TPFAP : I'm not ready!

[Depraved]

In Sklansky's TPFAP, he remarks in 'The System' section that AK (suited and offsuit), any pair, any suited ace, and any suited connector 45 and up comprises about 13% of the possible starting hands. He then states that if you don't know how that figure was computed, you're not ready for serious poker, and you deserve to lose.

1. I admit it - while being a decent mathematician, I am fairly weak in probability, and cannot perform this calculation. Could someone provide a proof of the statement?

2. I am not ready for the world series of poker main event, but I wonder how many people who enter can actually perform this calculation. I doubt it's very high. I suspect that less than 50% of consistent tournament winners can perform the calculation. Am I way off?

[Bozeman]

1) It is not a hard calculation, just a counting problem:
how many AK combos are there?
how many ranks of pairs? (13) How many of each? (6)
how many suited aces?
how many suited connector 45 through qk? (9*4)
how many hands are possible? (52*51/2)

2) my guesses: 50% at WSOP, 80% of consistent winners (unless you mean do it exactly in your head)

[Ed Miller]

This is a counting problem. You count the number of hands in the set of hands suggested by Sklansky, you count the number of hands possible, and then you divide.

I couldn't imagine someone possessing the "intellectual horsepower" (to borrow Microsoft's nauseating term to use in my nauseating post) to beat poker lacking the ability to perform this calculation.

[David Sklansky]

Until recently, less than 10% of good players could do this. Since I've written about it a lot, its probably up to 25%. I sometimes write things that are a bit over the top just to keep from getting bored. That doesn't mean they aren't true though.

[slider77]

I get this:

AK - 16
Pairs - 78
AXs - 48
Connectors - 32

Total: 178

Total hands: (52*51)/2 = 1326

178/1326 = 13.4%

[Depraved]

Maybe I'm wrong, but I think you got several numbers wrong since you're counting several combinations more than once, and I presume you can't do this. I have what you have with AK and pairs. However, with Axs, I have 44 since 4 AKs hands were already accounted for. On connectors, we can't use AKs either, but I got (4*9 = 36) which is larger than your 32 (not sure how you got less).

[slider77]

Yeah, I agree, I didn't look at the double counts and it should be 36 connectors...I guess the total ends up the same......178.

[George Rice]

[StevieG]

I get

any AK = 16
any pair = 78
any Ace suited = 44 (4 suits * 11 ranks other than K)
suited 45 and up = 32 (4 suits * 8 ranks up to QK)

total 170

170/1326 = 12.8%

[Schmed]

on an airplane.....

I love those kinds of statements from DS. He's an arrogant bottom liner and if you don't like it his attitude apprears to be tough schit.....ya gotta respect that..... (for the record that wasnt meant as a shot at DS more of a compliment)

[Schmed]

like I said above...keep doin it man....in every one of your books you come across something like that and it always makes me laugh...by the way thanks man..all the info matters....

[Bozeman]

connectors=36 (4*9 45,56,67,78,89,9t,tj,jq,qk)

[BruceZ]

I sometimes write things that are a bit over the top just to keep from getting bored. That doesn't mean they aren't true though.

I know just how you feel. I mean, how many times can you derive the odds of hitting a flush draw? After awhile, you have to start making over the top statements just for the challenge of trying to defend them, like claiming that straights should be worth more than flushes or something. /images/graemlins/grin.gif

[huzitup2]

offer the following "quiz" since everyone /images/graemlins/smirk.gif seemed to find this one so easy.

This is HIGHLY contrived, but it's 100% math so I don't believe that to be a relevant issue.

Limit tournament -

The SB has exactly enough to post his blind.

The BB also has exactly enough to post HIS blind.

UTG decides to raise blind (straddle), and yes, he too has exactly enough to make this raise.

All fold to you; you're on the button.

What percentage of your hands warrant a call based strictly on hot and cold results ?

*

Your stack size is irrelevant.

Or if you prefer, let's say this is a ring game meaning your stack size is totally irrelevant.

*

One request -

*

If you have written a book entitled "THEORY OF POKER" you must wait 24 hours before posting a reply.

Ditto for anyone who works for 2+2 and attended Va. Tech. /images/graemlins/wink.gif

*

Are these terms acceptable ?

If so, "let the games begin".

*

- H

[Copernicus]

oops wrong answer..i'll be back!

