A friendly place for friends who drink

[WesleyS]

Admiring my free schwag haul
View attachment ImageUploadedByHome Brew1457130816.410878.jpg

 

[drainbamage]

*krcrssch* Overwatch to PFC drainbamage, come in, over. *krrssch*

*krssch* New intel suggests a flask in the loadout for this mission, do you copy? High risk of failure at next waypoint. Mission critical, Foxtrot, Lima, Alpha, Sierra, Kilo necessary for success. Do you copy? Over. *krrscch*

Copy that, Command. Sidearm is loaded and ready to neutralize hostile soft drinks. Going comms dark during infiltration and initial recon.

PFC drainbamage, over and out.

Krrrssch

View attachment 1457131082242.jpg

 

[Hello]

Tasting some beers at Olde Hickory.
View attachment ImageUploadedByHome Brew1457131488.894670.jpg

 

[iijakii]

http://www.ktvu.com/news/100651691-story

 

[Psylocide]

Tasting some beers at Olde Hickory.
View attachment 341734

My, your palette looks sophisticated.....

 

[mhurst111]

Yes, but that's not including that leap year happens only once every 4 years, but rather assuming 366 days every year, or is it?

You are correct. Google says this:

BUT WHAT ABOUT LEAP YEAR?

The original problem can be solved with a slide rule, which is exactly what I did when I first heard it many, many years ago.

If we add February 29 to the mix, it gets considerably more complicated. In this case, we make some additional assumptions:

Equal numbers of people are born on days other than February 29.
The number of people born on February 29 is one-fourth of the number of people born on any other day.
Hence the probability that a randomly selected person was born on February 29 is 0.25/365.25, and the probability that a randomly selected person was born on another specified day is 1/365.25.

The probability that N persons, possibly including one born on February 29, have distinct birthdays is the sum of two probabilities:

That the N persons were born on N different days other than February 29.
That the N persons were born on N different days, and include one person born on February 29.
The probabilities add because the two cases are mutually exclusive.

Now each probability can be expressed recursively:

double different_birthdays_excluding_Feb_29(int n)
{
return n == 1 ? 365.0/365.25 :
different_birthdays_excluding_Feb_29(n-1) * (365.0-(n-1)) / 365.25;
}

double different_birthdays_including_Feb_29(int n)
{
return n == 1 ? 0.25 / 365.25 :
different_birthdays_including_Feb_29(n-1) * (365.0-(n-2)) / 365.25 +
different_birthdays_excluding_Feb_29(n-1) * 0.25 / 365.25;
}
A program to display the probabilities goes something like this:

void main(void)
{
int n;
for (n = 1; n <= 366; n++)
printf("%3d: %e\n", n, 1.0-different_birthdays_excluding_Feb_29(n) -
different_birthdays_including_Feb_29(n));
}
The result is something like this:

1: -8.348357e-18
2: 2.736445e-03
3: 8.194354e-03
4: 1.633640e-02
5: 2.710333e-02
***
20: 4.110536e-01
21: 4.432853e-01
22: 4.752764e-01
23: 5.068650e-01
24: 5.379013e-01
25: 5.682487e-01
***
As expected, the probabilities are slightly lower, because there is a lower probability of matching birthdays when there are more possible birthdays. But the smallest number with probability greater than 0.5 is still 23.

Of course, a mathematical purist may argue that leap years don't always come every four years, so the calculations need further modification. However, the last quadrennial year that wasn't a leap year was 1900, and the next one will be 2100. The number of persons now living who were born in 1900 is so small that I think our approximation is valid for all practical purposes. But you are welcome to make the required modifications if you wish.

 

[DisturbdChemist]

What's everyone's birthday here? We have like 40 friendly regulars, surely some of us have friendly birthdays because maths?

July 5th.

March 17. St. Patricks day woohoo

 

[mcbaumannerb]

Time to forget this week. Good batch this year. Now to wait for my jägerschnitzel.
View attachment ImageUploadedByHome Brew1457131690.070114.jpg

 

[iijakii]

Damn, super jealous of the schnitz'!

 

[ThirstyPawsHB]

Just went in for 1....

View attachment 1457131743833.jpg

 

[SupervisingChildren]

http://www.ktvu.com/news/100651691-story

Yup. She's from a town that I go fishing for inland trout in. I wonder if she has any ragerts about not using ms paint to do it instead of hand drawing it.

