# Thirsty work

January 12, 2012 12:11 PM Subscribe

Mathfilter. Bonus: involves football and beer

I asked a similar question a few years ago, responses were excellent, definitive and largely over my head (at least the more thorough ones) but I did learn a lot!

Here we go again, this one should be easier but is still beyond the meagre math skills I used to have..

Bud Light is including commemorative glasses for Super Bowl-winning teams in cases of their fine product (in Canada).

18 teams have won the Super Bowl, one glass per case. So assuming equal production and random distribution of glasses per team, how many cases SHOULD I have to buy to get one for my beloved Raiders?

Glasses for heathen teams that shall never touch my lips will be Freecycled. I keep the beer..

I asked a similar question a few years ago, responses were excellent, definitive and largely over my head (at least the more thorough ones) but I did learn a lot!

Here we go again, this one should be easier but is still beyond the meagre math skills I used to have..

Bud Light is including commemorative glasses for Super Bowl-winning teams in cases of their fine product (in Canada).

18 teams have won the Super Bowl, one glass per case. So assuming equal production and random distribution of glasses per team, how many cases SHOULD I have to buy to get one for my beloved Raiders?

Glasses for heathen teams that shall never touch my lips will be Freecycled. I keep the beer..

The first case gives you a new team with probability 1.

Once you have 1 team, you have a 17/18 probability of getting a new team with the next case; this means, on average, it will take you 18/17 cases to get your second team. (That the average cases required is the reciprocal of the probability of getting a new team may not be immediately obvious, but can be shown.)

Once you have 2 teams, you have a 16/18 probability of getting a new team with each case, and it will take you 18/16 cases, on average, to get the third team.

...

Once you have 17 teams, you have a 1/18 probability of getting the one remaining team with each case, and it will take you, on average, 18 cases to get the last team.

Thus, the average total number of cases required is 18/18 + 18/17 + 18/16 ... + 18/2 + 18/1, or

Note this is the

Happy drinking!

posted by DevilsAdvocate at 12:22 PM on January 12, 2012

Once you have 1 team, you have a 17/18 probability of getting a new team with the next case; this means, on average, it will take you 18/17 cases to get your second team. (That the average cases required is the reciprocal of the probability of getting a new team may not be immediately obvious, but can be shown.)

Once you have 2 teams, you have a 16/18 probability of getting a new team with each case, and it will take you 18/16 cases, on average, to get the third team.

...

Once you have 17 teams, you have a 1/18 probability of getting the one remaining team with each case, and it will take you, on average, 18 cases to get the last team.

Thus, the average total number of cases required is 18/18 + 18/17 + 18/16 ... + 18/2 + 18/1, or

**~62.912**cases.Note this is the

*average*, which may or may not be close to the*median*(the number of cases after which you have a 50% chance of having all 18 teams).Happy drinking!

posted by DevilsAdvocate at 12:22 PM on January 12, 2012

The probability of getting a Raiders glass is:

P = 1 - (17/18)^N

where N is the number of cases purchased. So in 5 cases, you have about a 25% chance of getting the glass, 18 cases will give you a ~64% chance, 30 cases gives you an 82% chance, and if you buy 100 cases, you'll find a Raiders glass with 99.7% probability.

posted by mr_roboto at 12:23 PM on January 12, 2012 [7 favorites]

P = 1 - (17/18)^N

where N is the number of cases purchased. So in 5 cases, you have about a 25% chance of getting the glass, 18 cases will give you a ~64% chance, 30 cases gives you an 82% chance, and if you buy 100 cases, you'll find a Raiders glass with 99.7% probability.

posted by mr_roboto at 12:23 PM on January 12, 2012 [7 favorites]

How much is each case singly vs. how much if you buy in bulk if cost is an issue? This could help the calculation of someone being able to tell you where the cutoff is for buying in bulk (when your odds drop for each new case or bulk purchase). If the cost is the cost is the same either way, it's moot because I'd buy one at a time and open them until you found the glass.

Or are you just concerned with the mathematics of when you reach close to mathematical surety? Your question seems more practical than that.

posted by OnTheLastCastle at 12:23 PM on January 12, 2012

Or are you just concerned with the mathematics of when you reach close to mathematical surety? Your question seems more practical than that.

posted by OnTheLastCastle at 12:23 PM on January 12, 2012

Ooops, sorry, misread the question. I was answering "how many cases would I have to buy to get at least one from each team?" which is not what you asked.

