Question about Problem in "Winning Low Limit Holdem"
us00...@interramp.com
In Lee Jones book "Winning Low Limit Holdem" he poses a question in one of
his quizzes that I do not understand the answer to.
On page 21 question 9 he asks what the odds are of making a full house
or better on the river with 88 and the board showing 6789. I understand the
first part of the answer that says there are 46 cards left and 10 of those
give you the full house or better which makes the odds 36:10. But after
that he writes:
If you want to squeak out the last detail, note that for your opponent to
have a straight, he must have either a five or a ten. Since that card is
one you don't want to see, you can remove it from your theoritical deck
and say you have 10 outs from 45 unseen cards make the odds 35:10.
I totally do not understand that part of it. Obviously I don't want him to have
the staight, but what does that have to do with my odds of drawing a full
house or better? And why does he remove the 1 card from the theoritical
deck?
Please help me understand.
Ben Clark
Tim Dierks
us00...@interramp.com wrote:
Since you'll beat anyone who doesn't have a straight, you only need to
fill up if someone has a 5 or a 10. Thus, the specific question you're
looking for an answer is: "Assuming someone has a straight, what are my
chances of improving to beat them?"
Because if they've got a straight, they've got to have a 5 or a 10, you
can assume that the chances of a 5 or 10 coming are a little lower; this
improves your chances of catching a 6, 7, 8, or 9. (to 35:10, as he says)
If, on the other hand, noone has a 5 or a 10, your chances of filling up
are a little worse (36:10); however, in this case, you're not as
concerned, because your set is likely to hold up (so you didn't need to
draw in the first place).
Note that this logic can be extended to drawing towards a full house when
there's a possible flush, as well. It's true any time that your opponent
would need to have a card other than the card you are drawing towards in
order to beat you.
--
Tim Dierks - Software Haruspex - t...@dierks.org
If you can't lick 'em, stick 'em on with a big piece of tape. - Negativland
Tom Weideman
>
>In Lee Jones book "Winning Low Limit Holdem" he poses a question in one of
>his quizzes that I do not understand the answer to.
>
>On page 21 question 9 he asks what the odds are of making a full house
>or better on the river with 88 and the board showing 6789. I understand the
>first part of the answer that says there are 46 cards left and 10 of those
>give you the full house or better which makes the odds 36:10. But after
>that he writes:
>
> If you want to squeak out the last detail, note that for your opponent to
> have a straight, he must have either a five or a ten. Since that card is
> one you don't want to see, you can remove it from your theoritical deck
> and say you have 10 outs from 45 unseen cards make the odds 35:10.
>
>I totally do not understand that part of it. Obviously I don't want him to have
>the staight, but what does that have to do with my odds of drawing a full
>house or better? And why does he remove the 1 card from the theoritical
>deck?
What he is doing is changing the odds from drawing to a full house to
the odds of winning the hand. He does this by noting that if your
opponent does not have the one card for the straight, then you don't
need the board to pair. It is not an accurate calculation, and I
personally think it is best to omit this part, since it leaves out so
much. If he wanted to compute the true odds of winning the hand, he
should go through ALL of the details, rather than make this confusing
approximation. Since this is a book for beginners, omitting this
entirely probably would have been the best course.
Even if Lee explains that this is an approximation at the true odds, the
effort required to explain to what extent this approximation is accurate
is daunting. My feeling is that it is highly unlikely that the 35:10 as
opposed to 36:10 odds will effect one's decision to fold, call, or
raise, and if this was done as an instructive mathematical exercise,
more detail is warranted.
Aloha,
Tom Weideman
Tom Weideman
>What he is doing is changing the odds from drawing to a full house to
>the odds of winning the hand.
I was very unclear here... Let me try again. He is giving an
approximation of the odds of improving your hand GIVEN that you need to
improve in order to win. The way I stated it above was very sloppy, as
it implies that Lee's calculation includes the possibility that you are
currently in the lead... it doesn't. Sorry about that. The remainder
of the post regarding approximating, etc. still rings true.
To calculate the true odds of improving *given that you need to
improve*, you must include all of the possibilities. Count the number
of hands your opponent can have where he currently stands better than
you and the number of outs in each case; then go from there. It's a
good exercise to see how close this comes to Lee's approximate odds.
