A Drink In The Ol' Saloon.

[Roger Kettle]

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...

[Mince]

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.

[madjock]

  I really do feel sorry for any of the pupils you teach Mince. 

[Tarquin Thunderthighs lll]

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.

[The Peepmaster]

And also he was highly unlikely to get a "Dead Man's Hand" again in his lifetime anyway. 

Happy July, everybody. (We had it here first).

[Mince]

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.

[Vulture]

Leave Mince alone, Tarks. Showing his 'superiority' and blinding us with maths keeps him happy!

[The Peepmaster]

Leave Mince alone, Tarks. Showing his 'superiority' and blinding us with maths keeps him happy!

I think Mince might have just scored.

[Fyodor]

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.

[Tarquin Thunderthighs lll]

Can we call it a bore draw? 

[The Peepmaster]

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! 

[Roger Kettle]

Unsurprisingly, I couldn't care less.

[Mince]

Unsurprisingly, I couldn't care less.

You make it sound as though this is choice on your part. 

[Mince]

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.

[Vulture]

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?