB Bet if hit, bet twice if miss
BBDBF Bet if hit, Bet once if miss, Draw (bet if hit, fold if miss)
BBDBB Bet if hit, Bet once if miss, Draw (bet if hit, bet if miss)

Player A Options:
AF All Fold
AC All Call
CAFFS Call any bet after first draw, fold to any bet second draw
CFFCS Call one bet only first round, Call all bets after second draw
CFFFS Only call first bet in first round
FFCS Fold to all first round bets and call all second round bets

Breaking the decision tree down to form you get:

_______AF______AC____CAFFS___CFFCS____CFFFS______F FCS
BDBF___-28____260_____132_____140______12________100
BDBB___100____-252____260_____-372_____140_______-412
BBB____100____-540___-540_____300______300_______100
BBDBF__100____132_____36_______12______-84_______100
BBDBB__100____-252____420______-372____300_______100

Each point of the grid is calculated by taking the value to B of the two strategies. AF V BDBF = .36*100-.64*100 = -28

Domination:
Strategy BBDBB eliminates BDBB (meaning BBDBB is always better than BDBB
Strategy AF Eliminates FFCS

Equalize all the strategies (ie., make it so it doesnt matter what strategy opponent chooses)

BDBF = 39.17%
BBB = 12.5%
BBDBF = 35.76%
BBDBB = 12.57%

AF = 46.67%
AC = 8.43%
CAFFS = 10.24%
CFFCS = 18.24%
CFFFS = 16.43%

The value of the game to B is $49.87. I am sure this is wrong as there is just too much math or I missed a viable option

[ML4L]

My mistake; I thought that I had made A's strategy clear. In terms of expectation, it doesn't matter whether A calls or folds. Player B checks and bets in specific proportions so as to make A completely indifferent as to whether he calls or folds. This is how player B attains the maximum edge; it is impossible for A to adjust his play to counter player B's strategy. You can double-check this; using player B's optimal strategy, calculate player A's expection if he always calls vs. if he always folds (and, if you want to make the effort, if he sometimes calls and sometimes folds). Player A should always have the same expectation.

Regarding Utah's matrix solution, the methodology is unfortunately too complicated for me to fully understand, but I can say that, generally, the solution to any problem such as this will involve complete indifference by player A. If player A can somehow affect player B's edge by playing a certain way, then player B is not playing his optimal strategy.

Again, someone please correct me if I am mistaken about any of this.

ML4L

[BB King's]

... but ...

Player A plays the same as in the other problem, flipping a coin as BB King put it.

... I strongly believe this statement is wrong !?!

[ML4L]

BB,

Could you show us why you don't agree? Again, using the strategy that a few of us agreed upon, calculate player A's EV under any mix of calling and folding. It should always be the same.

Whether it is ever correct for player A to bet or not, I'm less positive of, as I don't have math to back that up. That being said, I am still fairly certain that it would never be correct for him to bet...

ML4L

[ThePinkBunny]

thought that I had made A's strategy clear. In terms of expectation, it doesn't matter whether A calls or folds.

Yes it does matter what A does. Otherwise B could counter and gain a bigger expectation.

D

[ThePinkBunny]

player B is still bluffing at the same ratio as the other problem using game theory, wouldn't it make sense that player A still calls at the same ratio?

D

[Utah]

ML4L is correct.

If, B sets his strategy correctly (ie, chooses the correct percent to use each strategy), it makes zero difference what A does because the expected payout is exactly the same regardless. This is the Minimax Theorem and the principle of indifference.

Logically it makes perfect sense. If B is not making all strategies equal for A, then A can simply adjust his strategy to the path with the best payout for A.

This can also easily be proven mathematically. Take my matrix (and forget about the complexity) Take player Bs calculated % to use each strategy and calculate down each column of the matrix. You will find the answer comes out exactly the same for each column. Therefore, A is indifferent.

Player A does the exact same thing and makes B indifferent. This way you arrive at the optimal strategies.

Obviously, player B can adjust if A is not playing optimally.

Note: my answer below is slightly wrong because I had the wrong betting assumption (and I missed some options - A has 10 pure strategies, not 6). I thought you could bet twice on the last round if no betting took place on the first round, like the previous game theory question.

If you can only bet once on the last round if checked originally, the previous answer of $40.18 is absolutely correct and congrats to the 2 guys who solved it first. The game theory proof is straight-forward (I will try to post a simple version this week). I have also run the proof showing that A should never raise B.

[Utah]

Personally, I think it is important to use a standard game theory matrix as it gives you a model for solving all game theory problems and therefore makes both iteration analysis (e.g., evaluate if A raises) and future problems much easier.

