Probability question...

[J-K]

Given that I am holding pocket kings, and no other cards are known, what
are the chances that one or more aces hits the 5 card board? If you can,
explain the math behind the answer.

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[GrouchySmurf1002]

Post by J-K
Given that I am holding pocket kings, and no other cards are known, what
are the chances that one or more aces hits the 5 card board? If you can,
explain the math behind the answer.

50 unseen cards. 4 aces. It's easier to say "what is the chance no ace
hits the board?"

So that would be 46/50 * 45/49 * 44/48 = 77.45% So 22.55% of the time an
ace will fall.

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[Big Dave]

That is the odds of an ace on the flop, curious of answer to his question,
odds of an ace on board. (all 5 cards).

DB

Post by GrouchySmurf1002

Post by J-K
Given that I am holding pocket kings, and no other cards are known, what
are the chances that one or more aces hits the 5 card board? If you can,
explain the math behind the answer.

50 unseen cards. 4 aces. It's easier to say "what is the chance no ace
hits the board?"
So that would be 46/50 * 45/49 * 44/48 = 77.45% So 22.55% of the time an
ace will fall.

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[GrouchySmurf1002]

Post by Big Dave
That is the odds of an ace on the flop, curious of answer to his question,
odds of an ace on board. (all 5 cards).
DB

Post by GrouchySmurf1002

Post by J-K
Given that I am holding pocket kings, and no other cards are known, what
are the chances that one or more aces hits the 5 card board? If you can,
explain the math behind the answer.

50 unseen cards. 4 aces. It's easier to say "what is the chance no ace
hits the board?"
So that would be 46/50 * 45/49 * 44/48 = 77.45% So 22.55% of the time an
ace will fall.

Oops. I misread. You would just continue on though.

46/50 * 45/49 * 44/48 * 43/47 * 42/46 = 64.70% chance there is no ace on
board.

So there will at least one ace on board 35.30% of the time.

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[mb]

Post by GrouchySmurf1002

Post by Big Dave
That is the odds of an ace on the flop, curious of answer to his question,
odds of an ace on board. (all 5 cards).
DB

Post by GrouchySmurf1002

Post by J-K
Given that I am holding pocket kings, and no other cards are known, what
are the chances that one or more aces hits the 5 card board? If you can,
explain the math behind the answer.

50 unseen cards. 4 aces. It's easier to say "what is the chance no ace
hits the board?"
So that would be 46/50 * 45/49 * 44/48 = 77.45% So 22.55% of the time an
ace will fall.

Oops. I misread. You would just continue on though.
46/50 * 45/49 * 44/48 * 43/47 * 42/46 = 64.70% chance there is no ace on
board.
So there will at least one ace on board 35.30% of the time.

Actual probability will be slightly less, because for the game to reach
the river there should be other players in the game, and it's more likely
for them to continue to the river with having an ace in somebody's hand
than having random cards.

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[Jim C]

Post by GrouchySmurf1002
46/50 * 45/49 * 44/48 * 43/47 * 42/46 = 64.70% chance there is no ace on
board.

I think though, that you have to look at it from the perspective of
someone being able to beat you with that ace, meaning they have one.

In which case, the odds are a lot lower

Now we know 3 cards..and there are 3 aces left

46/49 * 45/48 * 44/ 47 * 43/46 * 42/45 = 71.8% chance no ace on the
board

[GrouchySmurf1002]

Post by Jim C

Post by GrouchySmurf1002
46/50 * 45/49 * 44/48 * 43/47 * 42/46 = 64.70% chance there is no ace on
board.

I think though, that you have to look at it from the perspective of
someone being able to beat you with that ace, meaning they have one.
In which case, the odds are a lot lower
Now we know 3 cards..and there are 3 aces left
46/49 * 45/48 * 44/ 47 * 43/46 * 42/45 = 71.8% chance no ace on the
board

Well, I'm sure that's why he's asking. But it isn't what he asked. You
can keep assuming to the point where 4 people all have an ace against your
kings, and thus the probability of an ace hitting the board is 0.

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[Jim C]

Post by GrouchySmurf1002
Well, I'm sure that's why he's asking. But it isn't what he asked.
You
can keep assuming to the point where 4 people all have an ace against your
kings, and thus the probability of an ace hitting the board is 0.

You are correct, HOWEVER, the most it can ever be is when there are
three aces out, so that would be worst case...and thus most applicable.

