A brief note on non-transitive dice

I've talked before about non-transitive dice. We say that given a pair of dice X and Y, X beats Y if more than half the time when the pair is rolled X has a larger number face up than Y. It turns out one can construct dice A, B and C such that A beats B, B beats C, but C in fact beats A. This is a neat and weird property.

During a recent discussion I used non-transitive dice as an example of a counter-intuitive aspect of mathematics, I was pointed to an even weirder variant. Consider the following set of dice: A has sides (5,5,5,2,2,2), B has sides (4,4,4,4,4,1) and C has sides (6,3,3,3,3,3).

Here A beats B, B beats C and C beats A. But here's the really cool part: Let's say I roll two copies of A, two copies of B or two copies of C. Now things actually reverse! That is, a pair of Bs beats a pair of As and a pair of As beats a pair of Cs and a pair of Cs beats a pair of Bs.

This is a much more sensitive property than just non-transitive dice. Most sets of non-transitive dice will not have this property. We can also describe this sensitivity in a more rigorous fashion. Suppose we have a strictly increasing function f(x). That is, a function such that f(x) is greater than f(y) whenever x is greater than y. Now suppose we take a set of non-transitive dice and relable each value x with f(x). Then they will still be non-transitive. But, given a set of non-transitive, reversable dice, reversibility is not necessarily preserved by the f mapping. This reflects the much more sensitive nature of the reversible dice.

Here's a question I have so far been unable to answer: Is it possible to make a set of die which do an additional reversal? That is, is there a set of dices such rolling three copies the dice results in another reversal direction?

Simpson's Paradox

Simpson's Paradox is a wonderful little statistical phenomenon which is counterintuitive to most people. Here is a simple, concrete example.

Consider two cancer drugs A and B. We do a study, Study #1, where we give some cancer patients drug A and some people drug B. Let's say we give 11 people drug A and 7 people drug B. Of the 11 people given drug A, 5 die and 6 survive. Of the 7 people given drug, B 3 die and 4 survive. So it seems that drug B is better than drug A since 4/7 is greater than 6/11.

Just to be sure we do another study, Study #2. Again we give some patients drug A and some patients drug B. Of those given drug A, 6 die and 3 survive. Of those given drug B, 9 die and 5 survive. Again it seems like drug B is better than drug A since 5/14 is larger than 3/9.

But wait! What happens if we look at all the data together? Now, of those given drug A, 11 died in total and 9 survived. For drug B, 12 died in total and 9 survived. So when we look at the combined data drug, A is better than drug B.

This paradox is known as Simpson's paradox. While one might think that this is the sort of thing that only comes up with cleverly picked numbers in the real world, there are actually many examples of actual data that exhibits this behavior.

Aside from being extremely counterintuitive, this result also plays havoc with our naive notions of what constitutes confirmation of a hypothesis. In particular, the fact that we can have two separate pieces of evidence which alone constitute confirming evidence but together constitute disconfirming evidence is jarring. Results like this one undermine naive Bayesian views of how science should function.

Non-transitice dice

Transitivity is a nice property and people like it when it occurs. Relationships that are transitive have the general property that A~B and B~C implies A~C. For example, if A is greater than B, and B is greater than C, then A is greater than B.

First, a definition: consider fair dice with some number of sides but the numbers are not necessarily the numbers one normally has on a die. So for example a die with sides( 1,2,3,4,5,6) on the sides would be ok, but so would one with (2,2,3,4) or even for our purposes (1,2) or even just (1) (this hypothetical dice just always lands on 1). For our purposes we will describe a die solely by what numbers it has on its sides. Now, a definition: we define die A to beat die B if when A and B are rolled more than half the time die A yields a higher number than die B. So for example die A (0,5,5,5,5) beats die B (1,2,3,4) since 4/5ths of the time A yields a higher result than B.

Now, one might think that if I have three dice, A, B and C and if A is beaten by B and B is beaten by C then A is beaten by C. However, as you likely guessed from the subject of this post, that is not always the case. And what’s more, we don’t need fancy die with many sides or anything. We can do a simple example with three six sided die. Consider the die A= (3,3,3,3,3,9) , die B=(1,4,7,7,7,7) and die C=(2,2,2,8,8,8). I claim that die A is beaten by die B, die B is beaten by die C but in fact die C is beaten by die A. Now, we could verify this by enumerating all the possible examples for each pair of dice rolls, but this isn’t tedious, so instead we’ll just think a little bit. First, consider the cases of A and B. 5/6^th of the time, A rolls a 3. 5/6^th of the time B rolls a 4 or a 7, both of which are greater than a 3. Thus, B has a higher number than A at least (5/6)^2=25/36>1/2 the time. Now, consider B and C. Half the time C rolls an 8 and so has a higher number than B automatically. Furthermore, when C rolls a 2 B sometimes rolls a 1. So more than half the time C yields a higher number than B. So C beats B. Now, the logic for the third set is almost identical to the previous one. Whenever C rolls a 2, A wins. But whenever A rolls a 9, A wins no matter what. So more than half the time A wins. Thus, A does in fact beat C.