Aloha,
Tom Weideman
Lee Jones
Tom Weideman <weid...@thomasj.iii.net> wrote:
>us00...@interramp.com wrote:
>>
>>In Lee Jones book "Winning Low Limit Holdem" he poses a question in one of
>>his quizzes that I do not understand the answer to.
>>
>>On page 21 question 9 he asks what the odds are of making a full house
>>or better on the river with 88 and the board showing 6789. I understand the
>>first part of the answer that says there are 46 cards left and 10 of those
>>give you the full house or better which makes the odds 36:10. But after
>>that he writes:
>>
>> If you want to squeak out the last detail, note that for your opponent to
>> have a straight, he must have either a five or a ten. Since that card is
>> one you don't want to see, you can remove it from your theoritical deck
>> and say you have 10 outs from 45 unseen cards make the odds 35:10.
>>
>>I totally do not understand that part of it. Obviously I don't want him to have
>>the staight, but what does that have to do with my odds of drawing a full
>>house or better? And why does he remove the 1 card from the theoritical
>>deck?
>
>opponent does not have the one card for the straight, then you don't
>need the board to pair. It is not an accurate calculation, and I
>personally think it is best to omit this part, since it leaves out so
>much. If he wanted to compute the true odds of winning the hand, he
>should go through ALL of the details, rather than make this confusing
>approximation. Since this is a book for beginners, omitting this
>entirely probably would have been the best course.
In retrospect, yes, it would have been better to leave out the entire
paragraph. By way of explanation (but not excuse) I'd like to plead that
it was a bunch of math.weenies that were reviewing the book, and we all
thrive on that sort of minutae. I distinctly recall one edit in the draft
where one my reviewers commented that he was waiting to see if I'd miss that
detail.
It is true that you can use this kind of logic to turn a close fold into a
close call, but it's a second order effect would have been best left to
a later chapter, or removed completely.
For everybody following at home:
To a first approximation, if you have a set, and believe your opponent has a
straight or flush on the turn, you're a 3.6:1 dog to improve to the best
hand on the river - play accordingly.
For more advanced students: If you're ahead on the turn and stay there on
the river, or improve on the river to pass a better hand, you win either way.
In my ill-chosen example in the book, I point out that for your opponent
to have a straight on the turn (with the board showing 6-7-8-9), he must have
a 5 or a 10. Given that, one of two things is true:
1. You're in front because he *doesn't* have a 5 or 10, so you don't need to
improve, or
2. He *does* have a 5 or 10, so you're drawing, but you can remove that card
from the deck from which you're drawing. So now you're just a 3.5:1 dog.
How much difference, you ask, does 3.5:1 make compared to 3.6:1? None,
except to a math.weenie. The flush example actually gets more interesting,
because there are times when you may want to give up your set if a
fourth card of the suit hits, and the board doesn't pair. Please
remember, however, that these are second order effects, and furthermore,
don't forget what (Benjamin Franklin?) said about the census, and the
town watchman, and so forth.
Regards, Lee
--
Lee Jones | "I like to dream - right between my sound machine.
le...@sgi.com | On a cloud of sound I drift in the night..."
415-933-3356 | -John Kay (Steppenwolf)
cw
>
>In Lee Jones book "Winning Low Limit Holdem" he poses a question in one of>
> If you want to squeak out the last detail, note that for your opponent to
> have a straight, he must have either a five or a ten. Since that card is
> one you don't want to see, you can remove it from your theoritical deck
> and say you have 10 outs from 45 unseen cards make the odds 35:10.
>
>
>Please help me understand.
>Ben Clark
> The 10 is what he will hold throw out one 10 for 44:10
Andy Bloch
half the pot, assuming he doesn't have a jack. If he has a ten, you have 3
outs to split. So you're much better off than a 3.5:1 dog.
In article <4cru54$k...@fido.asd.sgi.com>, le...@diver.asd.sgi.com says...
Tom Weideman
>2. He *does* have a 5 or 10, so you're drawing, but you can remove that card
> from the deck from which you're drawing. So now you're just a 3.5:1 dog.
> How much difference, you ask, does 3.5:1 make compared to 3.6:1? None,
> except to a math.weenie. The flush example actually gets more interesting,
> because there are times when you may want to give up your set if a
> fourth card of the suit hits, and the board doesn't pair. Please
> remember, however, that these are second order effects, and furthermore,
> don't forget what (Benjamin Franklin?) said about the census, and the
> town watchman, and so forth.
