Strange Probability

[JesFine]

The Series of Dice Rolls thread reminded me of some weird math paradoxes described in Martin Gardner's Colossal Book of Mathematics. Actually I'm not sure if they should be considered "paradoxes" just incredibly nonintuitive. Here's one:

Let's say you have two hats (black and gray) on a table. In the black hat are 5 colored chips and 6 white ones. In the gray hat are 3 colored chips and 4 white ones. Which hat would you pick from to maximize your chances of picking a colored chip? Obviously the black one -- your chances are 5/11 (~.45) vs 3/7 (~.43).

Now there is a 2nd table, also with a black and gray hat. In the black hat are 6 colored chips and 3 white chips. In the gray hat are 9 colored chips and 5 white ones. Now which hat do you choose to get a colored chip? Again, the black one -- your chances are 6/9 (~.67) vs 9/14 (~.64).

Easy so far... but now what happens if you combine the contents of the same colored hats? Now you have a black hat with 11 colored and 9 white chips, and a gray hat with 12 colored and 9 white chips. Now, if you want to maximize your chances of picking a colored chip, you pick from the gray hat 12/21 (~.57) vs 11/21 (~.52) ... what the heck happened?

This mathematical strangeness obviously has implications in research, especially when combining two independent studies. An example given in the book is that two independent investigations of the effectiveness of a drug may show that it is more effective on men than women while combining the two may show the opposite. There was even an example of this turning up in real life in an investigation to see if there was sex bias in student admissions at Berkeley (Science, "Sex Bias in Graduate Admissions: Data from Berkeley", February 1985)

This one is known as "Simpson's Paradox". There is another cool paradox that has deeper statistical implications that I will add later, but I wanted to see what you guys knew about this one first.

 

[Yahweh]

Your black hat/gray hat example isnt really entirely remarkable.

In very basic Geometry, you learn all about Trigonometry. Any given 2 sides of right triangle will always be larger than the 3rd side. Example: Side A is 3 inches, Side B is 4 inches, Side C is 5 inches.
A + B >= C 3 + 4 >= 5
C + A >= B 5 + 3 >= 4
B + C >= A 4 + 5 >= 3

Using this method, you can logically say a triangle with sides A = 3, B = 4, C = 8 could not exist.

 

[JesFine]

Your black hat/gray hat example isnt really entirely remarkable.

In very basic Geometry, you learn all about Trigonometry. Any given 2 sides of right triangle will always be larger than the 3rd side. Example: Side A is 3 inches, Side B is 4 inches, Side C is 5 inches.
A + B >= C 3 + 4 >= 5
C + A >= B 5 + 3 >= 4
B + C >= A 4 + 5 >= 3

Using this method, you can logically say a triangle with sides A = 3, B = 4, C = 8 could not exist.

I am not following this.

First, any given 2 sides of any triangle will be larger than the third side. Not just right triangles.

And I don't see the parallels between your example and the hat example.

I do agree that a triangle with sides 3, 4, and 8 could not exist, but I don't see what that has to do with anything...

Can you explain your position a little differently? I was pretty weirded out when I first read about Simpson's paradox, so I'm surprised you don't find it at least a little remarkable.

 

[Stimpson J. Cat]

One way to get an intuitive handle on this "paradox" is to remember that when you add the like colored hats together, you destroy any information about the grouping.

In the experiment, you have table 1:

black = ~ 0.45, gray = ~0.43

and table 2:

black = ~0.67, gray = ~0.64

That pairing is completely arbitrary, though. And when you add the contents of the like colored hats together, the pairing information is lost. Consider a different pairing of the same 4 hats.

Table 1:

black = ~0.45, gray = ~0.64

Table 2:

black = ~0.67, gray = ~0.43

In this case, the black hat would win one table, and the gray hat would win the other one. What's more, in this case Table 1 has 25 chips, and table 2 has only 16 chips. This gives more weight to table 1, in spite of the fact that the black hat won by a larger ratio on table 2 than the gray hat did on table 1.


This illustrates one of the ways that statistics can be used to mislead people. People will often be suspicious of the "count the hits, ignore the misses" trick, but may not realize that even if you include all of the subjects, by dividing them up into carefully selected different sized subgroups, one can easily produce whatever results they want.


