Beau Peep Notice Board
Beau Peep Notice Board => Saloon Bar => Topic started by: Roger Kettle on June 30, 2008, 09:07:09 PM
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Mighty quiet in here so I thought I'd mosey on in for a beer. Here are a few true tales from the Old West....
When Wild Bill Hickok was killed in a Deadwood bar, he was playing poker. He was holding two aces, two eights and a queen----now known, unsurprisingly, as "Dead Man's Hand".
This is the last day of June. The Sioux name for this month was "the moon of ripening berries".
Among his family, Custer's nickname was "Autie". As a boy, this was how he pronounced his own middle name, Armstrong. You have to say this was not a great attempt.
At the Battle of the Little Bighorn. there were 379 immigrant soldiers from 25 different countries.
The outlaw, John Wesley Hardin, was once sleeping in a hotel when the snoring from the adjoining room woke him. He fired two bullets through the wall, instantly killing the culprit and, at the same time, inventing the world's quickest cure for snoring.
Okay, I've finished my beer so it's time to mosey back out to the street...
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When Wild Bill Hickok was killed in a Deadwood bar, he was playing poker. He was holding two aces, two eights and a queen----now known, unsurprisingly, as "Dead Man's Hand".
But the chance of having that particular hand (assuming that all hands have equal probability) is:
ACE + ACE + EIGHT + EIGHT + QUEEN
4/52 * 3/51 * 4/50 * 3/49 * 4/48 * (13C4) = 0.0023
So 0.2% of all hands will be a Dead Man's Hand. So if he played at least 1000 games of poker, he would have obtained a Dead Man's Hand at least twice, and so the "unsurprisingly" loses some of its irony.
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..0 I really do feel sorry for any of the pupils you teach Mince. ..0
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When Wild Bill Hickok was killed in a Deadwood bar, he was playing poker. He was holding two aces, two eights and a queen----now known, unsurprisingly, as "Dead Man's Hand".
But the chance of having that particular hand (assuming that all hands have equal probability) is:
ACE + ACE + EIGHT + EIGHT + QUEEN
4/52 * 3/51 * 4/50 * 3/49 * 4/48 * (13C4) = 0.0023
So 0.2% of all hands will be a Dead Man's Hand. So if he played at least 1000 games of poker, he would have obtained a Dead Man's Hand at least twice, and so the "unsurprisingly" loses some of its irony.
Okay, it's late, and I may be missing something obvious here, but wasn't the hand named 'Dead Man's Hand' because Hickok was holding those cards when he was killed, not because he the hand resulted in his death? I saw no irony in Roger's use of the word "unsurprisingly", and therefore your sums were a complete and utter waste of time.
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And also he was highly unlikely to get a "Dead Man's Hand" again in his lifetime anyway. ..0
Happy July, everybody. (We had it here first).
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Okay, it's late, and I may be missing something obvious here, but wasn't the hand named 'Dead Man's Hand' because Hickok was holding those cards when he was killed, not because he the hand resulted in his death? I saw no irony in Roger's use of the word "unsurprisingly", and therefore your sums were a complete and utter waste of time.
So I misread it. Sue me.
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Leave Mince alone, Tarks. Showing his 'superiority' and blinding us with maths keeps him happy!
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Leave Mince alone, Tarks. Showing his 'superiority' and blinding us with maths keeps him happy!
I think Mince might have just scored.
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When Wild Bill Hickok was killed in a Deadwood bar, he was playing poker. He was holding two aces, two eights and a queen----now known, unsurprisingly, as "Dead Man's Hand".
But the chance of having that particular hand (assuming that all hands have equal probability) is:
ACE + ACE + EIGHT + EIGHT + QUEEN
4/52 * 3/51 * 4/50 * 3/49 * 4/48 * (13C4) = 0.0023
So 0.2% of all hands will be a Dead Man's Hand. So if he played at least 1000 games of poker, he would have obtained a Dead Man's Hand at least twice, and so the "unsurprisingly" loses some of its irony.
I don't think so. I can see where you're coming from, but the (13C4)
term doesn't make sense to me. The bit before that is the probability
of getting the 5 cards dealt in the order AA88Q, and it looks like the
(13C4) is an attempt to adjust for the different possible orders, but
it doesn't make sense.
What you need to do for the last term is to multiply the probabilities
by the number of different orders you could get AA88Q in. It doesn't
matter which suits they are, as you've already allowed for that in the
earlier calculation (4/52 refers to any ace).
