Family Wine Tasting

I hadn't thought that the answer could be derived as easily as that. My initial thought was that, sure, if there were 5 wines and you had to place one of them in its correct price range, the chances of doing that would be 1 in 5, but by having to put all of the wines in their respective price ranges, the probability would change.

Upon closer inspection, having to properly arrange all of the varietals appears to still result in that initial probability.

If we let x = # of types of varietals, there would be x!x "choices" with x! likely "correct" responses giving us

#correct/#total = x!/x!x = 1/x

So if we had 10 wines and we had to randomly pick the 3rd most expensive wine, the probability would be 1 in 10. If we had to "order" all 10 wines, we would still expect to get 1 in 10 correct (approximately). Is that right?

I didn't think so.

Hello test

1234

I tested 1234 and markr's sample space and solipsister's method work.
The probability of randomly choosing the most expensive wine is 1 in 4.
If you were trying to randomly pick the most expensive and the 2nd most expensive, you could expect to be right 25% of the time.*
If you were trying to randomly order all 4 wines, again, you could expect to be correct 25% of the time.

*This does not mean that you would expect to pick BOTH the most expensive and the 2nd most expensive 25% of the time.

I think JPB should get in on this question - though she's likely busy. I'll pm her in case it tweaks her fancy.

Me, I survived in math and its associations as I needed them, and very much like the taste of wine and the wine country Markr was talking about, but have close to no sense of smell (inherited, I take it). Went to some apparent important tasting high up in some building in Los Angeles (talk about serious people in suits) and figure I was uniformly shunned when I failed to do the sniffing business. Even then I was too old to fake it.

Forgive me Mac but in the hope it might be last I submitted that posting merely to turn off the flag for this thread

True, that would be more like 17%. But then shouldn't you have said, "…or the 2nd…."

dalehileman - I'm more than a little rusty regarding logic - it's been almost 50 years since my last class. If I used "or," would I have to explain that I was using the "inclusive or?" Is one of the "or"s the default?
I don't understand "turn off the flag for this thread."
Other than my lack of logic skill and turning off flags, am I on the right track - I'm still losing the family argument!
Thanks for help,
mrmac

To be sure, no moreso than I

Don't know if I ever attended one but it would have to be at least 60 years ago

Don't think necessary

Don't know, too technical for this old guy

My apologies. Entry in "My Topics," "My Posts," flagged in red denotes thread of interest

My apology Mac but to what end

With your lenience Mac, whose family where, and how

You're welcome but don't think I quite qualify

I've already answered the question and provided a distribution to compare the actual results against random chance which is more than you've conveyed. If the probability, x , that your sense of humour either died long ago, was never in existence to begin with, or is simply missing temporarily, with an equal probability and n people speculate about the proposition,then they will discover that it doesn't matter how many ways they can distribute the random guesses, the likelihood of the fate of your sense of humour will forever remain 1 upon x.

Your answer (1/3) happened to be correct for the expected number of correct assignments, but the method for arriving at that answer (if you followed your own link) was incorrect. Either you didn't read the problem carefully, or you don't understand that you don't understand the problem.

The problem is not one of independent sampling with replacement. It is an arrangement problem (permutations), and the events are not independent (nor is there replacement). Yes, the probability of a given bottle being correctly assigned is 1/3, but not because of the Binomial Distribution. And if you somehow manage to (mis)apply the BD to this problem, you'll end up with a non-zero probability of getting exactly 2 of 3 bottles placed correctly.


I merely provided a method that is relevant to the problem.

I await your snappy, biting retort that favors fancy prose intended to belittle over facts and relevancy.

markr - would you please evaluate my thinking as discussed above and encapsulated below:


If we let x = # of types of wines, there would be x!x "choices" with x! likely "correct" responses giving us

#correct/#total = x!/x!x = 1/x

I'm not sure why you're multiplying x! by x. I'd do it this way:

If you're trying to find the probability of getting one specific item (out of x items) in the correct place, then:

- There are x! total permutations of the x items.
- (x-1)! of those permutations have the specific item in the correct place (the number of ways to permute the remaining x-1 items).

Therefore, the probability is (x-1)!/x! = 1/x.

For two specific items out of x items:
(x-2)!/x! = 1/(x*(x-1))

And so on.

What is the family's position on this matter? What, specifically, are you asserting that they aren't buying, and what is their argument against it?

My thinking was that if you were making a sample space, x! would represent the number of rows (permutations), x would represent the number of columns, and x!x would represent the total number of cells. x! also represents the number of cells that have the specific item in the correct place.

x!/x!x = (x*(x-1)!)/(x!x) = (x-1)!/x! [which is what you have] = 1/x
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For two specific items out of x items:
(x-2)!/x! = 1/(x*(x-1))
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BUT that would be to have BOTH correct [AND], right? So if you had 4 wines and were trying to pick the least and most expensive, you would have a 1 in 12 chance of picking them both, a 10 out of 24 chance of getting at least something right, and a 12 out of 48 (1 in 4) chance of having a specific item in the right place.

I feel as though the argument shifted during the discussion. Here was the first report:
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"The good news was that overall, we picked a wine’s price point (low, medium or high) in slightly more than half of all tastings (53.5% of all answers correct).

However, as a group we were not statistically significantly better than random under several different statistical models – in local parlance that means: if we sat and happily drank all the wine, didn’t give a hoot about the scores, and ask Ranger and Ringo [OUR DOGS] to scribble “a”, “b” and “c” in the boxes, the results would have been indistinguishable from ours (assuming their writing was legible, and that they hadn’t been sneaking tastes when we weren’t looking)."

_______________
When I disagreed with this, saying that random selection should only produce 33% correct, there was this response:
_______________
"Given a series of "independent" tastings, the whole group including the canines, have a 2/3 chance of getting at least one wine correct in a sample of three. So, with 14 of us hard at work, at least 9 of us should have got 1 wine correct in each flight. Sadly for even the proudest wine tasters, we did statistically no better than that in any of the four flights."
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Judging in that manner, not enough weight is given when someone correctly identified all 3 wines. I'd be interested in your comments.

Yes, that's for both of two specific items.

Four wines:
- least and most expensive (two specific items): 1/12 (2 of 24 permutations)
- at least something: 5/8 (15 of 24 permutations because there are 9 derangements of 4 items)
- a specific item: 1/4

That's true. However, as I pointed out in my original post, they get more than one correct only 1/6 of the time. How often did that occur?

Can you post the data?

markr - unfortunately I have gotten to a point that any further "discussion" would probably be seen as argumentative, and since the holder of the data is my niece's fiance, I'll let it drop unless someone else brings it up. I appreciate your help!
mrmac

OK - I was going to run a chi square test on it.

C'mon, I, ignorant, am still interested.