Game Theory Question

[David Sklansky]

Playing a strange form of headup pot limit poker. Player A has a good unimprovable hand. Player B has a hand where his last card has a 24% chance to improve to the winnner. If he does, or if he bluffs, he can bet the pot. But if called he can do it AGAIN, (even though he receives no more cards).

Both players ante $100 and there is no betting until these last rounds. So Player B can bet $200 and if called then bet $600. Player A can call, raise, or fold, the first and second round. Both players have a billion dollars in front of them

A. What is the worst that Player A's expected value can be if he plays properly (eg positive $17 per hand, negative $5 per hand etc)?

B. How much better or worse (EV wise)would it be for Player B to be playing no limit with only one round of betting?

[Mike Gallo]

This question is way out of my realm of mathematics, however I would like to see the answer so I can at least attempt to understand game theory and the question.

Thanks

Michael

[Mason Malmuth]

Hi Everyone:

David just called me and there was an error in his original post which has now been corrected. So if you read his question before my post it has been changed.

Best wishes,
mason

[David Sklansky]

It took him ten minutes. So could Chris Ferguson, Mark Weitzman, Barry Greenstein, and Jimmy Warren. It's not a coincidence that they are all millions winner.

[Ed Miller]

Player B should make the first bet with frequency X such that, when called, Player A's equity in the called ($600) pot is equal to (1-X)*$200 + $200 (i.e. it doesn't matter whether Player A calls or folds to the initial bet).

In the called pot, Player B will have the best hand 0.24/X of the time, so his equity in the called pot is 1.5 * 0.24/X * $600 (assuming he bluffs appropriately). Therefore, Player A's equity in the called pot is (1 - 0.36/X) * $600.

Thus, (1 - 0.36/X) * $600 = (2-X) * $200
3X - 1.08 = X(2-X)
X^2+X-1.08 = 0

X = (-1 + (1 + 4.32)^1/2) / 2
X = 0.653

Thus, Player A has about a -$30 EV per hand in the first case.

In the no limit case, Player A has a little over 52% pot equity, or about +$4 EV per hand.

[Sarge85]

/forums/images/icons/confused.gif When I see posts/questions like this, I start to wonder - How successful can/will I be? I am not a math major or study and I'm wondering how much of this will I really need to know to be a solid player. I saw someone's reply and it looked like something I had in high school trig.

How much of odds/probablity, +/- EV, game theory will I really need to know? Tough to quantify knowledge I know, perhaps a better question is how much "base knowledge" is needed to be a solid player.

[ThePinkBunny]

A. Player A's expectation is -$8.

B. It is worse for player B, his expectation has gone from $8 to -$4.

I'll explain my thinking on part B first. Since player B can bet a nearly infinite amount compared to the pot, he's essentially giving player A 1:1 odds, so he can bluff at the same frequency he's betting a good hand. That's the same as him winning 48% of the time.

For part A, Player B will be laying 2:1 pot odds for Player A when he makes the $600 bet, so he needs to bluff half the time he value bets, so he'll end up betting $600 36% of the time. So obviously he's also going to bet $200 36% of the time. And he can also bluff that he's going to bet the $600. Again he's giving 2:1 odds so he's going to bluff half the time he's actually going to make the $600 bet. Therefor, he's going to make the $200 bet 54% of the time, which will end up being the equivalent of winning the $200 pot 54% of the time. So player B's expectation is $8.

I intuitively thought the no limit game would be better for player B. I guess this was a very good excercise for me.

D

[ThePinkBunny]

Many good players don't know diddly about math.

D

[Ed Miller]

This space intentionally left blank

[BB King's]

A. B has 24 % real outs and 30 % faked outs so 54 % total outs. EV(B)=$108. EV(A)=$92.

B. B has 24 % real outs and 24 % faked outs so 48 % total outs. EV(B)=$96. EV(A)=$104.

[BB King's]

... but did you do it in less than 10 minuttes ? /forums/images/icons/grin.gif

A very good explaination !

[Robk]

I think these are the minimums player A earns in each scenario if he plays correctly.

[Ed Miller]

If Player B bets as you describe: 54% of the time in the $200 pot and 36% of the time in the $600 pot, then Player A's correct strategy is to always call the first bet and always fold to the second bet. Using this strategy, his EV is:

EV = 0.54*0.36*(-300) + 0.54*0.64*(+300) + 0.46*(+100) = +$91.36 per hand which is much better than the -$8 per hand he gets by always folding to the first bet.

