3-card Guts

[RocketManJames]

In a 5-handed + Pot hand (6 hands total) 3-card Guts game...

Hand Ranks (these may not be 'probabilistically' ranked, but this is how friends play it)

3-card Straight Flush (regular straights and regular flushes are not counted as a power hand)
3 of a Kind
Pair
High Card

Anyone can choose to play his or her hand or fold it. Simultaneous declaration of play or fold by all players. If 2 or more players choose to play their hands, then the pot-hand is not considered. If this happens, all players show and winner wins what is in the pot and losers match what was is the pot.

If exactly one person chooses to play, he goes up against the pot hand.

My question is... how strong of a hand do you need to compete? Most people will choose to play any medium pair and higher (say 77x and above).

How likely is it that your 4 opponents hold a playable hand given that you've got nut no pair (AKQ). And is it worth going up against a random pot hand with such a hand, ever?

Any other interesting probabilities regarding hand strength in this game would be much appreciated. I'm sort of rambling in asking my questions, because frankly I'm not sure what questions I should be asking. Bottom line is... how strong of a hand do I need when the pot gets large? Large pot means that players' "playable" hands get more and more stringent, and it becomes increasingly likely that you will go up against a random pot hand, since no one dare contest you for fear of matching a large pot.

-RMJ

[bigpooch]

Interesting question!

Q: Is A /images/graemlins/spade.gif 2 /images/graemlins/spade.gif 3 /images/graemlins/spade.gif a "straight flush"?

Am I correct to assume that if there was a confrontation
with a loser (a player who wanted in but lost), no one antes
but the loser just makes up the total pot? Also, if there
are multiple losers, I suppose there will be a potsize of
LxS where L=number of losers and S=size of the pot.

Here is an outline of how you could solve the problem, but
there might be a rocket scientist out there that might find
a better way!

1) Make a continuous approximation for hand strength: a good
idea is to let q denote a variable for general hand strength
with a range 0<=q<1 where q=0 would be the "nuts" and q=1
would be the worst possible hand; also, P(q<alpha)=alpha is
the probability measure. Then determine for the continuous
version of the game the approximate q* which is the minimum
hand for which EV(q(1)=q*)>0 where your hand is denoted by
q(1), the "pot hand" is q(0) and your opponents' hands are
q(j) for j=2,3,4,5.

2) Having the value of q*, now try to find EXACTLY the worse
3-card hands that show a +EV by a computer program using the
added (intuitively clear) observation that the discrete
minimum hand should be a little stronger since you would
think the minimum hand is a pair and if you are to hold a
pair, it is more likely than normal that an opponent holds a
pair. You can now write some code to do some Monte Carlo
simulations on hands to see what hands make a little bit of
money and keep running the simulations until you have a set
of hands that are quite likely to be these minimum hands to
be able to compete with a +EV.

3) Now supply the mathematical proof that these minimal
hands do indeed show a positive EV and the hands which are
very slightly worse show negative EVs if you should decide
to compete with them. It may be the case that a hand XsXhYd
is the minimal hand whereas XsXhYs is -EV (more straight
flush possibilities for the latter hand).

4) As a more practical exercise, consider different numbers
of opponents and solve the same problem: solve it for one
to eight opponents as this game will probably not be played
with more than nine players!

5) You can also observe the mistakes made by your opponents
and determine approximately their error rate: i.e., the sum
total of their mistakes per hour. Of course, you won't be
the only beneficiary of these mistakes as there may be other
players that play almost as well as you! In any case, you
can more closely approximate your hourly rate to see if the
game is worth playing.

6) After you have made a few hundred (or thousand!) playing
this game, you can throw some money my way! /images/graemlins/smile.gif

[bigpooch]

The question is much easier than I expected! For the
continuous approximation, if q(1)=q* is the value of your
hand (q values must satifsy 0<=q<1 where 0=best hand and
1=worst hand), you will win the pot if none of the other
five hands are better, i.e., with a probability of (1-q*)^5.
On the other hand, you lose if any of them are better, so
all that is required is that (1-q*)^5<0.5 which gives
q*=0.1294494.

There are C(52,3)=22100 different guts hands:

4x12=48 "straight flushes" (here, I consider As2s3s valid)
4x13=52 trips
For each pair, there are 6x48=288 combinations.