The actual number of ways you can arrange AA88Q is 30. This comes from
the number of ways to arrange 5 cards = 5! = 5*4*3*2*1=120. Then since
you have 2 aces you have to divide this by 2! (which equals 2), to get
60, and then by 2! again for the eights to get 30.
So the actual probability is 4/52 * 3/51 * 4/50 * 3/49 * 4/48 * 30 =
0.0000554 , or approximately a one in eighteen thousand chance.
We can confirm this by thinking about it slightly differently. We can
calculate how many different poker hands there are, i.e. the number of
ways to choose 5 cards from 52. This is the calculation (52 C 5) which
can be calculated as 2598960. Then we need to work out how many of
these are the dead man's hand. There are 6 possible ways to choose two
aces, 6 possible ways to choose two eights, and 4 possible queens.
Multiply these together 6x6x4 = 144 possible dead man's hands.
So the probability of a dead man's hand is 144/2598960 = 0.0000554,
the same as above.
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Can we call it a bore draw? ..0 ..0
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When Wild Bill Hickok was killed in a Deadwood bar, he was playing poker. He was holding two aces, two eights and a queen----now known, unsurprisingly, as "Dead Man's Hand".
But the chance of having that particular hand (assuming that all hands have equal probability) is:
ACE + ACE + EIGHT + EIGHT + QUEEN
4/52 * 3/51 * 4/50 * 3/49 * 4/48 * (13C4) = 0.0023
So 0.2% of all hands will be a Dead Man's Hand. So if he played at least 1000 games of poker, he would have obtained a Dead Man's Hand at least twice, and so the "unsurprisingly" loses some of its irony.
I don't think so. I can see where you're coming from, but the (13C4)
term doesn't make sense to me. The bit before that is the probability
of getting the 5 cards dealt in the order AA88Q, and it looks like the
(13C4) is an attempt to adjust for the different possible orders, but
it doesn't make sense.
What you need to do for the last term is to multiply the probabilities
by the number of different orders you could get AA88Q in. It doesn't
matter which suits they are, as you've already allowed for that in the
earlier calculation (4/52 refers to any ace).
The actual number of ways you can arrange AA88Q is 30. This comes from
the number of ways to arrange 5 cards = 5! = 5*4*3*2*1=120. Then since
you have 2 aces you have to divide this by 2! (which equals 2), to get
60, and then by 2! again for the eights to get 30.
So the actual probability is 4/52 * 3/51 * 4/50 * 3/49 * 4/48 * 30 =
0.0000554 , or approximately a one in eighteen thousand chance.
We can confirm this by thinking about it slightly differently. We can
calculate how many different poker hands there are, i.e. the number of
ways to choose 5 cards from 52. This is the calculation (52 C 5) which
can be calculated as 2598960. Then we need to work out how many of
these are the dead man's hand. There are 6 possible ways to choose two
aces, 6 possible ways to choose two eights, and 4 possible queens.
Multiply these together 6x6x4 = 144 possible dead man's hands.
So the probability of a dead man's hand is 144/2598960 = 0.0000554,
the same as above.
You took the words right out of my mouth, Fido! ..0
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Unsurprisingly, I couldn't care less.
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Unsurprisingly, I couldn't care less.
You make it sound as though this is choice on your part. :)
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When Wild Bill Hickok was killed in a Deadwood bar, he was playing poker. He was holding two aces, two eights and a queen----now known, unsurprisingly, as "Dead Man's Hand".
But the chance of having that particular hand (assuming that all hands have equal probability) is:
ACE + ACE + EIGHT + EIGHT + QUEEN
4/52 * 3/51 * 4/50 * 3/49 * 4/48 * (13C4) = 0.0023
So 0.2% of all hands will be a Dead Man's Hand. So if he played at least 1000 games of poker, he would have obtained a Dead Man's Hand at least twice, and so the "unsurprisingly" loses some of its irony.
I don't think so. I can see where you're coming from, but the (13C4)
term doesn't make sense to me. The bit before that is the probability
of getting the 5 cards dealt in the order AA88Q, and it looks like the
(13C4) is an attempt to adjust for the different possible orders, but
it doesn't make sense.
What you need to do for the last term is to multiply the probabilities
by the number of different orders you could get AA88Q in. It doesn't
matter which suits they are, as you've already allowed for that in the
earlier calculation (4/52 refers to any ace).
The actual number of ways you can arrange AA88Q is 30. This comes from
the number of ways to arrange 5 cards = 5! = 5*4*3*2*1=120. Then since
you have 2 aces you have to divide this by 2! (which equals 2), to get
60, and then by 2! again for the eights to get 30.