[Copernicus]

I havent convinced myself that there is even a unique answer to this question, though I suppose there must be a "minimum" solution that I'm too lazy to determine right now.

An approximate answer to the question (which assumes independence of hands vs each of the 3 playing) is that you should be playing 24.7% of the 1326 hands that could be dealt, which represents the top 47 of the 169 hand combinations (eg pairs, AKs), and gives a heads up probability of winning just in excess of the 57.04% needed.

[huzitup2]

If you beat the "staddler" you always come out ahead, even if one or both of the blinds beats you.

You are being asked to RISK exactly 2 bets.

There are 5.5 bets at stake. (This includes your 2 bets).

There are THREE amounts you can COLLECT . . .

* 5.5 bets if you beat all 3 of your opponents.

* 4.5 bets if you beat only the BB and the straddler

* 3.5 bets if you beat only the SB and the straddler.

Remember, this is an all-in situation.

*

Hint: Try thinking in terms of a each player's hand having no "dependency" on the other two. In other words, each hand has a value of between 1 and 1,000,000 (with each value being equally likely).

In other words, ties are very unlikely; for the sake of simplicity you may assume that they NEVER occur.

[huzitup2]

You will lose to the straddler 50% of the time.

- (When you do, it doesn't matter how you do vs. the blinds).

*

You will beat all three 12.5% of the time. (W-W-W)

You will beat the straddler and the SB 12.5% of the time. (W-L-W)

You will beat the straddler and the BB 12.5% of the time. (W-W-L)

You will beat ONLY the straddler 12.5% of the time. (W-L-L)

*

Now, weight the results.

*

- H

[Copernicus]

I dont think this one works huzitup. I think you are making the same mistake I did the first time I tried it, and missed an important fact about how the game is mechanically played. (dont want to give too much away!)

[Ed Miller]

So if you are dealt a random hand, your random hand will beat three opponents with random hands only 1/8th of the time? I'm guessing the number is something more like, say, 1/4. /images/graemlins/tongue.gif

Might want to reevaluate your approach...

[Copernicus]

Your observation is right on, but doesnt address the fundamental problem in his approach.

It does, however, point out a flaw in my math which will cause me to revise my numbers! Of course things become much easier when silliness like that is pointed out.

The answer is that you play all hands.

[Copernicus]

Well....I have to qualify that last answer, because of the wording of the question..."what percentage of hands warrant a call".

That depends on your definition of "warrants". I still think the answer the poster is looking for is "call all hands", but that is the break even solution. If something better than breaking even is "warranted" (say you must have a Y% likelihood of being up $X after a certain period of time) then you can become more and more restrictive and improve your EV...I think, lol.

[Bozeman]

This is too easy. Exactly the same hands that you would call against just a single allin BB (no SB): those that figure to be +EV are those that win >50% against a random hand. Just that in this case you have 3 random hands that each give you 1:1 on their money. Admittedly, your win chances against the 3 are not independent, but that doesn't affect the choice, just makes it more likely that you scoop or lose instead of winning some of the side pots.

For holdem, these hands are: any pair, any ace, any king, queens above q6, suited queens, suited jacks above 5, suited t7 or better, j8+, and 98s. I had to check my memory on these, I had thought 78s and maybe 67s were in, but they are not, and I missed the txs cutoff and the jxo.

Craig

[Copernicus]

When you say you have 3 opponents who have each given you 1:1 on their money, you have missed the "trick" as well. Unless I am completely out of my mind today, it is easily shown that playing all hands is break even in the long run. If huz doesnt respond and the problem doesnt get any more action by 5, I'll post my reasoning.

[huzitup2]

"What percentage of your hands warrant a call", may have been confusing.

My appologies.

*

What pecentage of your hands would you call with if you wanted to MAXIMIZE your profit ?

Disregard varience; if you can earn one penny by playing a given hand over the long run it gets played.

[huzitup2]

22, the worst unsuited kings, and several of the other hands "suggested" do NOT win half of the time against a random hand.

I'm not saying that eliminates them from the group of hands you should play here.

Nor am I saying they SHOULD be included.

But they don't win half of the time vs. a random hand.

[Copernicus]

That is different than my interpretation then, which was a break even strategy that I am still quite convinced is "play all hands". Meanwhile, I assume that you understand what I am referring to as the "trick" in the problem? If not then I may be going down an entirely wrong path.