[GrouchySmurf1002]

Post by Jim C

Post by GrouchySmurf1002
Well, I'm sure that's why he's asking. But it isn't what he asked.
You
can keep assuming to the point where 4 people all have an ace against your
kings, and thus the probability of an ace hitting the board is 0.

You are correct, HOWEVER, the most it can ever be is when there are
three aces out, so that would be worst case...and thus most applicable.

Actually the worst case is still for all 4 aces to be live (your opponents
have none), as obviously an Ace is a very scary card when you're holding
pocket kings, cause you'll never know for sure if you're opponent has one.
So the more that are live, the more problem this causes for you.

Example: You hold your KK and the board comes AT9 rainbow. Your opponent
with QJ who originally bet and then called your raise, now makes a big bet
with his open-ender. You're in an awful position now, even though your
opponent doesn't hold an ace.

More importantly, I think the probability I gave for an Ace hitting just
the flop is much more appropriate in this case, as if one doesn't hit the
flop, you should be betting the hell out of your hand to drive out the Ax
hands, unless there is something like a 3-straight, 3 flush, or 3-straight
flush on board.

If his question was regarding going all-in preflop against someone with
Ax, then your answer is obviously more appropriate.

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[Gary Carson]

Post by GrouchySmurf1002
Actually the worst case is still for all 4 aces to be live (your opponents
have none)

And this is bad how?

Post by GrouchySmurf1002
as obviously an Ace is a very scary card when you're holding
pocket kings,

Well, an ace on the flop is even scarier to your opponents who don't have an
ace.>Your opponent

Post by GrouchySmurf1002
with QJ who originally bet and then called your raise, now makes a big bet
with his open-ender. You're in an awful position now, even though your
opponent doesn't hold an ace.

Are we playing no limit here (big bet)? You're playing against opponents
who'll call a big reaise preflop with QJ? If so then it's not a problem at
all, you'll get all his money eventually.

Post by GrouchySmurf1002
More importantly, I think the probability I gave for an Ace hitting just
the flop is much more appropriate

I'd think the appropriate number is the probability an active opponent has an
ace given that an ace hits the flop.


--
Gary Carson
You can buy an ebook manuscript copy of
The Complete Book of Hold'em Poker for $4.50 at
http://garycarson.com

[Daniel Austin]

This is a really HARD way of looking at the probability question.
In my opinion of course, but it is hard to interpret these formulas, when it
should be written in choose notation instead.
For the most part, these will be mathematically identical, but when the
formula is more eloquently written out, it is easier to pick up any errors.

Post by GrouchySmurf1002

Post by Jim C

Post by GrouchySmurf1002
46/50 * 45/49 * 44/48 * 43/47 * 42/46 = 64.70% chance there is no ace on
board.

I think though, that you have to look at it from the perspective of
someone being able to beat you with that ace, meaning they have one.
In which case, the odds are a lot lower
Now we know 3 cards..and there are 3 aces left
46/49 * 45/48 * 44/ 47 * 43/46 * 42/45 = 71.8% chance no ace on the
board

Well, I'm sure that's why he's asking. But it isn't what he asked. You
can keep assuming to the point where 4 people all have an ace against your
kings, and thus the probability of an ace hitting the board is 0.
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[Mike Bandy]

Post by Daniel Austin
This is a really HARD way of looking at the probability question.
In my opinion of course, but it is hard to interpret these formulas, when it
should be written in choose notation instead.
For the most part, these will be mathematically identical, but when the
formula is more eloquently written out, it is easier to pick up any errors.

What is choose notation? TIA.

[PacPalBuzz]

<< Subject: Re: Probability question...
From: Mike Bandy ***@hotmail.com
Date: Fri, Apr 30, 2004 23:20
Message-id: <***@4ax.com>
<< What is choose notation? TIA. >>

Mike - The number of different five card combinations possible in 50 cards is
given by the expression C(50,5).

Another way to write C(50,5) is "50 choose 5."

Solving, C(50,5) = (50!)/[(45!)(5!)] = 2118760.

"50 choose 5" is choose notation.

Buzz

[Mike Bandy]

Post by PacPalBuzz
<< What is choose notation? TIA. >>
Mike - The number of different five card combinations possible in 50 cards is
given by the expression C(50,5).
Another way to write C(50,5) is "50 choose 5."

Thanks, Buzz. I must've seen that hundreds of times. Oh, well.

[mrpoker]

Interesting post...