In my followup post, I mentioned that you were refering to probabilities
*assuming* that your are drawing. Unfortunately, the odds still aren't
correct, or at least they are not relevant. If you tell me: "You are
currently behind." on the turn, then there are 4 different possibilities
(assuming no flush possibilities) [memory refresher: board is 6789, I
have 88]
1. My opponent has a 5 and some other random card which is not a T.
2. My opponent has a T and some other random card which is not a J.
3. My opponent has a JT.
4. My opponent has pocket 9's.
In case 1, I have, in addition to 10 outs to win outright, 7 outs to
split the pot. In case 2, I have 3 additional outs to split the pot.
In case 3, I have no split-the-pot outs (but still have 10 winning
outs). And in case 4, I only have 1 out to win the pot, and none to split.
To incorporate the splits, it is really more appropriate to consider EV
than odds, and as I said, this is more complicated for this example
than stated in the text.
I realize it's a monumental task to write a book such as yours, and I
commend you on the job you have done. I also know that you are
well-aware of all that I have written above.
To anyone who may be excessively critical, I would say this: Teaching
any complex subject is very tricky. One must always balance the desire
to give all of the details (and therefore be complete) with the need to
teach a basic understanding. For example, someone knowledgeable about
physics could sit in a class I am teaching and say: "You are lying to
your students because you are not considering the effects of
relativity." Obviously if I do that, no one will learn anything. In
the case of a book, the restrictions are even greater... one cannot go
on forever. The trick is deciding where to draw the line.
Aloha,
Tom Weideman
Tom Weideman
>(assuming no flush possibilities) [memory refresher: board is 6789, I
>have 88]
>1. My opponent has a 5 and some other random card which is not a T.
>2. My opponent has a T and some other random card which is not a J.
>3. My opponent has a JT.
>4. My opponent has pocket 9's.
>
>In case 1, I have, in addition to 10 outs to win outright, 7 outs to
>split the pot. In case 2, I have 3 additional outs to split the pot.
>In case 3, I have no split-the-pot outs (but still have 10 winning
>outs). And in case 4, I only have 1 out to win the pot, and none to split.
Whoops, Of course a 5 or T splits the pot in case #4. Boy is my face red! :)
Aloha,
Tom Weideman
Chuck Johnson
(vs. *0*). Any 5 or T.
Chuck
Seth Breidbart
Tom Weideman <weid...@thomasj.iii.net> wrote:
>Tom Weideman <weid...@thomasj.iii.net> wrote:
>
>>(assuming no flush possibilities) [memory refresher: board is 6789, I
>>have 88]
>>1. My opponent has a 5 and some other random card which is not a T.
The second card can't be a J either, or 4 of your splits go away.
>>2. My opponent has a T and some other random card which is not a J.
>>3. My opponent has a JT.
>>4. My opponent has pocket 9's.
>>
>>In case 1, I have, in addition to 10 outs to win outright, 7 outs to
>>split the pot. In case 2, I have 3 additional outs to split the pot.
>>In case 3, I have no split-the-pot outs (but still have 10 winning
>>outs). And in case 4, I only have 1 out to win the pot, and none to split.
>
>Whoops, Of course a 5 or T splits the pot in case #4. Boy is my face red! :)
Seth
Tom Weideman
>In Lee Jones book "Winning Low Limit Holdem" he poses a question in one of
>his quizzes that I do not understand the answer to.
>
>On page 21 question 9 he asks what the odds are of making a full house
>or better on the river with 88 and the board showing 6789. I understand the
>first part of the answer that says there are 46 cards left and 10 of those
>give you the full house or better which makes the odds 36:10. But after
>that he writes:
>
> If you want to squeak out the last detail, note that for your opponent to
> have a straight, he must have either a five or a ten. Since that card is
> one you don't want to see, you can remove it from your theoritical deck
> and say you have 10 outs from 45 unseen cards make the odds 35:10.
>
>I totally do not understand that part of it. Obviously I don't want him to have
>the staight, but what does that have to do with my odds of drawing a full
>house or better? And why does he remove the 1 card from the theoritical
>deck?
To which at least part of my answer stated:
> It is not an accurate calculation...
>...If he wanted to compute the true odds of winning the hand, he
should go through ALL of the details...
In other words, the stated odds of 35:10 under the circumstances given are
*incorrect*. In my later posts, I left this alone and focused on the more
practical aspect of including split pots (and I did a very poor job of it, I
might add). As penance, I have decided to compute the correct exact odds and
demonstrate the method for those of you who may be interested.
