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Auto-ordered plates (10) Wed, Oct 01, 2008
If there was ever a doubt as to how much of a geek I am, this should make it clear. 

Our household owns several different colors of solid-color plates.  Two each of red, orange, yellow, and blue.  We use these plates in a random order, i.e. we don't intentionally use plates of the same color for a meal for two.  Furthermore, the order of the plates is constantly changing (and randomly, at that) because the order in which they're used isn't the order in which they're replaced, and sometimes newly washed plates are placed on top of the pile, while other times they're placed at the bottom.  If anything can be said about the usage of these plates, it's that it's completely random.  There's no order involved whatsoever.  Which makes it interesting when something like this happens: 

What we have here is a set of eight plates that have almost grouped themselves by color.  Two blues, one orange, two reds, two yellows, one orange.  There's most likely a way to calculate the probability of something like this happening, though I don't claim to understand probability.  What's equally interesting is that this isn't the first time it's happened.  It happened at least one other time in the last few months, which would make that probability even more amazing.  The day the plates completely auto-order themselves will most likely be the day the world ends.  If not, I'll post that picture too. #math

Dave Brown Wed, Oct 01, 2008
Ok: ultimate dork time:

Calculate the odds of getting a completely indexed stack.
Plate 1: Any plate will do. Probability = 1
Plate 2: Must match plate 1. Probability = 1/7
Plate 3: Any P=1
Plate 4: Match 3. P=1/5
Plate 5: Any. P=1
Plate 6: Match 5. P=1/3
Plate 7: Any. P=1
Plate 8: Only one choice P=1
Probability of whole configuration =
1 x 1/7 x 1 x 1/5 x 1 x 1/3 x 1 x 1 = 1/105.

The probability of the configuration in the picture =
1 x 1/7 x 1 x 4/5 x 1/4 x 2/3 x 1/2 x 1 = 1/105

Wait a minute! How can less indexed be just as likely as fully indexed?

Dave Wed, Oct 01, 2008
So you're telling me there's a chance.

I knew at least one of my nerd friends would chime in. 

Rich, what's your opinion?

Rich Wed, Oct 01, 2008
I never took a class that covered probability, or any statistics useful for describing... well, anything, really.  Dave Brown's explanation sounds good to me.  I'm curious what the probability distribution for the set of all possible configurations is, but not curious enough to actually calculate it.  Any takers?

Rus Thu, Oct 02, 2008
No on the distribution, and my calculator tells me that 1/7 * 1/5 * 1/3 = .009, or 1/105.  This pretty much confirms DB's math and equation setup.  Makes sense too.  (About 1 in 100 chances).  Keep putting dishes away Dave...eventually we shall have that apocalypse.

Dave Thu, Oct 02, 2008
Ok, DB, I think you're wrong on the probability of the configuration in the picture. 

I think it should be 1 x 1/7 x 1 x 4/5 x 1/4 x 1 x 1/2 x 1 = 1/70

The difference is that when there are three plates remaining and you have just chosen a matching red, there's no longer the opportunity to choose a matching red, so the probability would just be the probability of choosing any plate, which is 1.

Dave Brown Fri, Oct 03, 2008
I agree with you Dave as long as you're ok with the orange being below or in between the yellows.

Dave Fri, Oct 03, 2008
You're absolutely right.  But then I think we're talking about two different probabilities.  The one probability is of exactly matching the configuration in the picture, BBORRYYO, which would be: 

Plate 1:  Blue, P = 1/8
Plate 2:  Blue, P = 1/7
Plate 3:  Orange, P = 2/6
Plate 4:  Red, P = 2/5
Plate 5:  Red, P = 1/4
Plate 6:  Yellow, P = 2/3
Plate 7:  Yellow, P = 1/2
Plate 8:  Blue, P = 1

1/8 x 1/7 x 2/6 x 2/5 x 1/4 x 2/3 x 1/2 x 1 = 1/5040

The other probability is of matching the type of configuration in the picture, which, when written out, looks like this: 

Plate 1:  Any, P = 1
Plate 2:  Matching plate 1, P = 1/7
Plate 3:  Any, P = 1
Plate 4:  Not matching 3, P = 4/5
Plate 5:  Matching plate 4, P = 1/4
Plate 6:  Any, P = 1
Plate 7:  Matching plate 6, P = 1/2
Plate 8:  Any, P = 1

1 x 1/7 x 1 x 4/5 x 1/4 x 1 x 1/2 x 1 = 1/70

This second probability is color independent, i.e. the order could go BBORRYYO or RRYOOBBY or YYRBBOOR.  I'm not sure which most accurately represents what happened in my kitchen cabinet, but I'm definitely enjoying this nerd war.

Wendy Wed, Oct 08, 2008
Although I'm not as nerdy enough to post my probability comment, I am going through your calculations to agree/disagree. Ack, I'm one of you :)

Dave Brown Wed, Oct 29, 2008
Sorry Dave, but I still think we're wrong. There are several way to get the one off autoindexing:

Options with the misplaced plate on top:
aabbcddc p=1*1/7*1*1/5*1*2/3*1/2*1=1/105
aabccddb p=1*1/7*1*4/5*1/4*2/3*1/2*1=1/105
abbccdda p=1*6/7*1/6*1*1/4*2/3*1/2*1=1/84
Options with the misplaced plate third from the top:
abbccadd p=1*6/7*1/6*1*1/4*1/3*1*1=1/84
aabccbdd p=1*1/7*1*4/5*1/4*1/3*1*1=1/105
And options with the misplaced plate fifth from the top:
abbaccdd p=1*6/7*1/6*1/5*1*1/3*1*1=1/105

Total p= 1/105+1/105+1/84+1/84+1/105+1/105=13/210 = 6.2%

Now I just took this way too far, but what is a nerd war not taken way too far.

Dana Tue, Nov 04, 2008
I am proud to say I don't understand a thing of what you wrote above, Dave and company - nor do I want to, so I didn't try very hard.  It's official - you got all the math genes in the family.  And, I think we can safely say, you are WAY nerdier than I ever was.

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