 

[joshesmusica]

I got HBC 429, Sorachi Ace, Citra, and Amarillo on hand. Next beer needs to be with some combo of those. Anybody got any recommends? Don't even know what hbc 429 is supposed to be like, it and sorachi ace (which i've been wanting to try) were at this grab basket at the second biggest Norwegian national competition (but that one the public votes on their favorite).

 

[mcbaumannerb]

You are correct. Google says this:



BUT WHAT ABOUT LEAP YEAR?



The original problem can be solved with a slide rule, which is exactly what I did when I first heard it many, many years ago.



If we add February 29 to the mix, it gets considerably more complicated. In this case, we make some additional assumptions:



Equal numbers of people are born on days other than February 29.

The number of people born on February 29 is one-fourth of the number of people born on any other day.

Hence the probability that a randomly selected person was born on February 29 is 0.25/365.25, and the probability that a randomly selected person was born on another specified day is 1/365.25.



The probability that N persons, possibly including one born on February 29, have distinct birthdays is the sum of two probabilities:



That the N persons were born on N different days other than February 29.

That the N persons were born on N different days, and include one person born on February 29.

The probabilities add because the two cases are mutually exclusive.



Now each probability can be expressed recursively:



double different_birthdays_excluding_Feb_29(int n)

{

return n == 1 ? 365.0/365.25 :

different_birthdays_excluding_Feb_29(n-1) * (365.0-(n-1)) / 365.25;

}



double different_birthdays_including_Feb_29(int n)

{

return n == 1 ? 0.25 / 365.25 :

different_birthdays_including_Feb_29(n-1) * (365.0-(n-2)) / 365.25 +

different_birthdays_excluding_Feb_29(n-1) * 0.25 / 365.25;

}

A program to display the probabilities goes something like this:



void main(void)

{

int n;

for (n = 1; n <= 366; n++)

printf("%3d: %e\n", n, 1.0-different_birthdays_excluding_Feb_29(n) -

different_birthdays_including_Feb_29(n));

}

The result is something like this:



1: -8.348357e-18

2: 2.736445e-03

3: 8.194354e-03

4: 1.633640e-02

5: 2.710333e-02

***

20: 4.110536e-01

21: 4.432853e-01

22: 4.752764e-01

23: 5.068650e-01

24: 5.379013e-01

25: 5.682487e-01

***

As expected, the probabilities are slightly lower, because there is a lower probability of matching birthdays when there are more possible birthdays. But the smallest number with probability greater than 0.5 is still 23.



Of course, a mathematical purist may argue that leap years don't always come every four years, so the calculations need further modification. However, the last quadrennial year that wasn't a leap year was 1900, and the next one will be 2100. The number of persons now living who were born in 1900 is so small that I think our approximation is valid for all practical purposes. But you are welcome to make the required modifications if you wish.

Oooooh. Maybe we should throw a wildcard in the 50K game and double down if it happens to be the winners birthday?

 

[SupervisingChildren]

Just went in for 1....

Same here. I went to get a growler filled with bcbw for $25 and spent $90 between 5 bottles and the growler. Oops.

 

[Ridire]

You are correct. Google says this:

BUT WHAT ABOUT LEAP YEAR?

The original problem can be solved with a slide rule, which is exactly what I did when I first heard it many, many years ago.

If we add February 29 to the mix, it gets considerably more complicated. In this case, we make some additional assumptions:

Equal numbers of people are born on days other than February 29.
The number of people born on February 29 is one-fourth of the number of people born on any other day.
Hence the probability that a randomly selected person was born on February 29 is 0.25/365.25, and the probability that a randomly selected person was born on another specified day is 1/365.25.

The probability that N persons, possibly including one born on February 29, have distinct birthdays is the sum of two probabilities:

That the N persons were born on N different days other than February 29.
That the N persons were born on N different days, and include one person born on February 29.
The probabilities add because the two cases are mutually exclusive.

Now each probability can be expressed recursively:

double different_birthdays_excluding_Feb_29(int n)
{
return n == 1 ? 365.0/365.25 :
different_birthdays_excluding_Feb_29(n-1) * (365.0-(n-1)) / 365.25;
}

double different_birthdays_including_Feb_29(int n)
{
return n == 1 ? 0.25 / 365.25 :
different_birthdays_including_Feb_29(n-1) * (365.0-(n-2)) / 365.25 +
different_birthdays_excluding_Feb_29(n-1) * 0.25 / 365.25;
}
A program to display the probabilities goes something like this:

void main(void)
{
int n;
for (n = 1; n <= 366; n++)
printf("%3d: %e\n", n, 1.0-different_birthdays_excluding_Feb_29(n) -
different_birthdays_including_Feb_29(n));
}
The result is something like this:

1: -8.348357e-18
2: 2.736445e-03
3: 8.194354e-03
4: 1.633640e-02
5: 2.710333e-02
***
20: 4.110536e-01
21: 4.432853e-01
22: 4.752764e-01
23: 5.068650e-01
24: 5.379013e-01
25: 5.682487e-01
***
As expected, the probabilities are slightly lower, because there is a lower probability of matching birthdays when there are more possible birthdays. But the smallest number with probability greater than 0.5 is still 23.