[BB King's]

... that you misread the post !

There is 100 hands in the deck. 24 hands is good - 12 he will bluff with twice - 18 hands he will bluf only once.

Hope this helps !?

[ThePinkBunny]

I did it in about 20 minutes, but without the aid of pencil, paper, or a calculator.

D

[ThePinkBunny]

you're not taking into account the $100, he's already invested in the pot $91.36 - $100 = -$8.64, which would be -$8, except you've got some rounding error.

D

[BB King's]

... to mention - because it's so obvious - how A should play.

On both betting-rounds he should simply flip a coin.

[Ed Miller]

I didn't round once, and I did take into account the initial $100 invested. To be clear, there are three options... Player B checks first (and concedes to Player A), Player B bets first and checks second (again conceding to Player A), and Player B bets first and then bets again.

Assuming Player A always calls the first bet and always folds to the second bet, then Player A wins Player B's $100 ante when he checks first; Player A wins Player B's $100 ante and $200 bet when he bets first and checks second; Player A loses his $100 ante and $200 bet when Player B bets first and then bets again. Thus, Player A's EV is:

EV = 0.46*(+$100) + 0.54*0.64*(+$300) + 0.54*0.36*(-$300) which is +$91.36 as I said in my previous post (no rounding).

The error in your logic lies in your assumption that Player B has the best hand only 24% of the time when the first bet is called. Since Player B doesn't bet 100% of his hands, he will have the best hand much more often than 24% of the time after the first bet is called.

[ThePinkBunny]

I'll be honest, I just skimmed your first post and assumed you had typical rounding errors and such.

Instead what has happened is that you've misinterpreted what I've stated (maybe I didn't state it clearly enough).

When I state that player B should make the $600 bet 36% of the time. I'm saying 36% of ALL the hands and NOT 36% of the time he bets $200. In other words he's betting 2/3 of the time he bets $200. 2/3 * 56% = 36%.

So using your equation now corrected we have:

0.46(+$100)+0.54(1/3)(+$300)+0.54(2/3)(-$300) =

Now I haven't done the math, but I assume it equals -$8.

D

[ThePinkBunny]

This part is incorrect, but also isn't necessary to answer the original question.

D

[Ed Miller]

Oops.. Pink Bunny was right. The error I made in my approach was in the first line... Player A's equity in the called ($600) pot should be $200, not (1-X)*$200 + $200 (I knew that looked funny).

Going through the same math, then:

(1 - 0.36/X) * $600 = $200
3 - 1.08/X = 1
X = 0.54

and the EV is indeed -$8.

[Timer]

>>It took him ten minutes. So could Chris Ferguson, Mark Weitzman, Barry Greenstein, and Jimmy Warren.<<

Yeah, and all of these guys have PHDs or thereabouts, so I would be very unimpressed if they COULDN'T answer this in about ten minutes.

I can't answer it; I don't care about it; I don't want to think about it, because it gives me a headache.

Oh, in case you're interested. I hold my own quite nicely without knowing any of that game theory crap.

[BB King's]

... about the pot-limit-version of the game.

On both betting-rounds (in game A) he should simply flip a coin.

Solution to the no-limit-version of the game is:

q(P+B)-B=0 ...

where q is folding-frequency, P is the size of the pot, B is the bet.

Solving for q we get: q=B/(P+B).

That is - the bigger the bet - the more you should fold - no surprise.

[lorinda]

It's not a coincidence that they are all millions winner.

I hold my own quite nicely without knowing any of that game theory crap.

There is a subtle, but important difference between these two statements.

Lori

[Timer]

>> It's not a coincidence that they are all millions winner.

I hold my own quite nicely without knowing any of that game theory crap.

There is a subtle, but important difference between these two statements.

Lori<<

So, are you saying that if I become a game theory expert (or get a PHD) that I will become a millionaire too?

Sorry, doesn't work that way.

By the way, not all game theory experts are millionaires, and not all millionaire poker winners are game theory experts, nor do they have PHDs.

[Louie Landale]

Didn't read the responses.

[1] Pot Limit. On the 2nd round, B should bluff 12% of the time offering A 2:1 for a 2:1 bluff, so it wouldn't matter if A calls or not. A should call 2 times out of 3 to insure that B bluffs correctly. Of the 36% of the time B bets on the 2nd round, B wins 24% with the best hand and 2/3 or the final 12% he bluffs, or wins 32% of the time.