Dr. Stupid

 

[BillyJoe]

ANOTHER EXAMPLE

Black hat (50 coloured + 51 white) BEATS Grey hat (1 coloured + 2 white)

AND

Black hat (2 coloured + 1 white) BEATS Grey hat (99 coloured + 50 white)

BUT

Black hat (52 coloured + 52 white) EASILY LOSES TO Grey hat (100 coloured + 52 white)

 

[Yahweh]

I am not following this.

First, any given 2 sides of any triangle will be larger than the third side. Not just right triangles.

And I don't see the parallels between your example and the hat example.

I do agree that a triangle with sides 3, 4, and 8 could not exist, but I don't see what that has to do with anything...

Yeah, I had a hard time following myself also. I just wanted to show an example of a "rock-paper-scissors" event that exists in the real world. The Rock-paper-scissors thing is comparable (or not... I'm not entirely sure) to the gray hat thing.

Can you explain your position a little differently?

Honestly, I lost me also.

I was pretty weirded out when I first read about Simpson's paradox, so I'm surprised you don't find it at least a little remarkable.

Its easier to impress me with neat shiney things.

 

[69dodge]

Originally posted by JesFine
Let's say you have two hats (black and gray) on a table. In the black hat are 5 colored chips and 6 white ones. In the gray hat are 3 colored chips and 4 white ones. Which hat would you pick from to maximize your chances of picking a colored chip? Obviously the black one -- your chances are 5/11 (~.45) vs 3/7 (~.43).

Now there is a 2nd table, also with a black and gray hat. In the black hat are 6 colored chips and 3 white chips. In the gray hat are 9 colored chips and 5 white ones. Now which hat do you choose to get a colored chip? Again, the black one -- your chances are 6/9 (~.67) vs 9/14 (~.64).

Easy so far... but now what happens if you combine the contents of the same colored hats? Now you have a black hat with 11 colored and 9 white chips, and a gray hat with 12 colored and 9 white chips. Now, if you want to maximize your chances of picking a colored chip, you pick from the gray hat 12/21 (~.57) vs 11/21 (~.52) ... what the heck happened?

Here's what happened. Although the gray hat from each table is slightly better than the black hat from the same table, the hats from table 2 (both black and gray) are much better than the hats from table 1 (both black and gray). Furthermore, table 2's gray hat happens to be the biggest hat. The combined gray hat is the best because it contains the contents of the big gray hat from table 2, and the large advantage of its 'table 2'ness outweighs the small disadvantage of its grayness.

This mathematical strangeness obviously has implications in research, especially when combining two independent studies. An example given in the book is that two independent investigations of the effectiveness of a drug may show that it is more effective on men than women while combining the two may show the opposite.

The studies should not be combined. There was some unidentified factor at work in one of the studies that enhanced the effectiveness of the drug for everyone in that study, and it would be wrong to compare the women in that study with the men in the other study and conclude that they did better simply because they were women. The reason they did better was the unidentified factor that was absent from the other study.

 

[JesFine]

Thanks for the responses folks. There was a slight error in my original post -- I should have said the combined black hat odds were 11/20 (.55) instead of 11/21. Sorry about that.

Stimpy, your intuitive approach was very nice. One I hadn't thought of/seen before.

Yahweh, hysterical.... I actually thought you were trying to explain using vectors, which is a valid approach. I was trying to make a picture but had issues, so here is a site which shows the vector explanation. http://exploringdata.cqu.edu.au/sim_par.htm . It is using some mysterious activity called "cricket" to explain the paradox. Please don't pay attention to the terminology and only focus on the pictures. Sorry they are not shiny. Oh, and a better example of the "rock, paper, scissors effect" can be seen in the Series of Dice Rolls thread I mentioned in the original post.

BillyJoe and 69dodge, you both alluded to the fact that the absolute number of members of the groups outweighed the percentage values. This is also an important point.

I wonder what you meant, 69dodge, when you said the studies should not be combined due to some unidentified factor. The studies were, in fact, hypothetical, so we can hypothesize that there was no undetermined factor if we want. The numbers remain the same either way. In fact, I think if that were the case, then the combined study should be used instead of the separate independent study.