[huzitup2]

Playing all hands may show a profit.

However, it should be obvious that a BETTER stategy is available.

*

Example:

You have the nuts on the river and are first to act in a multiway pot.

Betting out certainly shows a profit; it's a no-lose situation.The worst that can happen is everyone folds - you can NEVER lose this bet.

Obviously, it is NOT always the optimal stategy.

* Assume you could see the other hands, and knew there were several strong hands out there - one or two held by aggressive players. Checkraising (or attempting to) is almost undoubtedly the better strategy, but it doesn't change the fact that betting out DOES show a profit.

-H

P.S. Please post your reasoning. I didn't calculate wheter or not playing every hand shows a profit - it very well may.

But it doesn't show the HIGHEST profit.

*

I'm a little suprised more posters haven't chimed in - glad you did, "C".

[huzitup2]

~ noon tomorrow at which time I will post the answer.

*

BTW, this was never meant as a contest; if anyone wants to take a crack at it between now and then, please do so.

*

Best wishes,

- H

[Ed Miller]

Playing all hands may show a profit.

Playing all hands obviously does not show a profit. I'm gonna pit my random hand against three others and somehow the money is going to slide my way?

[Copernicus]

If I post my reasoning for playing every hand being break even it will take away the possibility of making the mistake that other posters are making (and I made the first time I did it). I will PM it instead.

Meanwhile I am probably overthinking the "maximizing profit" scenario and missing something obvious, since I expect a much more elegant answer, but for now come up with playing 742/1296 hands. This may be off by a few because the simulation I used has some curious results at the critical points and 16 million hands might not be enough!

The hands I came up with are All pairs, Axs, Axo, Kxs, Kxo, Qxs, Qx except Q2, Jxs except J2s, T9s, T8s, T7s, T6s, 98s, 97s, 96s, 76s, (curiously skipping 87s), JT,J9,J8,J7. T9,T8,98.

[huzitup2]

I THINK I get it (the trick).

I'm eager to find out if I do.

*

- H

[huzitup2]

you're not, there is a table that deals with this - it's more acurrate than MOST sims.

www.gocee.com (http://www.gocee.com)

Click on "English Center", then look to your left for "Poker", finally click on "stategies".

There is a table entitled, "percentage of pots won".

- H

[huzitup2]

if you play all but the very worst of yours you should show a profit.

Playing EVERY hand should be a break even play.

*

I answered too quickly the first time. (Oops)

*

- H

[Copernicus]

I see percentage of pots won vs 9 players but not whats needed for this question.

Ooops it pays to read..i see it

[Copernicus]

Kong, playing all hands is break even, not profit making. Eliminating just the worst hand should tip it to profit in the long run. However, huz is asking for maximizing profit, which involves finding the optimum breaking point in the EV equation for this situation...one that eliminates losing elements of the equation without eliminating more beneficial winning elements of the equation.

Too bad he didnt ask for maximizing ROI..play AA only! I think. /images/graemlins/confused.gif

[Bozeman]

Yes, playing all hands is break even in the long run, but why play to break even. Don't balance out your +EV hands with -EV ones.

[Bozeman]

K2 0.5050872
22 0.5033402

Care to put your foot in your mouth again?

[Ed Miller]

Kong, playing all hands is break even, not profit making. Eliminating just the worst hand should tip it to profit in the long run. However, huz is asking for maximizing profit, which involves finding the optimum breaking point in the EV equation for this situation...one that eliminates losing elements of the equation without eliminating more beneficial winning elements of the equation.

I understand what he's asking for. My point is that he keeps posting things that are clearly wrong/not well thought out. Saying that playing all hands may be profitable is obviously wrong. Suggesting that a random hand wins only 1/8 times against three other random hands is also clearly wrong. The only reason I'm being picky is that I feel he is coming off as smug...

The answer to his question is non-trivial, and given his "hint" to finding the solution, I'm not convinced that he actually knows how to solve it.

[huzitup2]

n/m

[huzitup2]

n/m

[huzitup2]

"HE" believes he has calcuated the answer correctly; his reason for posting this is that he may have "stumbled" over the answer - i.e., he didn't use the correct formula.

I am asking for the assistance of those who believe they are CERTAIN they know HOW to solve this problem.