22100xq*=2860.8 and so it would seem that most of the top
2860 hands would be sufficient to contest: 2860-(48+52)=
2760 and when this is divided by 288, this is between 9 and
10, so let's check if 6s6h2d is sufficient:

If you are dealt 6 /images/graemlins/spade.gif 6 /images/graemlins/heart.gif 2 /images/graemlins/diamond.gif, the
distribution of opponent hands are now:

straight flushes: 48-(3+3+2)=40
quads: 11x4+1=45
pairs higher than 6s: for each pair there are
6x(48-3)=270 combinations
pairs of 6s: 47

Altogether there are C(49,3)=18424 combinations and if this
multiplied by q* this is just 2384.98. But the number of
hands as good as 662 is now 40+45+8x270+47=2292, so clearly
this hand is worth competing with.

Now, how about 5 /images/graemlins/spade.gif 5 /images/graemlins/heart.gif A /images/graemlins/diamond.gif ?

Going through the same calculation: there are still 40+45
"pat hands"; and for pairs of 6s up to Ks, there are 8x270
=2160 combinations; for a pair of aces, since an ace is not
available there are only 3x46=138 combinations. Adding all
of these gives 2383, very close to 2384.98! Nevertheless,
this hand is still worth a challenge!

Now, if there is a tie and the highest hands DON'T win, it
depends if your opponents will play 55A: if they don't, you
contest with it! If they all do, you simply don't!

Thus, a pair of sixes is sufficient and for 55A, it should
normally be worth playing. Also, the size of the pot should
not really influence your decision: you either win the whole
pot, or match the whole pot, right?

Assume your opponents will always play when they hold 55A or
better and don't compete with a worse pair. The question
now is whether you can sneak in with an AKQ. Say you have
A /images/graemlins/spade.gif K /images/graemlins/heart.gif Q /images/graemlins/diamond.gif. The number of hands for at
least one of your four opponents to compete is now:

straight flushes: 41
trips: 43
pair: AA/KK/QQ: each 3x46=138; altogether 3x139=414
pair of 6s to Js: each 270; altogether 6x270=1620
55A: 6x3=18

Thus, there are 2136 hands your opponents definitely play
out of C(49,3) hands. Approximately, there is a probability
of 0.6108511 that none of your opponents will play, so you
need only a 0.5/0.6108511 chance of beating the pot-hand for
this to be +EV, or there must be less than 3343.4
combinations of pot-hands that you lose to. So let's check
this: subtracting the 55A hands and just adding all of the
pairs 55 down to 22 back gives 2136-18+4x270= 3198 which is
clearly less than 3343.4.

Thus, it seems that you can try to steal with AKQ! Now,
suppose all of your opponents also play AKQ, then you may
not be able to play AKJ.

Suppose you have A /images/graemlins/spade.gif K /images/graemlins/heart.gif J /images/graemlins/diamond.gif.
Now, your opponents could also compete with AKQ or there are
9x4=36 more combinations or altogether 2136+36=2172 hands
out of C(49,3) hands. Now there is approximately a
probability of 0.6054685 that none of your four opponents
contest, but now you need a 0.5/0.6054685 chance of beating
the pot-hand and there must be less than 3209.3 such hands
that will beat your AKJ. There are now 3198+9x4=3234 such
combinations the pot-hand can win with and as this exceeds
3209.3, AKJ is unplayable.

Thus, it appears that you can contest with 55A or AKQ in the
five-handed version of this game. A pair lower than 55A is
unplayble and AKJ or worse are also losers.

Anyone care to check all this? /images/graemlins/smile.gif

[RocketManJames]

Very interesting, Pooch. I will read into this more carefully tomorrow, as it's very late now. I guess part of the trick is knowing your opponents 'playable' hands. I agree that the size of the pot doesn't matter, but the thing is that the size of the pot often dictates the strength of what your opponents hands. If the pot is $10, people will be much more willing to play 227. But, if the pot has grown to $1200, it's pretty safe to say that most of your opponents would muck 227 without a second thought.

Thanks for your analysis... will definitely read it more carefully.

-RMJ

[Phat Mack]

This is one of the more interesting posts I've read in a long time. Way to go, Pooch!