I was having trouble with the matrix math so I contacted a UCLA game theory professor on how to solve the math for a large matrix. Instead, he simplified the answer. His response is below. If you are interested, his website game theory information is at http://www.math.ucla.edu/~tom/math167.html

Tom Ferguson, UCLA Game Theory Professor, Answered (printed with his permission):

If Player I checks in the first round, both players know the position.
Therefore this branch of the game tree may be solved independently. In
addition this branch is just Basic Endgame in poker, and is easily seen to
have value -40. Player II calls with probability 1/2 and folds with
probability 1/2. With a losing card, Player I bets with probability 1/8
and folds with probability 7/8. (Player I always bets with a winning
card.) This branch of the game tree may be replaced by -40.

Similarly if Player I bets and draws on the second round, both players again know
the position. This branch also is just Basic Endgame with the same optimal
strategies and with value -120, 3 times that of the above paragraph. This
branch may be replaced by -120.

The remaining tree reduces to a three-by-three game.

Player I's pure strategies are
1. Check: check on the first round (with a losing card)
2. Bet,stand: bet on the first round, and bet on the second without drawing.
3. Bet,draw: bet on the first round, draw on the second round with a
losing card on the first.

Player II's pure strategies are
1. Fold: fold a bet on the first round.
2. Call,call: call on the first round, and on the second if I stands pat
and bets.
3. Call,fold: call on the first round, and fold the second round if I
stands pat and bets.

I find the game matrix of expected payoffs to be

Fold Call,call Call,fold
Check -12 148 28
Bet,stand 100 -540 300
Bet,draw 100 84 -36

This should probably be checked, but supposing it correct, I ran it through
the Matrix GameSolver and found

Value = 40.182
Optimal for I = (.53409, .12500, .34091)
Optimal for II = (.36364, .22273, .41364)

In the complete game, Player 1 has 5 pure strategies and I believe they
are the same as those you found. In your terms, the optimal strategy is:

P(Bet,Redraw[Bet,Fold]) = .53409 times 7/8 = .46733
P(Bet,Redraw[Bet,Bet]) = .53409 times 1/8 = .06676
P(Bet,[Bet,Bet]) = .12500
P(Bet,[Bet,Draw(Bet,Fold)]) = .34091 times 7/8 = .29830
P(Bet,[Bet,Draw(Bet,Bet)]) = .34091 times 1/8 = .04261

In the game as I have described it, Player II has more than 6 pure
strategies (I count 10) because she gets to hear if Player I draws on the
second round after betting on the first.

The first and the third probabilities above are the same as the
probabilities P you suggested in your e-mail.

[ThePinkBunny]

While everything you both have said is correct, A needs to use his correct optimal strategy or player B could theoretically adjust to him.

So while you don't need to find it to answer the expected value questions, it is important part of the whole answer to this problem.

D

[BB King's]

player B is still bluffing at the same ratio as the other problem using game theory, wouldn't it make sense that player A still calls at the same ratio? No !

Why don't you read my post Ugly Solution/Explanation ! where I explain A's calling-strategy ?

[BB King's]

A will only make money when B is not betting ...

EV(A)= (1-0.5727)*(1-0.3)*200 = 59.82

EV(B)= 200- EV(A) = 140.18

Notice: EV(B) is not +40.18

[ThePinkBunny]

Your solution is rather muddled, but let me try to point out where you might be going wrong.

This is UGLY: On the second round of betting (when there were a bet in the first round) A should only call 35 % of the time !??!?

Ok, here is where there might be a problem. There are two possibilities for the second round.

One where A draws again, and one where A stands pat. So does he call 35% in both cases?

I think if you do a little more thinking here, you'll figure it out.

D

[ThePinkBunny]

Notice: EV(B) is not +40.18

Yes is it. The two hundred dollars in the pot initially wasn't donated by Bill Gates. They each invested $100 in antes.

D

[ThePinkBunny]

BB King got me thinking. I thought A's strategy would be just to call 50% of the time on every round.

And I'm pretty sure this is correct, except for the case where B stands pat after the first draw.

A must call at a different frequency in this case in order to 'force' B to stand pat with a bluff a certain percentage of the time.

In other words, A needs to make sure standing pat and bluffing is worth the same to B as drawing to hit his hand.

Since drawing to hit his hand is worth a 30% equity (hitting 20% and bluffing 10%) for B. A can make the stand pat bluff worth the same for B by calling only 35% of the time B stands pat after his initial draw.

D

[ThePinkBunny]

See my 'rethinking...' post below.

I think see what part I was wrong, but your answer is so confusing I'm not sure if we now agree or not.

D

[BB King's]

Could be your explaination is better than mine ?!

Now check out the first round - hope I'm not sounding to bossy /forums/images/icons/laugh.gif

[Utah]

I think this is a little bit nonsensical. The reason is that you are making assumptions as to what strategies A and B employed earlier. This is different that solving the "drawing" branch because the "drawing" branch conveyed concrete information (i.e., B did not hit his hand on the first draw). This branch is therefore simple to solve.

