A dumb bar bet...
 

Hey all

I was hoping for some assistance from the statisticians to settle a dumb bar bet. I run a bar that carries 93 different liquors and cordials and 12 non-alcoholic mixers. We were debating how many drinks we could actually create (some of them being completely undrinkable not withstanding) assuming no drink would have more than 5 alcoholic and 3 non-alcoholic ingredients. Any way I can do this math?

My Thanks

AAeyes

Re: A dumb bar bet...
 

93C5 * 12C3, no?

I don't have a calculator on me so I don't know the answer or if that's the actual equation to use. I'm likely way off.

Re: A dumb bar bet...
 

correct equation or not, that's the best avatar in the world...

Re: A dumb bar bet...
 

Better than Jakethebake's:

http://img192.echo.cx/img192/7043/11164409248410ns.gif

Re: A dumb bar bet...
 

well before jumping into combinations lets think about it.

each alcoholic beverage can be mixed w/ either 1,2,or upt eo 3 mixers. so we have 93*3 immediately for the 1 alcoholic drink + all possible mixers.

now, for 2 alcoholic drinks we do 93C2 for the total # of 2 combinations out of 93 objects. so there's 4278 different ways to mix 2 alcoholic drinks. again we take this # and *3 to get the total # of 2 alcoholic drink mixes.

now for 3,4,5 its the same thing, ...

Total= 93C1*3 + 93C2*3 + 93C3*3 + 93C4*3 + 93C5*3=

165,075,465

so there you have it over 165 million different drinks.

unless i made a mistake which i may have.

Barron

Re: A dumb bar bet...
 

My answer is 16,123,871,372,127 drinks

94*94*94*94*94 7,339,040,224
x 13*13*13 2,197
= 16,123,871,372,128
- 1 (because a drink with no alcohol and no mixer isnt a drink)

Assumptions: 7-up with no alchohol is OK
Scotch with no mixer is OK
Double scotch with no mixer is not the above.

Re: A dumb bar bet...
 

That's what I was thinking at first, but I have to figure on the different combinations of mixers too and I don't know the equation. I think it's more than 165 million because I have to figure on each combo of liquor with every combo of mixer. So vodka cranberry + vodka oj + vodka cranberry oj + vodka soda + vodka cran soda + vodka cran OJ soda...... and I can't figure in all the various combinations. I think it should come out to billions of possibilities....

Re: A dumb bar bet...
 

Sorry, didn't see you weren't yet jumping into different combinations...

Re: A dumb bar bet...
 

So the formula is that simple? The number of liquors to the power of however many times I can use them * etc. etc.? somehow I thought it would be more difficult, but then again I'm not much of a statistician.

Many Thanks,

AAeyes

Re: A dumb bar bet...
 

That's the answer for drinks with exactly 5 liquors and 3 mixers. The OP asked for all drinks, which includes those with just 3 liquors and 2 mixers, etc. So the real answer is:

93C5*12C3 + 93C4*12C3 + 93C3*12C3 + 93C2*12C3 + 93C1*12C3 + 93C5*12C2 + 93C4*12C2 + 93C3*12C2 + 93C2*12C2 + 93C1*12C2 + 93C5*12C1 + 93C4*12C1 + 93C3*12C1 + 93C2*12C1 + 93C1*12C1 + 93C5*12C0 + 93C4*12C0 + 93C3*12C0 + 93C2*12C0
= 16452521345

Edit: I'm not counting any drink without alcohol as a drink.

COrrection...
 

[ QUOTE ]
Sorry, didn't see you weren't yet jumping into different combinations...

[/ QUOTE ]

no you were right. im pulling up excel to get it correct right now.

# is well over 100billion i think

its <93C1....93C5> * <12C3.....12C3> for each of the varying # of alcohols and mixers. im not writing it all out.

sorry for the earlier mistake
Barron

Re: COrrection...
 

[ QUOTE ]
[ QUOTE ]
Sorry, didn't see you weren't yet jumping into different combinations...

[/ QUOTE ]

no you were right. im pulling up excel to get it correct right now.

# is 12,105,534,100

so you can make 12 billion different drinks.

its <93C1....93C5> (vector) * <12C3.....12C3> (vector)

sorry for the earlier mistake
Barron

[/ QUOTE ]

shouldn't it be <12C0....12C3> which is exactly what I did? why are you ignoring all the drinks with 0, 1, or 2 mixers?

Re: COrrection...
 

yes. yours is right.

-Barron

Re: A dumb bar bet...
 

You guys are all ignoring different volumes of the same mixers and alcohols used to create different drinks. So, it's a lot more.

Re: A dumb bar bet...
 

Yes, it's very nearly infinite. Hell, you don't need all those different kinds of alcohol to make an infinite number of drinks. Just take gin and vermouth. You can serve 100% gin or 100% vermouth. You can make a martini with 50/50. A dry martini with 75% gin. A slightly less dry martini with 62.5% gin... you see where I'm going with this.

Eventually you get to the point where different drinks vary only by 1 molecule of gin and vermouth. This is functionally infinitely (i.e. the number of combinations is greater than the number of atoms in the universe), but let's try to estimate within a few orders of magnitude the actual number.

We'll say a drink is 200 mL and weighs about 200 g. 200 grams is about 11 moles of water, or 6.7 x 10^24 molecules. If there are only two types of liquor going into a drink, there are 6.7 x 10^24 drink combinations. With three types of liquor, it's ((6.7 x 10^24)^2)/2 = 2.2 x 10^49.

Perhaps some actuary can correct my statistics, but I believe with 9 liquors and 3 mixers, the number is very large.

Then you can add different drink sizes, etc. Whoever guessed the highest wins the bet.

Re: A dumb bar bet...
 

Vee have over 10^49 types of beer, vrom all over zee vorld. Call me vhen you are ready.

Re: A dumb bar bet...
 

[ QUOTE ]
You can make a martini with 50/50. A dry martini with 75% gin.

[/ QUOTE ]

You are really bad at making martinis.

Re: A dumb bar bet...
 

[ QUOTE ]
You are really bad at making martinis.

[/ QUOTE ]
You're not fk'n kidding. Can't stand em, even if they are made right.

I do like those florescent girl drinks they call martinis and serve in martini glasses, but then, I also like Boone's Strawberry wine coolers. It really impresses the ladies to see me drinking that straight from the bottle.

Re: A dumb bar bet...
 

[ QUOTE ]
You guys are all ignoring different volumes of the same mixers and alcohols used to create different drinks. So, it's a lot more.

[/ QUOTE ]

i'm quite sure that wasn't what the OP was getting at.

Re: A dumb bar bet...
 

This is easy.

First, consider all the alcohol that goes into the drink. There are 93 + 93C2 + 93C3 + 93C4 + 93C5 combinations.

Second, consider all the nonalcohol that goes into the drink. There are 12 + 12C2 + 12C3 combinations.

So multiply those two together.