What if, in a 5-handed game, one opponent played TTx or higher, another played 55A or higher, another 22x or higher, and the 5th played AKx. What could you play with +EV?

If this is a lot of work, I've got an old 3-card lowball simulator I could dust off and modify to play guts. I haven't been in a guts game for years. I might have to host a home game... /images/graemlins/smile.gif

[tpir90036]

[ QUOTE ]
all that is required is that (1-q*)^5<0.5

[/ QUOTE ]
why is this? are you assuming that the opponents are playing only the top 50% of hands. maybe i am misunderstanding what q* represents....

[bigpooch]

You're correct! It should read (1-q*)^5>0.5 but the
boundary value is right. At least someone is reading these
posts! /images/graemlins/smile.gif

Suppose hands are represented by q-values where 0<=q<1 and
q=0 represents the nuts (the best possible hand). You need
to win half the time at guts since you either win the pot
or pay the size of the pot (that's where the 0.5 comes from)
and so the probability that someone has you beat if you
choose to contest with a hand which is q* in strength is
(assuming independence, which isn't quite true in the actual
case of guts, but will be a good approximation):

1 - P(nobody has you beat)
= 1 - (1-q*)^5

So this needs to be <0.5 or 1-(1-q*)^5 < 0.5, i.e.,
(1-q*)^5 > 0.5.

[Phat Mack]

On the other hand, you lose if any of them are better, so
all that is required is that (1-q*)^5<0.5 which gives
q*=0.1294494.

I am having some problems grasping this. It seems that the exponent of 5 is there because we are facing five opponents, but we only have five when our opponents are playing with starting-hand standards--making the possiblity of playing against the dummy hand exist.

If you are dealt 6 6 2 , the
distribution of opponent hands are now:

straight flushes: 48-(3+3+2)=40
quads: 11x4+1=45
pairs higher than 6s: for each pair there are
6x(48-3)=270 combinations
pairs of 6s: 47

Altogether there are C(49,3)=18424 combinations and if this
multiplied by q* this is just 2384.98. But the number of
hands as good as 662 is now 40+45+8x270+47=2292, so clearly
this hand is worth competing with.

Here we are assuming that the opponents are playing anything, so don't we need a different q*, one that was solved for by using ^4?

[bigpooch]

Okay, my answer was badly mistaken! What I forgot is that
when you win the pot with zero challengers (the widow hand
isn't considered a challenger), you are up 4 antes or 4/5 of
the pot and when you lose, you only lose 4/5 of the pot (you
are just anteing for everyone and if you were to quit right
after paying the pot, instead of playing the hand, the fair
amount to take back is just 1/players x potsize, which in
this case, is 2/5 the original pot size). These were the
factors that I hadn't considered.

On the other hand, if you win with exactly one challenger,
you lose exactly the size of the pot = 5 antes. There are
also possibilities that you are among other losers: with
two challengers, you only lose 4/5 the pot (you lost the
ante and then the pot size but your equity is 2/5 the pot)
and with three challengers, you only lose 3/5 the pot and
with four, it's only 2/5 the pot (by pot, I mean the
original size of the pot).

Instead of using q*, let me use v for the value of the
worst legitimate hand to contest the pot with. As mentioned
before, we're just using a continuous approximation at first
to get a good idea of what hands are worthwhile playing.

As before v=0 is the nuts and the probability of an opponent
having a hand better than v is just v. To find which value
of v is worth contesting, you assume that all your opponents
also only contest with a hand which is v or better and give
up on the rest. Thus, the expectation for contesting with
hand v is (in terms of the original pot size):

(1-v)^4 x (4/5 x (1-v) - 4/5 x v) [zero challengers]
+4v(1-v)^3 x (-1) [one challenger]
+6v^2(1-v)^2 x (-4/5) [two challengers]
+4v^3(1-v) x (-3/5) [three challengers]
+v^4 x (-2/5) [everyone's in!]

But all that is required is that the above is positive, so
just multiplying by 5 and simplifying (lots of algebra!)
yields (after dividing by 2):

2- 22v + 46v^2 - 44v^3 + 21v^4 - 4v^5 > 0.