It implies pot odds have to be almost four to one to the holder of an
ace to play for the second ace, where one player calls (.28*4??)..

And can I assume that if the 28% applies when only one player calls,
that more callers than one suggests more than one ace has been
removed from the deck, and that the odds of catching an ace going
forward will therefore be lower than 28%, if any other caller also
holds an ace (something that happens quite often)..

IYO, would that reduce the chances of an ace falling to roughly 20% or
one positive occurrance for every five events? Can it be stated that
any ace-caller will require at least five to one pot odds in any event
where there are two or more callers?

Or am I all wet here?

Thank you in advance, I appreciate your analysis...

Post by Jim C

Post by GrouchySmurf1002
46/50 * 45/49 * 44/48 * 43/47 * 42/46 = 64.70% chance there is no ace on
board.

I think though, that you have to look at it from the perspective of
someone being able to beat you with that ace, meaning they have one.
In which case, the odds are a lot lower
Now we know 3 cards..and there are 3 aces left
46/49 * 45/48 * 44/ 47 * 43/46 * 42/45 = 71.8% chance no ace on the
board

[Chris]

Post by GrouchySmurf1002

Post by J-K
Given that I am holding pocket kings, and no other cards are known, what
are the chances that one or more aces hits the 5 card board? If you can,
explain the math behind the answer.

46/50 * 45/49 * 44/48 * 43/47 * 42/46 = 64.70% chance there is no ace on
board.
So there will at least one ace on board 35.30% of the time.

I'm going to read between the lines here a little and assume that the
OP wanted to know the chances of an A coming up against his KK because
he wants to know the odds of an opponenet holding Ax beating him. So I
would re-do Grouchy's calculations here as:

47/50 * 46/49 * 45/48 * 44/47 * 43/46 = 72.39%
So there would be an ace on the board 27.61% of the time with one
caller with an Ace.

48/50 * 47/49 * 46/48 * 45/47 * 44/46 = %80.81
So there would be an ace on the board 19.18% of the time with two
callers with an Ace.


-CJ

[PacPalBuzz]

<< Subject: Re: Probability question...
From: ***@mybluelight.com (Chris)
Date: Thu, Apr 29, 2004 08:28
Message-id: <***@posting.google.com>
<<"I'm going to read between the lines here a little and assume that the OP
wanted to know the chances of an A coming up against his KK because he wants to
know the odds of an opponenet holding Ax beating him. So I
would re-do Grouchy's calculations here as:">>

Chris - Yeah, that seems a better way to think. but in that case, don't you
want to put the card in the hand of that opponent with the AX? What I'm saying
is you can't just take an ace out of the deck. Don't you want to take another
card out as well? (Make it a six or something if you want to disregard
straights and flushes). But my point is, don't you want to think in terms of a
pack of only 48 cards including three aces?

And then, instead of
<<"47/50 * 46/49 * 45/48 * 44/47 * 43/46">>,
isn't <<45/48 * 44/47 * 43/46 * 42/45 * 41/44>>
even better?

Buzz

[PacPalBuzz]

<< Subject: Probability question...
From: "J-K" ***@gci.net
Date: Wed, Apr 28, 2004 11:27
Message-id: <CATjc.14186104$***@news.easynews.com> >>

J-K - Here's the plan of attack: We'll find the probability that no ace hits
the five card board and then find what we need to add to it to equal one. Do
you need an explanation for this?

Let's call the probability that no ace hits the board Q.
Let's call the probability that at least one ace hits the board P. Do you need
an explanation for this?

The sum of these probabilities is 1. Do you need an explanation for this?

Then P + Q = 1
is the equation for "The sum of these probabilities is 1." Do you need an
explanation for this?

Let's proceed to finding Q.
There are 50 missing cards, of which four are aces. From the viewpoint of the
player with the two kings, any five of the 50 cards could make up the five card
board. The number of different five card combinations possible in 50 cards is
given by the expression C(50,5). Do you need an explanation for this?
Another way to write C(50,5) is "50 choose 5."
Another way to write C(50,5) is 50!/5!/45!
Another way to write C(50,5) is
50*49*48*47*46/1/2/3/4/5.
You're probably more familiar with writing this as
(50*49*48*47*46)/(1*2*3*4*5).
Solving, C(50,5) = 2118760.

211870 is the number of possible five card boards when you hold two kings.

Now we need to find the number of possible five card boards that do not have
any aces when you hold two kings.

