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Family Wine Tasting

 
 
mrmac
 
  1  
Reply Fri 14 Dec, 2012 08:37 am
@solipsister,
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?
0 Replies
 
markr
 
  1  
Reply Sun 16 Dec, 2012 01:49 am
@markr,
Quote:
@solipsister,
Well then maybe you can explain how it applies to this problem.


I didn't think so.
dalehileman
 
  0  
Reply Sun 16 Dec, 2012 01:57 am
@markr,
Hello test

1234
mrmac
 
  1  
Reply Mon 17 Dec, 2012 03:13 pm
@dalehileman,
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.

ossobuco
 
  1  
Reply Mon 17 Dec, 2012 04:00 pm
@mrmac,
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.
0 Replies
 
dalehileman
 
  1  
Reply Mon 17 Dec, 2012 04:23 pm
@mrmac,
Quote:
I tested 1234 and markr's sample space and solipsister's method work.
Forgive me Mac but in the hope it might be last I submitted that posting merely to turn off the flag for this thread

Quote:
…...randomly pick the most expensive and the 2nd most expensive…...25% of the time.*…*….not mean…..BOTH…...
True, that would be more like 17%. But then shouldn't you have said, "…or the 2nd…."
mrmac
 
  2  
Reply Tue 18 Dec, 2012 06:37 am
@dalehileman,
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
dalehileman
 
  1  
Reply Tue 18 Dec, 2012 12:42 pm
@mrmac,

Quote:
dalehileman - I'm more than a little rusty regarding logic -
To be sure, no moreso than I

Quote:
it's been almost 50 years since my last class.
Don't know if I ever attended one but it would have to be at least 60 years ago

Quote:
If I used "or," would I have to explain that I was using the "inclusive or?"
Don't think necessary

Quote:
Is one of the "or"s the default?
Don't know, too technical for this old guy

Quote:
I don't understand "turn off the flag for this thread."
My apologies. Entry in "My Topics," "My Posts," flagged in red denotes thread of interest

Quote:
Other than my lack of logic skill and turning off flags, am I on the right track
My apology Mac but to what end

Quote:
- I'm still losing the family argument!
With your lenience Mac, whose family where, and how

Quote:
Thanks for help,
You're welcome but don't think I quite qualify
0 Replies
 
solipsister
 
  1  
Reply Thu 20 Dec, 2012 05:39 am
@markr,
Quote:
Quote:

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


I didn't think so.


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.
markr
 
  2  
Reply Thu 20 Dec, 2012 01:28 pm
@solipsister,
Quote:
I've already answered the question and provided a distribution to compare the actual results against random chance...

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.

Quote:
...which is more than you've conveyed.

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.
mrmac
 
  2  
Reply Fri 21 Dec, 2012 07:14 am
@markr,
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
markr
 
  2  
Reply Fri 21 Dec, 2012 06:14 pm
@mrmac,
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.
markr
 
  1  
Reply Sun 23 Dec, 2012 02:11 pm
@mrmac,
Quote:
I'm still losing the family argument!


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?
mrmac
 
  1  
Reply Mon 24 Dec, 2012 04:32 pm
@markr,
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
_______________________
For two specific items out of x items:
(x-2)!/x! = 1/(x*(x-1))
_______________________
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.
mrmac
 
  1  
Reply Mon 24 Dec, 2012 04:55 pm
@markr,
I feel as though the argument shifted during the discussion. Here was the first report:
______________
"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."
________________
Judging in that manner, not enough weight is given when someone correctly identified all 3 wines. I'd be interested in your comments.
markr
 
  1  
Reply Mon 24 Dec, 2012 06:08 pm
@mrmac,
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
0 Replies
 
markr
 
  1  
Reply Mon 24 Dec, 2012 06:35 pm
@mrmac,
Quote:
"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.

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?
mrmac
 
  1  
Reply Tue 25 Dec, 2012 05:38 pm
@markr,
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
markr
 
  1  
Reply Tue 25 Dec, 2012 07:48 pm
@mrmac,
OK - I was going to run a chi square test on it.
ossobuco
 
  1  
Reply Tue 25 Dec, 2012 07:51 pm
@markr,
C'mon, I, ignorant, am still interested.
 

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