Trying values around 0.129 (from the previous erroneous
analysis), one obtains v < 0.116155331. As before, looking
at the hand 772, there will be only 2022 hands just as good
out of C(49,3)=18424 and the ratio 2022/18424 = 0.10974 <
0.1161553 so 772 is worth playing.

How about a pair of sixes?
Consider 6 /images/graemlins/spade.gif 6 /images/graemlins/heart.gif J /images/graemlins/diamond.gif (initial guess)
There are now 48-(3+3+3)=39 straight flushes and 45 trips
for a total of 84 big hands. As before there are 270
combinations of pairs other than jacks and 138 pairs of
jacks. Altogether, there are 7x270+138+84=2112 pairs of 7s
or better. Also, there are the hand 66A, 66K and 66Q for
another 12 hands for 2124. Now, 18424 x 0.116155332 =
2140.05.

If the hand were instead
6 /images/graemlins/spade.gif 6 /images/graemlins/heart.gif 7 /images/graemlins/diamond.gif,
there would be just 2140 better hands.

Switching that 7 to a heart or a spade won't do as now that
would allow some more straight flushes. Is that the end of
the story? No! You may be familiar with the concept that
if you have a pair it's more likely than normal that other
players will have a pair (and especially if another player has a pair!).

This means that if you are beat with ANY 667, it's more
likely that you are beat in two or more spots when compared
with the continuous estimate. Thus, the equity reflected by
the continuous approximation underestimates the true equity
(see second paragraph above). Unless someone provides some
analysis to the contrary, it is likely that choosing to play
ANY 667 is +EV.

So I was clearly wrong in my original post about what the
minimum pair to play was. What about AKQ now? As before,
there are 1848 hands of 77 or better. There are now only
6x12=72 hands of 66 with an A/K/Q kicker and 6x20=120 hands
of 66 with a kicker between a 7 and Jack. Altogether, the
total is 2040 hands that win. The calculation is more
complicated for this than in my original analysis and as I
have to run now (good excuse! /images/graemlins/smile.gif ), perhaps someone else can
carry the ball! I'll be back to finish up if in the likely
event nobody takes the handoff.

[tpir90036]

[ QUOTE ]
To find which value of v is worth contesting, you assume that all your opponents also only contest with a hand which is v or better and give up on the rest. Thus, the expectation for contesting with hand v is...

[/ QUOTE ]
this is the part i don't understand. lots of people would probably stay in with any pair or even some dubious ace highs....there is no way to quantify this obviously...but it seems like your answer could end up being a little on the tight side against regular gambling opponents.

great work though....a very good read.

[Phat Mack]

After a showdown, do all losers match the pot with the winner dragging it all? Or does the pot build until one player beats the dummy hand?

[Phat Mack]

What about AKQ now? As before, there are 1848 hands of 77 or better.

I ran a sim last night with AcKcQd playing against four opponents playing 55A or better. AKQ got it heads up with the dummy 6152 times in 10000 hands. Of these, he won 5150 times for an overall 51.5% win rate. Pretty close to what was orignally predicted. More later.

[bigpooch]

When I had watched this played in my university days, the
winner took the current pot and if there were no losers,
everyone anted again; if there were losers, they contributed
to the construction of the new pot. Maybe the original
poster plays in a game where this is not the case!

[bigpooch]

True if you are playing against real gamblers. Playing in
such a way will still beat the gamblers, but won't win as
much as compared to if you loosen up a very tiny bit. As
Mike Caro had mentioned, if your opponents play too loose or
too tight, you can play looser; how apt for this game!

[Phat Mack]

This is how I remember playing.

[RocketManJames]

Yes, all losing players match the pot. But, the winner does not drag it all.

Example:

Pot = $20

3 challengers. The two losers match the pot = $40. The winner takes what WAS the pot = $20. Now, for the next round, the pot = $40 + whatever the antes are.

Example 2:

Pot = $20
1 Challenger (uncontested)

Because he's uncontested, he is up against the random "pot" hand. If he loses, the pot for next round is $40. If he wins, he takes the $20 pot. Now, the pot starts off at $0 + whatever the antes are.

---

Just so I can explain why I posted this question... we were playing $1 antes, and the pot had grown to around $80. I went in with ATT and there was another challenger. He had KJJ. He took the pot, and I had to match it for $80 or so.