I calculate 35% as well. However, your must assume both players are playing optimally. The reason being that you otherwise have no clue as to B's holding when he stands pat. For example, B could be playing a strategy where he only stands pat when he has made his hand. In that case, A would never call.

Since we are making this assumption, it is simple math to calculate the % and is not really a calculation on this particular branch but rather it is a calcuation taking into account the overall strategy - which you had already solved.

[ThePinkBunny]

Not sure if I understand what you're trying say.

The point was that we're suppose to solve for A and B's strategy, which by each of them using game theory, it doesn't matter what the other does, his expectation will remain the same.

A flipping a coin when B stands pat after the first draw was exploitable. A calling 35% of the time is not.

D

[Utah]

How do you know it is exploitable? What if B stood pat with a miss equal to the amount of time he stood pat with a made hand? How do you know the true optimum percent? Because, you solved the overall strategy.

Simply put - you cannot solve this discreetly (whereas you can solve the problem of the "drawing" branch discreetly regardless of the preceeding A/B actions employed).

Certainly, you can calculate the percentage that A folds or raises - as you did. But this is a simple exercise in math - not strategy).

For example, lets say the EV of the drawing branch was positive $40 instead of negative $40. Almost certainly, B would check more when he missed versus the current optimum strategy. Since he would check more if he missed, he is more likely to have a made hand if he stands pat and bets. If he is more likely to have a made hand, A will fold more often when B stands Pat.

Do you see in this case how the other branch affects this branch? Additionally, the stand pat branch affects the "drawing" branch. Therefore, you need to solve the entire strategy right up front in this case.

[ThePinkBunny]

How do you know it is exploitable?

Because it is easy to see how B can adjust from his game theory strategy if A were to call fifty percent of the time.

What if B stood pat with a miss equal to the amount of time he stood pat with a made hand?

It doesn't matter if A calls 35% of the time B stands pat. Their expectation won't change.

I'm not solving the problem trying to find either A or B's optimum strategy depending on what the other does. I'm solving it according to game theory, meaning I'm trying to find A's best strategy which is independant of what B does.

D

[BB King's]

EV(B)= 140.18

[Utah]

Okay, I thought this was mind-blowing! I contact a complete stranger on the internet because I found his game theory notes to be oustanding and I needed help with a matrix. I figured there was about a 10% chance that this guy would respond.

He does respond and I exchange several interesting emails with Professor Tom Ferguson, Game Theory Professor at UCLA. Then I get an email from him yesterday saying that he is Chris "Jesus" Ferguson's father.

Totally cool. Helps explain why Chris "Jesus" is so awesome at this stuff (as you might recall, David Sklansky said Chris got the first problem in under 10 minutes).

Here is the copy of the email he sent telling me he was his father. He talks about some very cool game theory stuff he solved with his son (printed with the permission of both Tom and Chris "Jesus" Ferguson):

Hi! It's me again.

It had slipped my mind that I had worked on one of
these problems before with my son, Chris "Jesus".
(You may have heard of him. He won the 2000 World
Poker Championship.) The problem in question is:

_________________________________________________


>This only adds a little more complexity. The
>interesting thing is that with a two round betting
>scheme with no draw, a single round bluff (i.e, bet
>once and give up if called) increases player 1's
>expected value. I am trying to think through and
>really game application of this.

This is an important practical problem, especially in
a game like limit hold'em. After the flop, you have a
very good hand but he may have the nuts. He bets the
limit and if he has the nuts, he can do this two more
times if you call. With what probability should you
call once, twice or three times? One should be able
to solve a theoretical version of this problem with an
unlimited number of rounds to get an idea of what the
players should do in real situations.

__________________________________________________ ___

We solved this problem with n rounds when the sizes of
the bets for the various rounds are fixed. This would
allow treatment of limit and pot-limit games, by
finding the optimal sizes of the bets on the various
rounds. (If the players are infinitely wealthy,
this would be to bet the maximum at each round.)
We applied it to the treatment of no-limit
(table-stakes) games. One interesting thing we
found was as follows. If there are just two rounds,
how much should Player I bet on the first round?
In a no-limit game, Player I will go all-in in the
second round. The answer is: SquareRoot(M+a) - a.
Here a = ante = (pot size)/2, and M = the minimum of
the sizes of the fortunes of the two players at the
start of betting. (So M is essentially the maximum
total amount that can be bet.) One interesting
thing is that this bet size is independent of the
probability that Player I gets a winning card!

We did a few other things such a look at the problem
with a continuum of rounds, and also the problem where
Player II's hole card gives him some information on
whether or not Player I got a winning card.

Unfortunately. I've never found the time to polish it
up to have it published. Maybe I now have an extra
incentive.

Tom.

[BB King's]

My mistake; I thought that I had made A's strategy clear. No - you have not !!!