So the actual probability is 4/52 * 3/51 * 4/50 * 3/49 * 4/48 * 30 =
0.0000554 , or approximately a one in eighteen thousand chance.
We can confirm this by thinking about it slightly differently. We can
calculate how many different poker hands there are, i.e. the number of
ways to choose 5 cards from 52. This is the calculation (52 C 5) which
can be calculated as 2598960. Then we need to work out how many of
these are the dead man's hand. There are 6 possible ways to choose two
aces, 6 possible ways to choose two eights, and 4 possible queens.
Multiply these together 6x6x4 = 144 possible dead man's hands.
So the probability of a dead man's hand is 144/2598960 = 0.0000554,
the same as above.
Yes, I know, but Roger would not have known the difference.
The answer would have been better if you had stuck either to permutations or to combinations.
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When Wild Bill Hickok was killed in a Deadwood bar, he was playing poker. He was holding two aces, two eights and a queen----now known, unsurprisingly, as "Dead Man's Hand".
But the chance of having that particular hand (assuming that all hands have equal probability) is:
ACE + ACE + EIGHT + EIGHT + QUEEN
4/52 * 3/51 * 4/50 * 3/49 * 4/48 * (13C4) = 0.0023
So 0.2% of all hands will be a Dead Man's Hand. So if he played at least 1000 games of poker, he would have obtained a Dead Man's Hand at least twice, and so the "unsurprisingly" loses some of its irony.
I don't think so. I can see where you're coming from, but the (13C4)
term doesn't make sense to me. The bit before that is the probability
of getting the 5 cards dealt in the order AA88Q, and it looks like the
(13C4) is an attempt to adjust for the different possible orders, but
it doesn't make sense.
What you need to do for the last term is to multiply the probabilities
by the number of different orders you could get AA88Q in. It doesn't
matter which suits they are, as you've already allowed for that in the
earlier calculation (4/52 refers to any ace).
The actual number of ways you can arrange AA88Q is 30. This comes from
the number of ways to arrange 5 cards = 5! = 5*4*3*2*1=120. Then since
you have 2 aces you have to divide this by 2! (which equals 2), to get
60, and then by 2! again for the eights to get 30.
So the actual probability is 4/52 * 3/51 * 4/50 * 3/49 * 4/48 * 30 =
0.0000554 , or approximately a one in eighteen thousand chance.
We can confirm this by thinking about it slightly differently. We can
calculate how many different poker hands there are, i.e. the number of
ways to choose 5 cards from 52. This is the calculation (52 C 5) which
can be calculated as 2598960. Then we need to work out how many of
these are the dead man's hand. There are 6 possible ways to choose two
aces, 6 possible ways to choose two eights, and 4 possible queens.
Multiply these together 6x6x4 = 144 possible dead man's hands.
So the probability of a dead man's hand is 144/2598960 = 0.0000554,
the same as above.
Yes, I know, but Roger would not have known the difference.
The answer would have been better if you had stuck either to permutations or to combinations.
Who wears combinations in the middle of summer?
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Happy July, everybody. (We had it here first).
I'm not really one for oneupmanship, Peepmaster, but Malc and I had it a couple of hours before you!
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Happy July, everybody. (We had it here first).
I'm not really one for oneupmanship, Peepmaster, but Malc and I had it a couple of hours before you!
That thought had occurred to me, Joan!
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That thought had occurred to me, Joan!
Next time it occurs to you to think, perhaps you might try it out.
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When Wild Bill Hickok was killed in a Deadwood bar, he was playing poker. He was holding two aces, two eights and a queen----now known, unsurprisingly, as "Dead Man's Hand".
But the chance of having that particular hand (assuming that all hands have equal probability) is:
ACE + ACE + EIGHT + EIGHT + QUEEN
4/52 * 3/51 * 4/50 * 3/49 * 4/48 * (13C4) = 0.0023
So 0.2% of all hands will be a Dead Man's Hand. So if he played at least 1000 games of poker, he would have obtained a Dead Man's Hand at least twice, and so the "unsurprisingly" loses some of its irony.
I don't think so. I can see where you're coming from, but the (13C4)
term doesn't make sense to me. The bit before that is the probability
of getting the 5 cards dealt in the order AA88Q, and it looks like the
(13C4) is an attempt to adjust for the different possible orders, but
it doesn't make sense.
What you need to do for the last term is to multiply the probabilities
by the number of different orders you could get AA88Q in. It doesn't
matter which suits they are, as you've already allowed for that in the
earlier calculation (4/52 refers to any ace).