The pot got that big, because the previous hand, with the pot just under $40, someone had gone in with AAx, and the pot hand had a straight flush. Pretty ridiculous.

-RMJ

[bigpooch]

Okay, so there are antes even if there was at least one
loser!

Unfortunately, that just means the pot gets a little bigger;
hopefully, there aren't any recalculations! /images/graemlins/frown.gif

[Phat Mack]

What if, in a 5-handed game, one opponent played TTx or higher, another played 55A or higher, another 22x or higher, and the 5th played AKx. What could you play with +EV?

To determine this, I wrote a five-handed-plus-dummy guts sim. I had 1 player sit out and had the other players hands that were at least TTx, 55A, 22x and AKx respectively. I ran 10,000 deals and found the median winning hand to be a JJ3. To test the result, I had the absent player now play JJ3 against the same line-up and dealt another 10,000 hands. JJ3 won 73%! There is obviously something wrong with my methodology. More later.

[Phat Mack]

I looked at the results from the first sim and found that the program was not counting deals where the four players were folding to the dummy. I included these deals in the results, found the median hand to be 33K, and re-ran the second sim. 33K won 53.58% of the time, much closer to the prediction.

[Gator]

Can I add some modifications to reflect the way we play so I can more closely approximate optimal play?

We play three of a kind beat a straight flush (not that it matters - in that one would play either of these hands)
We play with straights and flushes -- a straight beats a flush which beats a pair.
We play with a maximum burn (generally $20) thus creating opportunities to risk $20 and win $40 or more.
The community hand is always dealt (even if 2 or more players are in).
The community hand gets four cards!

My gut feel is to play pair of tens or better -- loosening somewhat when the pot is laying odds (i.e. risk $20 to win $40).

Comments on this modified version.

Thanks.

[Phat Mack]

I ran some showdown sims where four opponents played the same hand requirements along with a dummy hand playing random cards. I listed the winning hand percentiles along with the percentage of hands played by the dummy. I guess that these percentages are accurate +/- 3%. Not of these sims considers pot size, future antes, etc.

<font class="small">Code:</font><hr /><pre>
Opps. 50%ile 60%ile 70%ile 80%ile 90%ile dummy
Axx 559 77A TT3 QQ2 KKJ 14.97%
AKx 55T 883 TT5 QQ3 KKT 39.10%
22x 55J 884 TT5 QQ3 KKJ 43.64%
33x 557 77A TT4 QQ3 KKT 50.14%
44x 554 882 TT3 QQ3 KKJ 53.26%
55x 554 77J TT2 JJA KKJ 56.62%
66x 332 77Q TT2 JJA KKT 60.67%
77x AK2 774 TT3 QQ3 KKJ 63.24%
88x AJ8 882 TT2 QQ2 KKJ 66.19%
99x AT7 559 TT6 QQ4 KKQ 69.41%
TTx A75 AK4 TT3 QQ3 KKT 74.35%
</pre><hr />

It looks as though your opponents need to play 66 and above before loosening up. If you want a 60% chance, you might wait til they play 99. There is probably some interesting gamesmanship that could be considered as to getting opponents to tighten up to the point where you could loosen your requirements.


Out of curiousity, I tried to see if there was an "algebra" that could be used to average opponent's hand strengths; in other words, I wanted to know if having four opponents playing 772 was the same as having two opponents playing 99x and two playing 552. Initial results show that it is very close, but I am skeptical about whether I've proved anything.

<font class="small">Code:</font><hr /><pre>
Four opponents playing __ AKQ
772, 772, 772, 772 0.5211
662, 772, 772, 882 0.5210
552, 662, 882, 992 0.5206
552, 552, 992, 992 0.5203
</pre><hr />

I also wanted to get a general idea of how differing numbers of opponents effected the numbers.

<font class="small">Code:</font><hr /><pre> Opponents AKQ win rate

1 0.7362
2 0.6551
3 0.5848
4 0.5211
5 0.4645
6 0.4139
7 0.3689
8 0.3288
</pre><hr />

[James Boston]

These games where the loser or losers match the pot don't seem to be something that can produce long term +EV results. Lucky timing is too much of a factor. The luck of the cards you're dealt is a factor in all forms of poker. In hold'em you could be dealt the best (AA) or the worst (72o) at any given time. The dieffernce is that in most games you can expect the pot size to be consistent when you have the best of it and when you have the worst of it. I KNOW, some pots stay small and some get huge, but over one session of poker an average pot size can be determined. This way you don't suffer huge swings.