I will start by simplifying the question some. I'll get rid of the possible
over set and make it so there is only one card for the straight. Assume the
board has on it AKQJ rainbow, and you hold AA. You are heads up. Consider
the following two scenarios:
1. Your opponent shows you that he holds a ten by turning it over, but does
not show you the other card and makes no claims about it.
2. Your opponent truthfully tells you that he has a ten (i.e. at least one ten)
in his hand without showing one to you.
Question: Are the probabilities of your winning the hand (disregarding splits)
the same in these two cases?
Answer: NO!
I will do the calculations in a moment, but first let me point out that the
situation Lee describes is equivalent to #2 in that you proceed on the
assumption that he has the straight. But the calculation of odds done by Lee
is equivalent to #1, since the ten shown *really is* taken out of the deck of
cards available for the other two slots (opponents other card and river).
Here is how the calculations are done: Imagine three slots into which cards
are placed: Opponent's card #1, opponent's card #2, and river card. Count the
number of card placements that will satisfy the criteria given for each case.
Then count the number of card placements that result in a win for you in each
case and divide by the previous number. This is the probability of winning in
each case. Note that we will have to account for the fact that card order is
not important. Let's do some counting...
Case #1: Total number of possibilities
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Slot 1 (opp #1) Slot 2 (opp #2) Slot 3 (river) Total
T (1 card, shown) 2 thru 9 (32 cards) remaining (44 cards) 1408
T (1 card, shown) T (3 cards) remaining (44 cards) 132
T (1 card, shown) J,Q,K (9 cards) remaining (44 cards) 396
T (1 card, shown) A (1 card) remaining (44 cards) 44
total outcomes: 1980
Case #1: Number of ways to win
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Slot 1 (opp #1) Slot 2 (opp #2) Slot 3 (river) Total
T (1 card, shown) 2 thru 9 (32 cards) outs (10 cards) 320
T (1 card, shown) T (3 cards) outs (10 cards) 30
T (1 card, shown) J,Q,K (9 cards) outs (9 cards) 81
T (1 card, shown) A (1 card) outs (9 cards) 9
total outcomes: 440
Case #1: Winning probability = 440/1980 = 0.222222 (= 10:35)
Case #2: Total number of possibilities
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Slot 1 (opp #1) Slot 2 (opp #2) Slot 3 (river) Total
T (4 cards) 2 thru 9 (32 cards) remaining (44 cards) 5632
2 thru 9 (32 cards) T (4 cards) remaining (44 cards) 5632
T (4 cards) J,Q,K (9 cards) remaining (44 cards) 1584
J,Q,K (9 cards) T (4 cards) remaining (44 cards) 1584
T (4 cards) A (1 card) remaining (44 cards) 176
A (1 card) T (4 cards) remaining (44 cards) 176
T (4 cards) T (3 cards) remaining (44 cards) 528
total outcomes: 15312
total outcomes (disregarding card order): 15312/2 = 7656
Case #2: Number of ways to win
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Slot 1 (opp #1) Slot 2 (opp #2) Slot 3 (river) Total
T (4 cards) 2 thru 9 (32 cards) outs (10 cards) 1280
2 thru 9 (32 cards) T (4 cards) outs (10 cards) 1280
T (4 cards) J,Q,K (9 cards) outs (9 cards) 324
J,Q,K (9 cards) T (4 cards) outs (9 cards) 324
T (4 cards) A (1 card) outs (9 cards) 36
A (1 card) T (4 cards) outs (9 cards) 36
T (4 cards) T (3 cards) outs (10 cards) 120
total outcomes: 3400
total outcomes (disregarding card order): 3400/2 = 1700
Case #2: Winning probability = 1700/7656 = 0.222048
You can see that these odds do not differ by very much, and that in fact we
get close to the actual odds by making the approximation of case #1. But
case #1 does *not* give the exact odds.
Where does the difference come from? When the opponent shows the ten,
it is more restrictive to the total number of hands than when he just states
that he has a ten (or equivalently, when we assume he has one). This is
because there are 6 hands in which any two tens can appear, but only 3
hands in which a *specific* ten (the one shown) can appear. But the number
of outs against TT remains the same either way, so the probability of
winning is higher in the more restrictive case. At its core, this is somewhat
similar to the "Monty Hall problem" discussed not too long ago in rgp. This
probably explains why it is hard to grasp intuitively.
