Family Wine Tasting

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I've tried this in two other places with no interest being shown. Perhaps it's too easy (but not for me), so I would appreciate any help.
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Recently a group of 14 participated in a wine tasting. Wines were divided into 4 different categories (Zinfandel, Shiraz, Malbec, Cabernet Sauvignon) with 3 wines in each group. In a blind tasting, we were to determine whether a wine cost less than $10, $20 or more, or in between - one wine of each varietal in each price range. We weren't allowed to put, say, two Malbecs in the less than $10 category.

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As a group, we picked a wine’s price point (cheap, mid, expensive) in slightly more than half of all tastings (53.5% of all answers correct). What would we have expected from a fictional group that made all of its selections randomly?

With respect to the 3 different categories (cheap, mid, expensive), we correctly identified the mid-priced wine only 45% of the time. We did better with the cheap and expensive categories. Is there anything of significance that can be deduced from that?

There is no money riding on this - only family honor.

Dick

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Type: Question • Score: 1 • Views: 5,115 • Replies: 43

View best answer, chosen by mrmac

@mrmac,

Don't know about wine, but

http://www.realbeer.com/discussions/showthread.php?t=23994

http://www.realbeer.com/discussions/showthread.php?p=264242#post264242

OT but anyone wishing to discuss The Brew, I am [email protected],, recording Secretary of The Greater Southwest Beer-Tasting Society

We do however respect wine tasters

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@mrmac,

Assuming the varietals are known when tasted (i.e. you couldn't guess that the cheap Zin was an expensive Cab), a random assignment for a group of 3 would break down like this:
3 for 3: 1/6 of the time
1 for 3: 1/2 of the time
0 for 3: 1/3 of the time
Note: Can't go 2 for 3.

(3 * 1/6) + (1 * 1/2) + (0 * 1/3) = 1

Which means the expected number of correct guesses would be 1 (out of 3). You did better than random.

Regarding the poorer performance with the mid-priced wine:
You either get them all right, all wrong, or swap two of them
I think your results make sense since the least common errors would likely be mistaking a cheap for an expensive and vice versa. Therefore, you're probably most likely to mistake an expensive for a mid (and vice versa) or a cheap for a mid (and vice versa). In both cases, the mid-priced wine was involved, but each of the others was only involved in some fraction (summing to one) of those errors.

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@mrmac,

http://en.wikipedia.org/wiki/Binomial_distribution

Talk about using a sledge hammer to crack a nut.

3 random equal choices denotes a probability of 1/3 or the long way round:

Quote:

Assuming the varietals are known when tasted (i.e. you couldn't guess that the cheap Zin was an expensive Cab), a random assignment for a group of 3 would break down like this:
3 for 3: 1/6 of the time
1 for 3: 1/2 of the time
0 for 3: 1/3 of the time
Note: Can't go 2 for 3.

(3 * 1/6) + (1 * 1/2) + (0 * 1/3) = 1

Which means the expected number of correct guesses would be 1 (out of 3). You did better than random.

Given you were blind the normal approximation to the binomial is fraught to say nothing of the gallons of poisson distribution. Suffice to say the central limit theorum and a null hypothesis suggests you are a family of tipplers but significant evidence of taste buds is scant with 95% confidence.

1 Reply

@solipsister,

Beats using a screw driver. What problem are you solving such that your link to the binomial distribution is relevant?

1 Reply

@markr,

markr - thank you very much for your prompt reply. My argument was virtually identical to yours. I made the case that poor performance with the mid-priced wine was a positive, all other things being equal - better to switch two adjacent price range wines than to mistake the cheap for the expensive and vice versa.

My family members have not been convinced by my logic, so if you would just confirm that you are Chairman of the Mathematics Department at MIT, I'll be all set.

The only thing that I haven't tried is to get them to set up an experiment with playing cards and get their results empirically. That's how I was able to convince my 7th grade students "to switch" when confronted with the Monte Hall problem.

1 Reply

@mrmac,

Cal Tech, but close enough...

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@markr,

graduate, graduate student, ???

1 Reply

@mrmac,

I'm joking (since I thought you were, too). I do have a degree in math from UCSD several decades ago.

2 Replies

@markr,

I'm just a reader, dumb re math. I lived in La Jolla in '64: good grand rounds at the then Scripps Institute, where I was a peon.

Keep on with the talk, eh?

1 Reply

@ossobuco,

We moved to SD from the east coast in '69 when I was a kid. I graduated in '82, and left the city in '91 - too crowded, which is a shame because the weather is superb. I'm now in San Luis Obispo county.

1 Reply

@markr,

But I like that area.. alway have.

I can't talk math to you, but perhaps wine. I've been to a lot of the area vinyards but I don't talk wine too well either. Hey - I think the countryside there is beautiful, shoulder settling. I'm basically from Venice, and my husband and I sometimes used to set off for a few days drive, usually ending in your area or the paso robles area.

I've some friends with a property up near 46 - they were hit with a big
earthquake when first there.

Anyway, nice to talk. I'll be quiet for a while as I'm talking in a math thread.

2 Replies

@ossobuco,

I'm just south of Paso Robles.

0 Replies

@ossobuco,

Ah - I don't suppose Templeton? I liked Templeton when I was there. Not my business of course, where you are, but I liked the town. We stopped at a somnulent cafe for a burger (or what) and they had good short wine by the glass list. That's not usual, even in the spiffy parts of Los Angeles.

1 Reply

@markr,

Now, now dear, just because I opprobriated your prolix proof is no reason to go all probe and ballistic on me.

The question was:

Quote:

What would we have expected from a fictional group that made all of its selections randomly?

I was alluding to comparing the actual results with a random expectation of 1/3 by way of a normal approximation to the binomial, in the absinthe of the complete results, but shirley you knew that.

1 Reply

@ossobuco,

Bingo!

Well, this is wine country...

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@solipsister,

Pointing someone to the binomial distribution article doesn't explain the solution. It doesn't apply to this problem. If you're going to diss my explanation, at least provide an appropriate (and preferably better) one.

1 Reply

@markr,

Quote:

If you're going to diss my explanation, at least provide an appropriate (and preferably better) one.

I already did. 3 choices randomly means a probability of 1 in 3.

Quote:

Pointing someone to the binomial distribution article doesn't explain the solution.

I wasn't seeking to explain your long-winded method of arriving at 1/3, I was referring to analysis of the statistics.

2 Replies

@solipsister,

Well then maybe you can explain how it applies to this problem.

1 Reply

@markr,

I was joking - but then hoped that you weren't! I do appreciate your help and your clear explanation.

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