A dumb bar bet...

[AAeyes]

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

[tbach24]

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.

[AAeyes]

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

[tbach24]

Better than Jakethebake's:

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

[DcifrThs]

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

[NutzyClutz]

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.

[AAeyes]

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....

[AAeyes]

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

[AAeyes]

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

[DougOzzzz]

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.

[DcifrThs]

[ 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

[DougOzzzz]

[ 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?

[DcifrThs]

yes. yours is right.

-Barron

[Ulysses]

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

[parttimepro]

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.

[drudman]

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

[Ulysses]

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

[/ QUOTE ]

You are really bad at making martinis.

[parttimepro]

[ 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.

[SossMan]

[ 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.

[AlphaWice]

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.