So, going into the 1st betting round A knows that B will win 32% of the time. Since A is getting 2:1 to call a pot bet B should bluff 16% of the time. It doesn't matter if A calls that bet or not, so lets say he folds. Thus, B wins 48% of the time.

A wins the Antes 52 and B wins 48% resulting in a net +EV for A of $4/hand.

[2] No Limit. On the 2nd bet B can bet a bunch offering A only 1:1 and so should bluff 24% of the time. A should call half the time resulting in B winning 24%+12% or 36% of the time.

So on the 1st betting round A knows B is going to win 36% of the time. Likewise B can bet a bunch and should bluff 36% of the time. It woun't matter if A calls so B will win those 72% of bets.

A wins 28% and B wins 72% resulting in a net -EV for A of $44/hand.

- Louie

I sound confident, but I'm not. But that's my guess.

[George Rice]

When figuring EV problems it’s important to keep in mind that you are playing for the money in the middle, so that the optimal strategy will always have an expectation between that amount and zero. So in this problem, A’s expectation will be between $0 and $200 (if B could some how offer A less than $0, A would just fold 100% of the time). Also, it doesn’t matter how the $200 got in the pot, because we are figuring expectation from a certain point on. Moreover, in problems like these B’s objective is to bet and bluff in a ratio that guarantees the same profit no matter what A does. A’s objective is to make sure that B does this, and not try for extra EV. That said, let’s address the problem at hand.

The first thing to recognize is that A can never be the aggressor. He can’t bet or raise because only B knows if B improved. A’s bet can never gain, as B will simply fold if he doesn’t improve, and raise when he does. In fact, B will bluff raise sometimes, taking even more equity from A. So A’s decision will be to call or fold. Both A and B recognize this.

If B bets, then A will be getting 2:1 on his calls ($200 dead money + $200 B’s bet vs. $200 A’s call). B’s strategy is to bet his improved hands and enough of his unimproved hands so that A’s action is irrelevant. So B will offer A a situation where A is getting 2:1 on a call which will win 1:2 times. Since B will improve 24/100 times, he needs to bluff 12 out of the remaining 76 times he doesn’t improve. If there was only one round of betting, B’s EV would be ($400 x 24) - ($200 x 12) for every 100 hands, which is (9600-2400)/100 = $72. A’s expectation is the difference between that and $200, so A’s EV=$128. Notice that if there was no betting A’s EV would be $152. The round of betting gave B an opportunity to gain EV.

Adding a second round of betting give B a second chance to gain EV. Again, A will be getting 2:1 on his call ($600 dead money + $600 B’s bet vs. $600 A’s call). B’s strategy will be the same, his bluffs will occur half as much as his legitimate bets. Since he already did this in the first round of betting, the only hands he has left are the 24 improved hands and the 12 bluffs. What a coincidence! So B will be betting all of his hands. B’s expectation will therefore be ($800 total bets + $200 antes) x 24 winners minus the 12 losers ($800 x 12) for every hundred hands, which is 24000-9600/100 = $144. A’s expectation is therefore $56 if he plays optimally.

Notice how A’s EV went from $152 with no betting, to $128 with one round of betting, to $56 with two rounds of betting.

With only one round of betting but at no limit instead of pot limit the problem is a little different. B now can bet so much ($1 Billion) that A’s odds on calling are even money (the $200 is too small to factor in on the calling odds, but is still a factor on EV). B can now bluff as many times as he bets his improved hands. He will bet the 24 improved hands and bluff 24 out of the remaining hands. Those hands will be a wash for both players and A’s EV will simply be the 52 remaining hands multiplied by the $200 which is $104. B’s will be $96

So, the no limit situation gave B an opportunity to gain more EV than the pot limit situation for one round of betting. But that extra round of betting in the pot limit situation was worth more still. In fact, if there was an unlimited number of rounds of betting, B’s EV would approach $200, and A’s would approach $0! That would not be the case in the no limit situation.

Answers:

A) $56 is the worst player A’s expectation will be if he plays optimally in the pot limit situation.

B) $$48 worse for player B to be playing no limit with only one round of betting

[Fishy]

If A plays correctly he would have +3.5687940000000 EV.

He would be better off at potlimit because the boy just can't lose! But he would be better off by only +688.90000EV.

Hope this helps you,

Fishy

PS: Does EV stand for "Every Vay" he plays???

[ThePinkBunny]

You better recheck your pot limit explanation, and I'll leave that at that.