I am having computer problems, but once I fix them, I'll post the other "paradox" I talked about earlier.

 

[Yahweh]

Yahweh, hysterical....

Thanks, you've just made my day .

I actually thought you were trying to explain using vectors, which is a valid approach. I was trying to make a picture but had issues, so here is a site which shows the vector explanation. http://exploringdata.cqu.edu.au/sim_par.htm . It is using some mysterious activity called "cricket" to explain the paradox.

I was trying a vector approach. I tried to go about it by using very common easy to understand laymens terms, but the example I cited wasnt very "good", and I lost myself when I forgot what it was I getting at. I gave up, and rather than try to explain something that is out of my area of expertise, I'd rather impress everyone with how much I know about 8th grade Geometry... oh, and I thought maybe someone else could make a little more sense out of what I was getting at.

Please don't pay attention to the terminology and only focus on the pictures.

Are they shiney?

Sorry they are not shiny.

Oh, and a better example of the "rock, paper, scissors effect" can be seen in the Series of Dice Rolls thread I mentioned in the original post.

I thought it was a very good and interesting example.

 

[JesFine]

OK, time for a new one. This one is known as Blyth's paradox, and it can be modeled using the spinners attached to this post.

Spinner A is not really a spinner. When you spin the dial, it will always land on 3. Spinner B has a .56 chance of landing on 2, .22 chance of landing on 6, and .22 chance of landing on 4. Spinner C has a .51 chance of landing on 1 and .49 chance of landing on 5.

Imagine that these spinners represent the effectiveness of a drug in combatting a particular illness, on a scale of 1-6, higher numbers are more effective. If drugs A & B are the only ones on the market, A beats B by a probablity of .56. If A & C are the only ones on the market, A beats C by a probability of .51. But what happens if all 3 are on the market? A will beat both B & C, right?

NOPE. A will have a probability of .56 x .51 = .2856 of beating both, B has (.44 x .51) + (.22 x .49) = .3322 of beating both, and Drug C will have the best probability of all .49 x .78 = .3822! (For purposes of this example, "beating" means "maximizing the best chance of getting the best drug". )

So imagine you are a doctor who has to decide which drug to give his patient. Your choices are A & C, and you pick A. Now somebody introduces Drug B into the market, and, just like that, statistics say you have to pick Drug C! The same drug that was previously considered worse than Drug A.

So there you go, another weird one. Who invented statistics anyway?

 

[Rayn]

Wow. When I first read through your "Blyth Paradox", I couldn't crunch the numbers, but I get it now. It doesn't seem so much of a paradox as more of a convoluted word problem. I never really liked word problems though, so that's my bias. Though, I did almost have to write some things down to understand this one.

 

[69dodge]

Originally posted by JesFine
I wonder what you meant, 69dodge, when you said the studies should not be combined due to some unidentified factor. The studies were, in fact, hypothetical, so we can hypothesize that there was no undetermined factor if we want. The numbers remain the same either way.

I assumed the hypothetical studies had the same numbers as the tables with the hats. The two tables are the two studies; the black hats are men, the gray hats are women. In each study, the men did slightly better than the women. However, everyone in study 2 did much better than everyone in study 1. Surely there is some reason for this difference between the studies, even if the reason hasn't been identified. We can't hypothesize that there's no unidentified factor, if the numbers show that there is one.

 

[angard]

The drug problem,

I get the "twist/paradox".

However, I would recommend Drug B first, then drug A, lastly drug C.

Over time, if I got it right, B has the highest number (average), followed by A (3) and C is under 3.

Also, drug B has the highest potential (6).

 

[Brown]

Here's a column from Prof. Paulos that discusses this issue and some practical problems that could (and do!) result from it.

Such odd results can surface in a variety of contexts. For example, a certain medication X may have a higher success rate than another medication Y in several different studies and yet medication Y may have a higher overall success rate. Or a baseball player may have a lower batting average than another player against left-handed pitchers and also have a lower batting average than the other player against right-handed pitchers but have a higher overall batting average than the other player.