*

2 PLUS 2 = 4

2 TIMES 2 = 4

*

In spite of this, I'm fairly sure that "+" and "x" sign are not interchangeable.

- H

[Bozeman]

Ok, this isn't quite as easy as I thought. The non-independence of your wins is a factor. Let x be the chance you scoop, y be the chance you beat straddler and BB, and z be the chance you beat straddler. Obviously, x<y<z, and z-y is the chance you beat straddler but lose to BB, and y-x is the chance that SB beats you, but you take the side pot.

You make money if x*5.5+(y-x)*3.5+(z-y)*2=2x+1.5y+2z>2

The answers are very close to my original (in which I forgot two, q6o and t9o). If z>1/2, y>1/3 and x>1/4, the hand will definitely be worth playing. To get x and y, I had to use gocee, which has random errors because they used simulation instead of enumeration (as the table at jazbo (http://www.jazbo.com/poker/huholdem.html) does). So there may be a few debateable marginal hands. For all the ones I mentioned except 22, k2o-k5o, q6o-q7o, q2s-q3s this is true. 22, k2o, k3o, q6o, q2s fail the equation, while the rest pass.

In addition, there are two new hands that pass because they play decent multiway (4 handed), though they are headsup dogs: 97s, 87s.

But I'm guessing you had a different solution in mind (though this one is the correct one),
Craig

PS Would someone like to do a more accurate (longer) simulation of 3way (and maybe 4way) against random hands? I suppose one could take the enumeration route for a few hands.

[Bozeman]

BTW, this comes out to 600/1326 hands. (How did you get 1296=36^2 hands, copernicus?)

[Copernicus]

Bozeman, the 1296 was a brain cramp of some sort. The 742 was the numerator I came up with though, because I remember noting that 724/1326 = 55.96%.

You and I arrived at the same EV formula, although by different routes. What I missed the first time is the side pot set up, whereby you have to beat all 3 to win the 2 bet main pot, beat 2 to win the 1.5 bet side pot, and beat 1 to win the other 2 bet side pot. The EV formula falls directly out of that noting that P(beating 3)=1/4, P(2)=1/3 and P(1)=.5. (I originally looked at it as playing the small blind for his .5, the BB for his 1, and the straddler for 2, which is obviously incorrect but I think it was missed by other posters too.)

My method of choosing hands is a bit different than yours, but I wouldnt think they were 124 hands apart.

The difference is you have restricted it to hands where ALL of the conditions apply..that is x>1/4, y>1/3 AND z>1/2. That is a sufficient condition but I think it is too restrictive. I attempted to includ hands where only one or two of those conditions are true, but the +EV margin of those that are true exceeds the -EV margin of those that arent true. I just eyeballed the EV margins, so i might have included a few extra hands, but again 124 seems like a lot.

I used the default simulation provided with Hold Em Analyzer, which was based on 1.69 billion trials, not the 16 million I mentioned earlier, each of the 169 hands being run 1 million times for each number of opponents up to 10.

[Bozeman]

"The difference is you have restricted it to hands where ALL of the conditions apply..that is x>1/4, y>1/3 AND z>1/2. That is a sufficient condition but I think it is too restrictive. I attempted to includ hands where only one or two of those conditions are true, but the +EV margin of those that are true exceeds the -EV margin of those that arent true. I just eyeballed the EV margins, so i might have included a few extra hands, but again 124 seems like a lot."

If you read my post, you will see that this is not true. I merely used the restrictive condition to limit the # of hands I needed to evaluate completely. While I started with the condition that all three are satisfied, since there is little multiway variation for holdem hands between 2,3,&4 players, I then added the ones that satisfy the equation but not the restrictive case. This means that some hands in the original (z>1/2) list that didn't satisfy the 3 way restriction still passed, and (only) two hands that don't satisfy z>1/2, 97s and 87s, are +EV. Interestly, less hands satisfy the equation than satisfy z>1/2, because marginal hands that play well multiway are typically suited.

While both of us use simulation, I bet that your simulation has larger error bars. I can't imagine that 76s and j3s for example are +EV here.

Craig

[Copernicus]

Yes, the difference between the GoCee chart and the HeA chart accounts almost exactly for the differences in our lists. When I get to my computer I'll run a few hands in question through 10 million sims and compare.