And your no limit explanation is wrong because if player A calls player B 50% of the time, then Player B should only bet his winning hands and then he'd win a billion dollars 12% of the time.

D

[onlinechamp]

Fishy I think your answer is wrong although your thought process is right need to rework some of the math to find your errors!

P.S. EV= Estimated Value!!!!!!!!!!!

once a fish always a fish!

[BB King's]

A very nice post w/ a lot of good explanations !!! But this is wrong:

Adding a second round of betting give B a second chance to gain EV. Again, A will be getting 2:1 on his call ($600 dead money + $600 B’s bet vs. $600 A’s call). B’s strategy will be the same, his bluffs will occur half as much as his legitimate bets. Since he already did this in the first round of betting, the only hands he has left are the 24 improved hands and the 12 bluffs. What a coincidence! So B will be betting all of his hands. B’s expectation will therefore be ($800 total bets + $200 antes) x 24 winners minus the 12 losers ($800 x 12) for every hundred hands, which is 24000-9600/100 = $144. A’s expectation is therefore $56 if he plays optimally.

[Fishy]

"Fishy I think your answer is wrong although your thought process is right need to rework some of the math to find your errors!"

I used a random number generator, the latest and best!! It is totally random even in the billionth billionth confidence level.

So you think player A will lose more in potlimit?? What was your math??

Please elaborate and communicate, onlinechamp.

[onlinechamp]

Ok fishy math is!

CHI SQUARE Statistics would work here!!

[onlinechamp]

OC laughs at this and says champ knows the answer!

Also OC goes on to say some play some 1-1 with the champ and we can test it out I see this as useless info!

once a champ always a champ!

[Fishy]

but does he doesn't know that player B has 24% chance of improving like you would when you play champ, so your answer could be different.

Player B has 3 to 1 to win, if he hits he invest $200 more to win a potential $1000.
If he missed and not bluff he loses the $200.

When the pot is $600, Player B can bluff $600. If A thinks he can win over 50% of the time he will call.

No limit, Player A can move in and not let Player B draw. Even if B draws, it would be a big negative EV 1 to 1 money odds vs 3 to 1 card odds.

In a real poker game, each player will be thinking what the other player has with different probability from game theory assuming randomness.

I say most of the time the OnlineChamp will do better than game theory against everybody in da world:)

What was the question again?? Too lazy to look back.

[onlinechamp]

Thank you fishy and sklansky this is useless ?

[Hung]

I'm lousy at Math, does that mean I'm a lousy player?

[Louie Landale]

What I don't get about this is how B's expectation can POSSIBLY go down if he has MORE betting options in the No-Limit case. If betting MORE than the pot is bad for B, then he won't bet more than the pot. Also, if betting more than the pot is bad for B than we should consider the possibility that betting LESS than the pot is BETTER for B in the Pot Limit case. It also may be that betting some number like 1.5 the size of the pot is the best B can do.

- Louie

[David Sklansky]

Louie, he doesn't have more options in no limit because he has only one betting round.

[onlinechamp]

Well hung that would make you a fish!

[DiamondDave]

Let's start the analysis with the second betting round.
The pot contains $X (never mind how it got there).
A will never raise because B will never call when he is beaten.
If B bets, the pot offers A 2:1 odds.
B bluffs often enough that the odds against a given bet being a bluff are also 2:1.
That is, he ends up betting for value .24 of the time, bluffing .12 of the time, and mucking .64 of the time.
Player A must "regulate" B by calling often enough. (If A never called, B would bet every time. If A always called, B would never bluff.)
He can do this by calling half of the time when his opponent bets.
B breaks even on his bluffs. 1/2 the time he puts in $X and loses it to A, 1/2 of the time he wins the pot ($X) immediately.
Since B bluffs sometimes, A ends up calling B when he has the best hand sometimes.

So the possible results, associated probabilities, and (payoff for A, payoff for B) are
1. B mucks and concedes .64 of the time (X, 0)
2. B bets for value and A calls .12 of the time (-X, 2X)
3. B bets for value and A folds .12 of the time (0, X)
4. B bluffs and A calls .06 of the time (2X, -X)
5. B bluffs and A folds .06 of the time (0, X)

E(payoff for A) = .64X - .12X +.06(2X) = .64X
E(payoff for B) = .12(2X) + .12(X) +.06(-X) + .06X = .36X

E(payoff for A) + E(payoff for B) = X = pot size makes sense, since betting after the initial stake is established is a zero-sum game and the initial stake at the start of the betting round is the pot.

