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 |

Comments:Dave BrownCalculate 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?

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

Rich, what's your opinion?

RichRusDaveI think it should be 1 x 1/7 x 1 x 4/5 x 1/4 x

1x 1/2 x 1 = 1/70The 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 BrownDaveexactlymatching 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

typeof 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.

WendyDave BrownOptions 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