This process can be repeated (though it is a bit trickier) for the open-ended
4-straight on the board, and the divergence between Lee's approximation and
the correct probability is greater in that case, but still it is not significant.
If you include the possibility of an over set in the original problem, it
diverges a little more. Sorry, I *do* have enough of a life that I don't have
the time to compute the exact expected return for this problem (including
split pots). I encourage anyone who has the time to do so, however, as I
would be interested in the result. It would be useful to know for pot and no-
limit considerations.
Never was my sig more appropriate than in this case. :)
Aloha,
Tom Weideman | "Things should be made as simple as possible,
| but not any simpler." -- Albert Einstein
Andrew Bloch
>I will start by simplifying the question some. I'll get rid of the
>possible over set and make it so there is only one card for the
>straight. Assume the board has on it AKQJ rainbow, and you hold AA.
>You are heads up. Consider the following two scenarios:
>
>1. Your opponent shows you that he holds a ten by turning it over,
>but does not show you the other card and makes no claims about it.
>
>2. Your opponent truthfully tells you that he has a ten (i.e. at
>least one ten) in his hand without showing one to you.
>
>Question: Are the probabilities of your winning the hand
>(disregarding splits) the same in these two cases?
>
>Answer: NO!
Actually, correct answer is: it depends!
Similar the Monty Hall problem, which Tom refered to, the correct
answer depends on the exact algorithm your opponent uses for exposing
one card.
Does Monty Hall always open a door without the prize behind it? Or
does he open a door at random, and if there happened to be the prize
behind that door, the contestant just loses? Or, does he only offer
you the chance to switch when you've picked the right door? [I think
he actually used a mixed strategy, sometimes opening a door that had
the prize behind it, or not giving you a chance to switch, but I
haven't seen a rerun of Let's Make a Deal in a long time so I'm not
sure.] Most of the time that people pose this problem, they don't make
it clear how Monty chooses a door to open.
Back to poker: Does your opponent in scenario 1 always pick one card
and expose that card, whether or not it is actually a ten? Or, does
your opponent in scenario 1 look at both his cards, and only expose
a card if it is a ten? If your opponent follows the latter algorithm
your probability of winning the hand is the same as in scenario 2.
Andy
Tom Weideman
>In article <4d3scp$l...@malasada.lava.net>, weid...@lava.net says...
>>possible over set and make it so there is only one card for the
>>straight. Assume the board has on it AKQJ rainbow, and you hold AA.
>>You are heads up. Consider the following two scenarios:
>>
>>1. Your opponent shows you that he holds a ten by turning it over,
>>but does not show you the other card and makes no claims about it.
>>
>>2. Your opponent truthfully tells you that he has a ten (i.e. at
>>least one ten) in his hand without showing one to you.
>>
>>Question: Are the probabilities of your winning the hand
>>(disregarding splits) the same in these two cases?
>>
>>Answer: NO!
>
>
>Similar the Monty Hall problem, which Tom refered to, the correct
>answer depends on the exact algorithm your opponent uses for exposing
>one card.
>
>Does Monty Hall always open a door without the prize behind it? Or
>does he open a door at random, and if there happened to be the prize
>behind that door, the contestant just loses? Or, does he only offer
>you the chance to switch when you've picked the right door? [I think
>he actually used a mixed strategy, sometimes opening a door that had
>the prize behind it, or not giving you a chance to switch, but I
>haven't seen a rerun of Let's Make a Deal in a long time so I'm not
>sure.] Most of the time that people pose this problem, they don't make
>it clear how Monty chooses a door to open.
>
>Back to poker: Does your opponent in scenario 1 always pick one card
>and expose that card, whether or not it is actually a ten? Or, does
>your opponent in scenario 1 look at both his cards, and only expose
>a card if it is a ten? If your opponent follows the latter algorithm
>your probability of winning the hand is the same as in scenario 2.
Very nice. I didn't want to get into the "Monty Hall" portion of this
problem in my original post (though I did allude to it), and you do a
great job of explaining it. I was tempted to delve into it deeper in a
later post by asking the following brain teaser: "Suppose he tells you
he has a ten (case #2) and *then* exposes a ten (case #1). Which
probability is correct?" Your comments have short-circuited my mischief
by making it clear that the probability for case #2 is still correct.
Hopefully the answer to the original question is clear: The method
described in Lee's book does give an approximation of the odds (though
for practical purposes it is not necessary), but not the exact odds.