So, since A has the best of it, he prefers a larger pot. It says that he can only call, raise, or fold after B acts. B knows that A will raise if he bets, so B checks. (My reading of the puzzle was that A cannot bet on the first betting round if B checks. Anyone else read it differently)?

Assuming that A cannot bet on the first round if B checks, the size of the pot will be X=$200 at the time the last card is dealt.

Player A expects to drag in an average of .64($200) = $128 per hand. This yields an expected per-hand profit of $28 since each player must ante $100. Player B obviously loses $28 in expectation on each deal.

If A can bet after B checks on the first round and B calls, the pot will contain X=$600 at the start of the second round. B can expect to take an average of .36($600) = $216 at the end of the hand. This gives him a (negative) expected profit of -$84 on average, so he will call the bet rather than surrendering the antes immediately and accepting a profit of -$100. Player A gets .64 of the $600 in expectation, which is $384. His chip position increases by an average of $84 per deal in this case.

Not sure how to answer the no-limit question. Intuitively, it seems that A could steal the antes every time by moving all-in.

[Louie Landale]

He can bet up to the size of the pot just like in Pot Limit. He can also bet MORE than the pot which is why I said he has more options.

- Louie

[George Rice]

You claim that paragraph's wrong but don't say what you think is wrong about it. Care to explain?

[George Rice]

If player B bets 54% of his hands on the first round he will be betting the 24 improved hands plus 30 bluffs (out of a 100). More than half of his bets will be bluffs. Player A will be getting 2:1 on calling a hand that he's favorite to win! This can not be optimal.

Also, don't figure the dead money into the expectation calculations, it only confuses things. Expectation should be figured from the point where the players have a choice of what to do. You can't just subtract the $100 ante because when player A anted, he didn't know he would find himself in this situation. It was a possibility, but so was the reverse. If player A had to ante knowing this would be his hand, he would simply refuse to play, rendering expectation zero. Keep in mind we are talking OPTIMAL strategy.

In the no-limit situation, B will be losing 52% of the times, and break even the other 48%.

[George Rice]

It's been a long time since I've tried one of these problems and I over-simplified it.

Player A's expectation will be no worse than $92 if he plays optimally.

Player B's expectation goes from $108 in the pot-limit situation to $96 in the no-limit situation.

I still think ev should not take into account the dead money in the pot. The ev before the antes is zero, otherwise Player A would refuse to play.

[George Rice]

Player B gets more bluffing opportunities because of the extra betting round. Specifically, he can bluff 30 times instead of 12 times. Player A will be getting way the best of it on the first round, but overall will be getting only 5:4 on all money put in the pot.

We know Player B will bluff 1/3 of the time on the second round (12 times). Therefore, $1000x24 - $800x12 - $200(X-12)=(24+X)x$200, 12000=400X, 30=X where X is the number of bluffs in the first round.

[Rushmore]

I note that one poster states his inability to do this math, and his seeming curiousity as to the response.

I further note another poster railing against "this game theory crap."

Most notably, I see David's words indicating Howard Lederer's (et al) proven track record (s), relative (we may infer) to his ability to tackle such a problem.

I cannot do this type of problem. I cannot imagine myself ever inclined to adapt to this sort of thinking, nor do I see myself actually capable of this adaptation.

My question is sincere, and it is THIS:

Must we assume that these "Game Theory Brainiacs" are successful poker players because of their abilities in this arena, or may I be permitted to believe that this talent is indicative of the fact that these are highly intelligent people, and, as such, they might logically be expected to excel at the poker table?

Again, this is not intended as a refutation of Game Theory necessity. I suppose it's intended as an attempt to validate my belief that I might be able to promote myself from "Consistent Winner" to "Monstrosity Of Ostentatious Success" without the ability to master the Rubic's Cube.

[Utah]

Can someone check my math? I disagree with the answer.

Using Standard game theory – 2x2 matrix, competitive zero sum game with no saddle point

B has two strategies.

Strategy 1: Bet if he hits, Bet if he misses
Strategy 2: Bet if he hits, fold if he misses

All other strategies are dominated

A has two strategies:
Strategy 1: Call
Strategy 2: fold

The value matrix:
___Call Fold
BB -660 100
BF 140 -52

-660 = .76*-900 + .24*100
140 = .24*900 + .76*-100
100 = B bets and C folds
-52 = .24*100-.76*100

P = probability of strategy BB for B
Q = probability of strategy Call For A
V = Value of the game for B

Using a grid:

a b
d c


P = (c-d)/(a-b+c-d) = (-52-140)/(-660-100+52-140) = 192/952 = 20.2% time use strategy BB

Q = (c-b)/(a-b+c-d) = 152/952 = 16.0% use strategy Call if B bets

V = (ac-bd)/a-b+c-d) = (34320-14000)/-952 = -$21.3

B can be expected to lose -$21.3 dollars per hand

Any help is appreciated

[ccwhoelse?]

what is the solution?