Here's an example of why I hate "match the pot" games. Suppose there's a $10 pot in your 3-card guts game. You find 3 jacks. Clearly a playable hand. 4 people stay in. You, a guy with 3 queens, a guy with 3 kings, and a guy with 3 aces. Obviously this is unlikely. But know you've just lost $10 and from now on you have to risk losing $30 to get involved in a hand. A few hands later somebody beats the pot hand with a pair. He just happened to have a playable hand when the pot was big, while your even more playable hand got cracked when the pot was small. Winning money in a game like this comes down to who has the best cards when the pot is huge, not who plays consistently over time.

[UTGunner]

[ QUOTE ]
These games where the loser or losers match the pot don't seem to be something that can produce long term +EV results. Lucky timing is too much of a factor. The luck of the cards you're dealt is a factor in all forms of poker. In hold'em you could be dealt the best (AA) or the worst (72o) at any given time. The dieffernce is that in most games you can expect the pot size to be consistent when you have the best of it and when you have the worst of it. I KNOW, some pots stay small and some get huge, but over one session of poker an average pot size can be determined. This way you don't suffer huge swings.

Here's an example of why I hate "match the pot" games. Suppose there's a $10 pot in your 3-card guts game. You find 3 jacks. Clearly a playable hand. 4 people stay in. You, a guy with 3 queens, a guy with 3 kings, and a guy with 3 aces. Obviously this is unlikely. But know you've just lost $10 and from now on you have to risk losing $30 to get involved in a hand. A few hands later somebody beats the pot hand with a pair. He just happened to have a playable hand when the pot was big, while your even more playable hand got cracked when the pot was small. Winning money in a game like this comes down to who has the best cards when the pot is huge, not who plays consistently over time.

[/ QUOTE ]

You are wrong. This is a great +EV game. I played it in college and made a ton of money.

Without any statistical analysis, i just dealt out a thousand hands to get an idea of where each hand fell on a 0 to 100 percentile scale. If there were 5 players I stayed with top 20% of hands, with 4 players top 25% etc. (There was no house hand in my game). Not sophisticated, but very profitable.

The size of the pot does not affect the EV of this game. How others play can affect it, loose players mean many mega-pots to win - but you have to play appropriately tight.

[Bozeman]

"You find 3 jacks. Clearly a playable hand. 4 people stay in. You, a guy with 3 queens, a guy with 3 kings, and a guy with 3 aces."

This is a complete strawman. Are you claiming that you wouldn't lose a ton with these exact starting hands in a seven card stud game? Are you claiming that this will happen enough to have any impact on your EV?

These games do have a fair amount of variance, but they can have great +EV. If you believe what you have written, you need to have your poker mind examined.

Craig

[bigpooch]

You also need to know that if a hand loses, how often does
it encounter multiple challengers: if so, the hand doesn't
quite lose the amount contributed to the next pot (other
losers also contribute!). Nevertheless, your simulations
should prove helpful for players that play the game.

It also seems that AKQ is unplayable with five or more
opponents, unless they are playing much too tight! But for
three or fewer opponents, it's definitely worth trying to
steal with; of course, that begs the question as to what
weaker hands are ever playable for heads up, three handed,
etc.

[James Boston]

I admitted that my example was extreme. Here's why I question the EV in these games. I will admit I have done NO statistical analysis of this.

In my experience with these games, when the pot hand gets beat, the game starts over with a fresh, smaller pot. Tight play at this point in the game will probably give you +EV. However, a bad beat when the pots are huge will completely cancel that out. Then, you're back to trying to win small pots. In games where the pot sizes are more consistent, one bad beat isn't going to negate the other pots you've won nearly as much.

Likewise, in a "non-match the pot game" an idiot who picks up a monster won't show +EV for the session. This is possible in Guts. These games, to me, are only profitable if you have the right hand at the right time, not play the best hands over time.