One thing which i didnt bother with, which could be the problem in comparing the charts, is the chances of ties.

[huzitup2]

I will post the METHOD (formula) I used as soon as I can.

* I hate it when life interferes with poker.

*

Playing the best 82.3% of your hands will maximize your e/v in this case.

*

1. Playing all but the worst hands will show a profit.

2. Playing only the best will maximize your "ROI".

*

However, the best 82.3% will maximize your e/v.

*

With or without (preferrably with) showing WHAT formula "you" used . . .

Does anyone wish to confirm (or quarrel with) this figure ?

PLEASE RESPOND; there is undoubtedly one or more of you out there who can explain this better than I can.

I did the calculations in the best manner in which I knew how; I do have a solid backround in calculating (and interpreting) statistics, however, said backround may not have prepared me for "problems" such as this one.

* When I was in school, "statistics for future poker players" was an elective (lol); I received my diploma without being required to take this class.

*

Be back soon,

- H

[huzitup2]

-

[Bozeman]

I think you are wrong, I can guess why, and my post (as well as Copernicus's results) shows the correct answer, which is not nearly so simplistic.

[huzitup2]

an answer that was not even in the same ballpark, or are we separated by 1 or 2 per cent.

I THOUGHT I read all of your (and C's) posts; I saw some good stuff but the only figure I saw as C's . . .

and I am certain that 57.4% is not close.

I must have missed something; methods aside (although I do want to know how you arrived at your answer) what figure did you come up with ?

I did get some help from a friend - a math geeky friend :-); "he" came up with a different (but very close) figure.

He also used a different formula than I did.

However, he did say that my formula - while not the best available - was usable in this instance.

Best wishes,

- H

P.S. One of the first things they teach lawyers is to never ask a question (of a witness) unless they already know the answer.

I am NOT an attorney; I do think I know the answer but that is NOT why I posed this question.

I want to know what the best way to "tackle" problems of this type is; I also want to know if my [simpler] approach is (I hate this expression, but), "close enough for government work".

[Copernicus]

Based on the GoCee tables we both came up with 600 hands out of 1326. The EV formula is as posted before, is very simple, and is clearly right. It not only makes mathematical sense but "logical" sense. It is:

x*2 + y*1.5 + z*2 >= 2 where x, y and z are the probabilities of beating 3, 2 and 1 players respectively. The payoffs are obvious, because they are the main and sidepots, and the probabilities applied to each of them are the numbers of players contending with you for those sidepots.

It makes logical sense, since everyone is playing under the same conditions, and should have break even EV with random cards. When x, y, and z are purely random (25%, 33 1/3% and 50%) the entire equation breaks even for you and the straddler, and each individual term is break even for the participants in that side pot.

Thus, the only possible reasons for a difference are

1) Bozeman and I dont understand what you mean by "maximizing EV".
2) We understand but somehow are missing that the total EV somehow increases from breakeven to a peak and then declines.
3) Our procedure is correct, but we are using vastly different hand simulations for random hands vs 1, 2 and 3 opponents.

[huzitup2]

Gotta do this quick - plane leaves in 2 hrs.

*

It is a mistake to view this as 3 separate heads-up mathchups - it isn't.

*

Matchup #1 is a 3-way matchup with 1 small bet at stake.

Matchup #2 is a 3 way matchup that may or may not involve a side pot.

Matchup #3 is a 3 way matchup that may involve one or even 2 sidepots.

*

Using a method wherby you consider your hand to be pitted against 3 other hands in 3 INDEPENDENT headsup matches will not work.

*

Perhaps someone can pick up the "red" phone and summon the "caped crusader" (is Chief O'hara still alive :-) and have him put this to bed.

If Batman is unavailable, I promise (cross my heart) to post my calculations a/s/a/p.

*

- I type at the blistering speed of 15-20 wds/minute AND (yikes) don't know how to "cut and paste"; if I knew how I would have done so already.

I have the work typed out; I'm going to ask a friend to paste it onto this forum, but it may have to wait 'till I can type it out if "he" is unavailable.

- Wish me a safe (and uneventful) flight,

Best wishes,

- H

[Bozeman]

Apparently you read slowly too, since I showed how accounting for the non-independence of the 3 pot matchups does not change the result. I guess Majorkong was right,
Craig