Assuming an equal distribution by team (not by number of super bowl wins) you will, on average, need to buy 18 cases to get a Raiders glass.

posted by DevilsAdvocate at 12:24 PM on January 12, 2012

Assuming an equal distribution by team (not by number of super bowl wins) you will, on average, need to buy 18 cases to get a Raiders glass.

posted by DevilsAdvocate at 12:24 PM on January 12, 2012

Also, the

posted by DevilsAdvocate at 12:27 PM on January 12, 2012

*average*is 18 cases, but the*median*, per mr_roboto's formula, is only 13 cases.posted by DevilsAdvocate at 12:27 PM on January 12, 2012

*if you buy 100 cases, you'll find a Raiders glass with 99.7% probability.*

You can probably get it a lot cheaper on eBay, is what I think they're saying.

posted by empath at 12:27 PM on January 12, 2012 [2 favorites]

While it's technically a compound event, as each case you buy subtracts one from the total number made, the numbers are high enough that it's easier to treat it as an independent event, akin to the probability of rolling a six in a certain number of dice rolls.

In that case (pardon the pun) the formula you're going to need, probability-wise, is

posted by holgate at 12:31 PM on January 12, 2012

In that case (pardon the pun) the formula you're going to need, probability-wise, is

`1 - (17/18)^n`where`n`is the number of cases, so 24 cases gives you a 75% chance of striking it lucky.posted by holgate at 12:31 PM on January 12, 2012

Excellent questions sir! Quick answer is let's assume 1 glass/team rather than SB (it says on the cases 1/18 glasses).

Didn't want to taint/convolute the question with too much info but I bought two cases today. Both happened to have Chicago Bears glasses. Not helpful but I'll chalk that up to chance. The glasses have "Chicago Bears, Champions" with a Super Bowl XX engravement. Also not helpful that Chicago only won one SB; not sure what a Steelers glass would look like.

Or Raiders for that matter. I pledge to keep all posted if anyone is actually interested. And tangentially, yes all SBs count including the next one the Raiders win based back in LA or Mexico City or the moon. Oakland is preferred but as long as they wear them colours..

posted by raider at 12:34 PM on January 12, 2012

Didn't want to taint/convolute the question with too much info but I bought two cases today. Both happened to have Chicago Bears glasses. Not helpful but I'll chalk that up to chance. The glasses have "Chicago Bears, Champions" with a Super Bowl XX engravement. Also not helpful that Chicago only won one SB; not sure what a Steelers glass would look like.

Or Raiders for that matter. I pledge to keep all posted if anyone is actually interested. And tangentially, yes all SBs count including the next one the Raiders win based back in LA or Mexico City or the moon. Oakland is preferred but as long as they wear them colours..

posted by raider at 12:34 PM on January 12, 2012

Can you find some way to peek inside the case? Do you know anyone who has a borescope? That would raise your odds significantly.

posted by procrastination at 12:41 PM on January 12, 2012

posted by procrastination at 12:41 PM on January 12, 2012

How do they pack the glass in the case? I one beer missing from each of these special cases?

posted by exphysicist345 at 12:54 PM on January 12, 2012

posted by exphysicist345 at 12:54 PM on January 12, 2012

Thanks. Of course you're right and it seems exactly the same as rolling a die.

Not sure how my wife would feel about filling the garage with 100 cases of beer and still possibly coming out a loser.

posted by raider at 12:57 PM on January 12, 2012

Not sure how my wife would feel about filling the garage with 100 cases of beer and still possibly coming out a loser.

posted by raider at 12:57 PM on January 12, 2012

*The glasses have "Chicago Bears, Champions" with a Super Bowl XX engravement.*

So I'm not sure what exactly this tells us, because the Bears have won only one Super Bowl. But if you get any glasses from a team that's won more than one, then it might be possible to tell whether there's one glass per winning team or one glass per Super Bowl.

By the way, one glass per Super Bowl is better for you: if it's one glass per team that has won you have a 1/18 chance of winning on each glass, and if it's one glass per win you have a 3/45 = 1/15 chance of winning on each glass.