If there are to be no aces, then there are only 46 cards that can be on the
five card board. Do you need an explanation for this?

The number of different five card combinations with no aces is given by the
expression C(46,5). Do you need an explanation for this?
Solving, C(46,5) = 1370754.

The probability of no aces is C(46,5)/C(50,5) =
1370754/211870. Do you need an explanation for why we're dividing here?

If not, the result of the calculation is 0.647.

Then, from P+Q = 1, doing a little algebra,
P + 0.647 = 1
P = 1 - 0.647
P = 0.353, which is the probability we seek.

You ask <<"what are the chances">>

Assuming "chances" mean "probability," the answer is 0.353.

If you want odds, they are 647 to 353 against, or
1.83 to 1 against, or
close to 2 to 1 against. Do you need an explanation for this?

1.83 to 1 is actually a bit less than 2 to 1. If you're getting at least two to
one for your money, then you have a good bet (favorable odds).

I'm not a mathematician. (But I do like to solve math problems that relate to
things that interest me). If you need any explanations along the way, perhaps
they might be better provided by one of the excellent mathematicians who
sometimes post here.

Lastly, in the famous words of Jerrod Ankenman, who took me apart in the 2004
ESCARGOT Chowaha contest (I was the chip leader with three of us left),
remember

"Math is hard."

Buzz

[rbdavis3]

Post by J-K
Given that I am holding pocket kings, and no other cards are known, what
are the chances that one or more aces hits the 5 card board? If you can,
explain the math behind the answer.

EVERY FREAKING TIME!

No explanation necessary.

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[J-K]

Thanks for the replies so far...
Here is the reason for the question. I have a bet with someone on the
probability of an ace hitting the board when he has pocket kings. I took
35.3% and every way I know mathematically comes to 35.3% of the time. He
is adamant that the correct answer should be 40%, and in all my efforts to
convince him that it is 40% he fails to grasp the concept. I don't know
if there is an explanation that can be worded just right to convince him,
but thanks for everybody's help.

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[PacPalBuzz]

<< Subject: Re: Probability question...
From: "J-K" ***@gci.net
Date: Wed, Apr 28, 2004 15:37
Message-id: <7fXjc.14197454$***@news.easynews.com>>>

<<"He is adamant that the correct answer should be 40%, and in all my efforts
to convince him that it is 40% he fails to grasp the concept. I don't know
if there is an explanation that can be worded just right to convince him...">>

J-K - It's very simple.

You put up sixty dollars and he puts up only forty dollars. Take a 52 card deck
and remove two kings. Then shuffle, cut, burn, flop, burn, turn, burn, river.

If there is an ace in the five cards, he gets your sixty dollars, but if there
isn't, you get his forty dollars.

If the correct answer is 40% and you do this 100 times, then you and he should
break even.

If the correct answer is 35.3% and you do this 100 times, then you figure to
win $470.

Here's the math:
$40.00*64.7 - $60*35.3 = $2588 - $2118 = $470.

That should do the trick.

($40*60 - $60*40 would = break even).

But if you're trying to win an argument with someone who refuses to admit when
he is wrong, then he won't go for the bet, and he won't admit he's wrong
either.

Buzz

[MM]

I like your way of persuading the guy's friend. But in case he gets
unlucky early and loses a few hundred before the law of averages and
probabilities turn around, try using cents instead of dollars. After a
while increase it to dimes. He'll finally understand. :)

[J-K]

Just to show you how stubborn he is, this is what I explained to him. To
get to his 40% number that you need to add the 4 in 50 chance of an ace
five times over for a board. But the reason that is incorrect is it is
not an accurate mathematical formula for this particular instance. If we
were to use his math to figure out the probability of an over card coming
against two jacks, we would get 12 of 50 five times over which would be 60
out of 50, meaning that there is 120% chance of an over card against two
jacks, a mathematical impossibility. So, what can you do. Thanks for the
response so far.

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[Beldin the Sorcerer]

Post by J-K
Thanks for the replies so far...
Here is the reason for the question. I have a bet with someone on the
probability of an ace hitting the board when he has pocket kings. I took
35.3% and every way I know mathematically comes to 35.3% of the time. He
is adamant that the correct answer should be 40%, and in all my efforts to
convince him that it is 40% he fails to grasp the concept. I don't know
if there is an explanation that can be worded just right to convince him,
but thanks for everybody's help.

The correct math is
1-(46/50x45/49x44/48x43/47x42/46)



do it on a calculator in front of him