Of course, a mathematical purist may argue that leap years don't always come every four years, so the calculations need further modification. However, the last quadrennial year that wasn't a leap year was 1900, and the next one will be 2100. The number of persons now living who were born in 1900 is so small that I think our approximation is valid for all practical purposes. But you are welcome to make the required modifications if you wish.

.

View attachment 1457132132118.jpg

 

[lumpher]

Because some other dummy had it locked up. I've only met two other people born on 7/7 but never on the same year

My wife's birthday is 7/7, but not 1977

 

[Qhrumphf]

You are correct. Google says this:

BUT WHAT ABOUT LEAP YEAR?

The original problem can be solved with a slide rule, which is exactly what I did when I first heard it many, many years ago.

If we add February 29 to the mix, it gets considerably more complicated. In this case, we make some additional assumptions:

Equal numbers of people are born on days other than February 29.
The number of people born on February 29 is one-fourth of the number of people born on any other day.
Hence the probability that a randomly selected person was born on February 29 is 0.25/365.25, and the probability that a randomly selected person was born on another specified day is 1/365.25.

The probability that N persons, possibly including one born on February 29, have distinct birthdays is the sum of two probabilities:

That the N persons were born on N different days other than February 29.
That the N persons were born on N different days, and include one person born on February 29.
The probabilities add because the two cases are mutually exclusive.

Now each probability can be expressed recursively:

double different_birthdays_excluding_Feb_29(int n)
{
return n == 1 ? 365.0/365.25 :
different_birthdays_excluding_Feb_29(n-1) * (365.0-(n-1)) / 365.25;
}

double different_birthdays_including_Feb_29(int n)
{
return n == 1 ? 0.25 / 365.25 :
different_birthdays_including_Feb_29(n-1) * (365.0-(n-2)) / 365.25 +
different_birthdays_excluding_Feb_29(n-1) * 0.25 / 365.25;
}
A program to display the probabilities goes something like this:

void main(void)
{
int n;
for (n = 1; n <= 366; n++)
printf("%3d: %e\n", n, 1.0-different_birthdays_excluding_Feb_29(n) -
different_birthdays_including_Feb_29(n));
}
The result is something like this:

1: -8.348357e-18
2: 2.736445e-03
3: 8.194354e-03
4: 1.633640e-02
5: 2.710333e-02
***
20: 4.110536e-01
21: 4.432853e-01
22: 4.752764e-01
23: 5.068650e-01
24: 5.379013e-01
25: 5.682487e-01
***
As expected, the probabilities are slightly lower, because there is a lower probability of matching birthdays when there are more possible birthdays. But the smallest number with probability greater than 0.5 is still 23.

Of course, a mathematical purist may argue that leap years don't always come every four years, so the calculations need further modification. However, the last quadrennial year that wasn't a leap year was 1900, and the next one will be 2100. The number of persons now living who were born in 1900 is so small that I think our approximation is valid for all practical purposes. But you are welcome to make the required modifications if you wish.

Thank you for Googlethinking for me. Obviously I didn't expect a significant difference, but clearly there's a marginally insignificant difference.

My mind is too flustered with both stress for my upcoming exam next weekend, and stress for tomorrow morning's possibly most important football match of my lifetime. I will be in the pub drinking at 7:45 am with the local Yiddos, undoubtedly next to a sea of red, and hopefully by then end our boots soaked in their tears. COYS!

 

[Ridire]

Carb sample of my blonde, in bottles one week, no fridge time.

View attachment 1457132569010.jpg

 

[DisturbdChemist]

Having this now.

View attachment 1457132640148.jpg

 

[BeerMeDuffMan]

You are correct. Google says this:

BUT WHAT ABOUT LEAP YEAR?

The original problem can be solved with a slide rule, which is exactly what I did when I first heard it many, many years ago.