The actual number of ways you can arrange AA88Q is 30. This comes from
the number of ways to arrange 5 cards = 5! = 5*4*3*2*1=120. Then since
you have 2 aces you have to divide this by 2! (which equals 2), to get
60, and then by 2! again for the eights to get 30.
So the actual probability is 4/52 * 3/51 * 4/50 * 3/49 * 4/48 * 30 =
0.0000554 , or approximately a one in eighteen thousand chance.
We can confirm this by thinking about it slightly differently. We can
calculate how many different poker hands there are, i.e. the number of
ways to choose 5 cards from 52. This is the calculation (52 C 5) which
can be calculated as 2598960. Then we need to work out how many of
these are the dead man's hand. There are 6 possible ways to choose two
aces, 6 possible ways to choose two eights, and 4 possible queens.
Multiply these together 6x6x4 = 144 possible dead man's hands.
So the probability of a dead man's hand is 144/2598960 = 0.0000554,
the same as above.
Yes, I know, but Roger would not have known the difference.
The answer would have been better if you had stuck either to permutations or to combinations.
I gave 2 methods of arriving at the correct answer. You gave one method
of arriving at the wrong answer. I think we may be able to agree which
is better.
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COMING SOON - a mathematical explanation of The Offside Law.
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I gave 2 methods of arriving at the correct answer. You gave one method
of arriving at the wrong answer. I think we may be able to agree which
is better.
How are you defining "better"? My only purpose was to convince Roger of something mathematical, and quite frankly I could have written the solution to a quadratic equation and he would have been none the wiser. Your method, on the other hand, was rather overkill. I just could not be bothered to look up the equation for combinations and permutations.
Or are you proposing maths duels at dawn? ;D
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I gave 2 methods of arriving at the correct answer. You gave one method
of arriving at the wrong answer. I think we may be able to agree which
is better.
Or are you proposing maths duels at dawn? ;D
If it involves terminating your existence, I'm all for it. ..0
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I gave 2 methods of arriving at the correct answer. You gave one method
of arriving at the wrong answer. I think we may be able to agree which
is better.
How are you defining "better"? My only purpose was to convince Roger of something mathematical, and quite frankly I could have written the solution to a quadratic equation and he would have been none the wiser. Your method, on the other hand, was rather overkill. I just could not be bothered to look up the equation for combinations and permutations.
Or are you proposing maths duels at dawn? ;D
Last word. Word.
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He don't like it when you stick it up him, Fido!
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Does this mean we're heading for a high noon showdown?
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I'm not really one for oneupmanship, Peepmaster, but Malc and I had it a couple of hours before you!
Joan, kindly stop blabbing to people every time you and I have it.
Are you back yet? I thought we might have it again. I brought the stuff.
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Not back yet, Malc - three more weeks to go. Thought it was my turn to bring the stuff.
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Oh, ok. Could you get it in tubes this time? Fourteen should do it. It rots through the paper cups. I don't know how you women manage to keep it all in. My eyes water after a couple of minutes.
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We're made of stern stuff, Malc ... and what are you doing up at 3.10am? Been indulging in the leftovers from last time?
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Big tubes or little tubes?
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Little tubes! I'm not made of stone, woman!
I was up watching the Man U game and decided to post a couple of messages afterwards. Another huge drawback to living in Oz is the time that EPL games are broadcast live.
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Little tubes! I'm not made of stone, woman!
When this little conversation started I wondered how long it would take for me to be out of my depth - not long! I'm well in over my head and if I try to reply to that, I have the feeling the water I'm in will be hot.
Oh, what the heck ...
What are you made of then? Where's your mettle, man?
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My mettle's here under this cushion. I had it made into a weighty upper garment with short sleeves.
It's a heavy mettle t-shirt.
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My mettle's here under this cushion. I had it made into a weighty upper garment with short sleeves.
It's a heavy mettle t-shirt.
<-
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So, definitely not hard rock then?
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So, definitely not hard rock then?
You should be so lucky!
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:o
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VULTURE! Really! :o :o :o
I was referring to the music genre. (http://www.clipartof.com/images/emoticons/xsmall2/2242_angelic.gif) (http://www.clipartof.com)
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For a minute it was like being in the dressing room with my ladies team. They are getting very used to me now. After the last game, during my post-match talk in the changing rooms, two of them turned modestly away and took their bras off.
One more season and I reckon I can get us all showering together.
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You were counting, Malc?
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One more season and I reckon I can get us all showering together.
He wasn't counting - he was dreaming....!
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I don't suppose he's got a YouTube clip of that... :(
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I don't suppose he's got a YouTube clip of that... :(
Possibly.... but for private viewing only! ..0
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I wasn't counting, no, but I suspect they had two each.