[Ed Miller]

Read ThePinkBunny's posts in this thread.

[J_V]

Johnny Chan, Hellmuth, Brunson, Nguyen probably couldn't get it...I wonder how they could possibly win without this knowledge.

[J_V]

David picked five guys who are PHD's and essentially said because they understand this, they win. Im going to let others elaborate on why this statement is not only arrogant...but flawed.

[cferejohn]

[ QUOTE ]
It's not a coincidence that they are all millions winner.

I hold my own quite nicely without knowing any of that game theory crap.

There is a subtle, but important difference between these two statements.

Lori


[/ QUOTE ]

Yes, one of them does not mangle the english language... /images/graemlins/grin.gif

[bsk1999]

Disagree, here's why. The practical limit player B can ever loose is $52/hand. This is because if he is ever loosing more, he will only bet when he makes his hand (24%) and muck when he doesn't (76%). Knowing this, Player A will never call a bet, leaving Player B winning 100*.24 and loosing 100*.76, for loosing a net $52/hand.

[oddjob]

[ QUOTE ]
<img src="/forums/images/icons/confused.gif" alt="" /> When I see posts/questions like this, I start to wonder - How successful can/will I be? I am not a math major or study and I'm wondering how much of this will I really need to know to be a solid player. I saw someone's reply and it looked like something I had in high school trig.

How much of odds/probablity, +/- EV, game theory will I really need to know? Tough to quantify knowledge I know, perhaps a better question is how much "base knowledge" is needed to be a solid player.

[/ QUOTE ]

i am no math whiz at all and cannot even fathom a guess, but when i see that half my opponents can't even count the right amount of chips to call a bet or raise, that always makes me feel that i don't have to be a math whiz to be successful at the low limits i play at.

[bsk1999]

EV(A) = +4

He accomplishes this by folding to a raise 50% of the time, and re-raising 50% of the time.

EV(A) = .24*100*50 + .24*500*.50 + .76*X*100*.5 - .76*X*300*.5 - .76*(1-X)*100
(X = percentage of time B bets unimproved hand)
= 72 + 38*X - 114*X +76*X - 76
= 4
Mathematical proof to follow soon, but the arguement follows that Player B can "react" to any strategy player A's takes if he is given the chance to raise again, and therefore can obtain a positive EV for himself. But, if Player A only folds or raises, he no longer plays Player B's strategy, but merely takes advantage of his superior hand position (ie winning 76% of the time). Player A must eliminate Player B's options, and make Player B make decisions based solely on the whether his hand was improved or unimproved.

[bsk1999]

Sorry, got some signs wrong, and missed a decimal. Corrected Formula below

EV(A) = -.24*100*.50 - .24*500*.50 - .76*X*100*.5 + .76*X*300*.5 + .76*(1-X)*100
(X = percentage of time B bets unimproved hand)
= -72 - 38*X + 114*X - 76*X + 76
= 4

[thomastem]

Given everything even would you make more or less profit having this skill at the poker table? This is the only question that matters regarding if it's beneficial or not to have this ability in your arsenal.

[adios]

" Both players have a billion dollars in front of them "

Ah don't think I'd be playing no limit poker under these circumstances.

[johnd192]

So what was the correct answer?

[tpir90036]

kind of late to the party...but what is the actual correct answer? there are about ten different "correct" answers in this thread. methinks that they can't all be right. a decent explanation would be interesting as well.

[RydenStoompala]

The answer is nearly none. I like being surrounded by Phd's here at the office. It's incredibly satisfying making six times their money and not having to spend all day with my nose in a notebook, reinventing ballistics equations. We need solid salaried math heads to design things. Now and then they bankrupt the odd company or blow up a shuttle, but they like playing poker and that's a good thing.

[Deadbuddha]

i dont think you need to be a HUGE math buff to be decently successful, you must know some basic odds though, like how to calculate your outs and such. You dont even have to think about it too much, either let it become second nature or try to give your best guestimate.
although players with deep probabily knowledge will have an edge.