If we add February 29 to the mix, it gets considerably more complicated. In this case, we make some additional assumptions:

Equal numbers of people are born on days other than February 29.
The number of people born on February 29 is one-fourth of the number of people born on any other day.
Hence the probability that a randomly selected person was born on February 29 is 0.25/365.25, and the probability that a randomly selected person was born on another specified day is 1/365.25.

The probability that N persons, possibly including one born on February 29, have distinct birthdays is the sum of two probabilities:

That the N persons were born on N different days other than February 29.
That the N persons were born on N different days, and include one person born on February 29.
The probabilities add because the two cases are mutually exclusive.

Now each probability can be expressed recursively:

double different_birthdays_excluding_Feb_29(int n)
{
return n == 1 ? 365.0/365.25 :
different_birthdays_excluding_Feb_29(n-1) * (365.0-(n-1)) / 365.25;
}

double different_birthdays_including_Feb_29(int n)
{
return n == 1 ? 0.25 / 365.25 :
different_birthdays_including_Feb_29(n-1) * (365.0-(n-2)) / 365.25 +
different_birthdays_excluding_Feb_29(n-1) * 0.25 / 365.25;
}
A program to display the probabilities goes something like this:

void main(void)
{
int n;
for (n = 1; n <= 366; n++)
printf("%3d: %e\n", n, 1.0-different_birthdays_excluding_Feb_29(n) -
different_birthdays_including_Feb_29(n));
}
The result is something like this:

1: -8.348357e-18
2: 2.736445e-03
3: 8.194354e-03
4: 1.633640e-02
5: 2.710333e-02
***
20: 4.110536e-01
21: 4.432853e-01
22: 4.752764e-01
23: 5.068650e-01
24: 5.379013e-01
25: 5.682487e-01
***
As expected, the probabilities are slightly lower, because there is a lower probability of matching birthdays when there are more possible birthdays. But the smallest number with probability greater than 0.5 is still 23.

Of course, a mathematical purist may argue that leap years don't always come every four years, so the calculations need further modification. However, the last quadrennial year that wasn't a leap year was 1900, and the next one will be 2100. The number of persons now living who were born in 1900 is so small that I think our approximation is valid for all practical purposes. But you are welcome to make the required modifications if you wish.

I have nothing cool to add to this, I just wanted to add to the clutter of those watching at home. Happy scrolling, ********!

 

[iijakii]

Well I know what I need to try soon

http://i.imgur.com/AK21zRp.gifv

 

[mcbaumannerb]

Now the chocolate RIS with my jägerschnitzel.
View attachment ImageUploadedByHome Brew1457133233.247003.jpg

 

[bob1852]

What's the consensus on DEM recipes?

I normally use Reaper's Mild, but would be interested in tinkering with it a bit more.

Kegerator is almost empty, which in and of itself is unacceptable. Need to get 10g of DEM going soon...

 

[Fedora]

What's everyone's birthday here? We have like 40 friendly regulars, surely some of us have friendly birthdays because maths?

July 5th.

Psy and I have the same birthday.

 

[Ridire]

I was born on Mother's Day. Narrows it down to a few dates.

 

[Psylocide]

What's the consensus on DEM recipes?



I normally use Reaper's Mild, but would be interested in tinkering with it a bit more.



Kegerator is almost empty, which in and of itself is unacceptable. Need to get 10g of DEM going soon...

Q's is pretty great.

View attachment ImageUploadedByHome Brew1457133345.372772.jpg


[ame]https://www.youtube.com/watch?v=O-jOEAufDQ4[/ame]

 

[Fedora]

Oh yea. Lagunitas IPA.

 

[Qhrumphf]

Now the chocolate RIS with my jägerschnitzel.
View attachment 341741

I would kill for some Jägerschnitzel right now.

 

[iijakii]

What's the consensus on DEM recipes?

I normally use Reaper's Mild, but would be interested in tinkering with it a bit more.

Kegerator is almost empty, which in and of itself is unacceptable. Need to get 10g of DEM going soon...

The first one I brewed, I think largely based off of @Qhrumphf 's, is still my favorite iteration.

Did:
70% MO
14% C60
9% Pale Chocolate
7% Biscuit

Fuggles and Wyeast 1318.

1.035 OG, 1.010 FG, 18 IBU or so.

Reaper's looks good but I think you'll be appreciative of what the biscuit malt brings to the table. Ive still never had a commercial DEM so can't say how accurate it is but damn it's tasty.

 

[BGBC]

NHC beers dropped off. Having Jan at Singlecut, all Brett C beer.

View attachment ImageUploadedByHome Brew1457134069.592627.jpg