*Not sure how my wife would feel about filling the garage with 100 cases of beer and still possibly coming out a loser.*

If you mean that literally, it's quite unlikely that you'd come out a loser in 100 cases: the chance of losing is (17/18)

^{100}or about one in 300.

posted by madcaptenor at 1:01 PM on January 12, 2012

Procrastination and Exphysicist, not sure how familiar you are with Canadian beer packaging/marketing but the cases are "24s" (6x4 bottles) with only 20 in them. The glass is in a box with bubble wrap, occupying the middle four spots. So no cheating (I became a master of that with T-shirts they'd stick on the top)

And no I am not paying for 24 beers..

posted by raider at 1:03 PM on January 12, 2012

And no I am not paying for 24 beers..

posted by raider at 1:03 PM on January 12, 2012

When you get the NY Football Giants glass(es) let me know.

posted by JohnnyGunn at 2:10 PM on January 12, 2012

posted by JohnnyGunn at 2:10 PM on January 12, 2012

The formulas above which answer "what's the probability I get Raiders glasses with X cases?" are correct.

It's worth mentioning, too, though, that your phrasing of the question "How many cases should I have to buy to get one?" sounds very much like the related question concerning Expected Value.

e.g., If I have a 10% chance of winning $10 and a 90% chance of losing $3, what's my

Back to your question: If the probability that you get the glasses in one case of beer is 1/18, then the

One way to interpret this is that if 1,000 people were all (independently) trying to get Raiders glasses, some would get them after 5 cases and some after 30, but overall you'd expect about 18,000 cases to be bought.

posted by losvedir at 2:33 PM on January 12, 2012

It's worth mentioning, too, though, that your phrasing of the question "How many cases should I have to buy to get one?" sounds very much like the related question concerning Expected Value.

e.g., If I have a 10% chance of winning $10 and a 90% chance of losing $3, what's my

*expected*outcome? It's .1 * 10 + .9 * (-3) = a loss of $1.70. In other words, if you played that game over and over, say 100 times, then you'd expect to be down about $170 afterwards.Back to your question: If the probability that you get the glasses in one case of beer is 1/18, then the

*"expected"*number of cases you'd need to buy is 18.One way to interpret this is that if 1,000 people were all (independently) trying to get Raiders glasses, some would get them after 5 cases and some after 30, but overall you'd expect about 18,000 cases to be bought.

posted by losvedir at 2:33 PM on January 12, 2012

When you get an Eagles glas...aw, dammit.

posted by Holy Zarquon's Singing Fish at 2:43 PM on January 12, 2012

posted by Holy Zarquon's Singing Fish at 2:43 PM on January 12, 2012

*Not sure how my wife would feel about filling the garage with 100 cases of beer and still possibly coming out a loser.*

Probably like a Bills fan.

Although to the question, I think it is extremely unlikely that production is really random and that teams are evenly distributed. Happy drinking.

posted by true at 5:50 PM on January 12, 2012

**I would be more than willing to pay you if you happen to get a New Orleans Saints glass. Please, please, please. Oh please.**

posted by govtdrone at 6:14 PM on January 12, 2012

Haha, very nice True.

I fear we're treading into MeTa territory with glass requests so could anyone who wants one please shoot me a MeMail. Not making promises but that way I'll have on file as I nobly pursue this challenge..

posted by raider at 8:25 PM on January 13, 2012

I fear we're treading into MeTa territory with glass requests so could anyone who wants one please shoot me a MeMail. Not making promises but that way I'll have on file as I nobly pursue this challenge..

posted by raider at 8:25 PM on January 13, 2012

Mathowie asked me for an update/resolution on this. I'm sure he was fascinated. Anyway after getting in 4 x Chicago Bears and 1 x St Louis Rams I was forced to abandon the project as obviously either the production of glasses was skewed for inexplicable reasons or I am a legendary loser. Also I was no longer thirsty. Apologies to those who requested glasses for their teams..

posted by raider at 4:35 PM on February 14, 2012

posted by raider at 4:35 PM on February 14, 2012

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This thread is closed to new comments.

If it is 1 glass per Super Bowl win, do you include the 1983 Super Bowl win for your beloved Raiders when they were in LA?

posted by Mister Fabulous at 12:21 PM on January 12, 2012