Tag Archives: non-transitive

Mathematical abstraction and the robustness to assumptions

By Posted in decision making, Probability, Rstats, Teaching, uncertainty Tagged , , , , , 3 Comments

I’ve been showing my new favourite toys to just about anyone foolish enough to actually engage me in conversation. I described how my shiny new set of non-transitive dice work here, complete with a map showing all the relevant probabilities.

All was neat and tidy and wonderful until fellow ecologist, Aaron Ball, tried to burst my bubble.

Nope. I couldn’t find the error. Fortunately, he works across the hall so I just went and asked him.

The problem he found, it turns out, was not with my calculations but with my assumptions. Aaron told me that dice constructed with rounded corners and hollowed out pips for the numbers on the faces tend to be biased in the frequency at which each face rolls up. I had assumed, of course, that each side of each of the five dice would roll with the same probability (ie. 1 in 6).

As with any model of a real world system, the mathematics were carried out on a simplified abstraction of the system being modelled. There are always, by necessity, assumptions being made. The important thing is to make these assumptions as explicit as possible and, where possible, to test the robustness of the model predictions to violations of the assumptions. Implicit to my calculations of the odds of the non-transitive Grime dice was the assumption that the dice are fair.

To check the model for robustness to this assumption, we can relax it and find out if we still get the same behaviour. Specifically, we can ask here whether some sort of pip-and-rounded-corner-induced bias can lead to a change in the Grime dice non-transitive cycles.

It seems a natural place to look would be between the dice pairings which have the closest to even odds. We can find out what level of bias would be required to switch the directionality of the odds (or at least erase the tendency for one die to roll higher than the other). Lets try looking at Magenta and Red, which under the fair dice assumption have odds p(Magenta > Red)=5/9. What kind of bias will change this relationship? The odds can be evened out by either Magenta rolling ones more often, or red rolling nine more often. The question is then, how much bias would there need to in the dice in order to even out the odds between Magenta and Red?

Lets start with Red biasing toward rolling nine more often (recall that nine appears on only one face). Under the fair dice hypothesis, Red can roll nine (1/6 of the time) and win no matter what Magenta rolls, or by rolling four (5/6 of the time) and win when Magenta rolls one (1/3 of the time).

P(Red > Magenta) = 1/6 + 5/6 * 1/3, which is 4/9.

If we set this probability equal to 1/2, and replace the fraction of times that Red rolls nine with x, we can solve for the frequency needed to even the odds.

x + 5/6 * 1/3 = 1/2

x = 2/9

Meaning that the Red die would have to be biased toward rolling nine with 2/9 odds. That’s equivalent to rolling a nine 1 and 1/3 times (33%) more often than you would expect if the die were fair!

Alternatively, the other way the odds between Red and Magenta could be evened is if Magenta biased towards rolling ones more often. We can do the same kind of calculation as above to figure out how much bias would be needed.

1/6 + 5/6 * x = 1/2

x = 2/5

Which corresponds to Magenta having  a 20% bias toward rolling ones. Of course, some combination of these biases could also be possible.

I leave it to the reader to work out the other pairings, but from the Red-Magenta analysis we can see that even if the dice deviated quite a bit from the expected 1/6 probability for each side, the edge afforded to Magenta is retained. I couldn’t find any convincing  evidence for the extent of bias caused by pipping and rounded corners but it seems unlikely that it would be strong enough to change the structure of the game.

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A quick guide to non-transitive Grime Dice

By Posted in decision making, Probability, Rstats, Uncategorized, uncertainty Tagged , , , , , , , , 5 Comments

A very special package that I am rather excited about arrived in the mail recently. The package contained a set of 6-sided dice. These dice, however, don’t have the standard numbers one to six on their faces. Instead, they have assorted numbers between zero and nine. Here’s the exact configuration:

red<-c(4,4,4,4,4,9)
blue<-c(2,2,2,7,7,7)
olive<-c(0,5,5,5,5,5)
yellow<-c(3,3,3,3,8,8)
magenta<-c(1,1,6,6,6,6)

Aside from maybe making for a more interesting version of snakes and ladders, why the heck am I so excited about these wacky dice? To find out what makes them so interesting, lets start by just rolling one against another and seeing which one rolls the higher number. Simple enough. Lets roll Red against Blue. Until you get your own set, you can roll in silico.

That was fun. We can do it over and over again and we’ll find that Red beats Blue more often than not. So it seems like Red is a pretty good bet. Now lets try rolling Olive against Red. I’ll wait.

Hey, look at that, the mighty Red has fallen. Olive tends to roll a higher number than Red more often than it doesn’t. So far, we have discovered this relationship:

Olive > Red > Blue

All hail the dominant Olive! Out of these three dice, if we want the best chance of winning, we should always pick Olive right? No dice, as they say. When we roll Olive against Blue, we find that Blue wins more often!

For any one of these three dice, there is another that will roll a higher number more often than not.

Olive > Red > Blue > Olive > Red > Blue > Olive > Red > Blue..

This forms a chain of dominance relationships that is a closed cycle. This property is called intransivity, and you can use it to win riches beyond your wildest dreams, er, well, at least to impress your friends.

Neat, right? But there’s more! We can do the same trick with Yellow, Magenta, and Red (Red > Magenta > Yellow > Red > …). With all five dice, there is a chain for which the order is given by that length of the word for each colour.

Red > Blue > Olive > Yellow > Magenta > …

Awesome. But that’s not it, either! You may have noticed from our three way comparisons that there is another five way chain. This time, the chain order is given by the alphabetical order of the words for each of the colours.

Blue > Magenta > Olive > Red > Yellow > …

What are the odds?

So far I’ve just asked you to take my word for it that the dominance relationships are as I described. Working out the odds of winning for any given pairing of dice as actually quite straightforward. Start by looking at the number on each side of the first die, one at a time. Count how many sides on the opposing die are less than the current number and divide by six. Since each side on the first die has a 1/6 chance of appearing, divide by 6 again. Sum these values for all six sides and you will have the probability that the first die will roll a higher number than the second.

For example, P(Red > Blue) = 5/6 x 1/2 + 1/6, which is 7/12.

Here I’ve worked out all of the pairwise odds:

Grime_dice

So, you can always win in this game as long as you get to be second to choose a colour. The odds are strongest in your favour when your opponent either chooses Magenta or Red, and you choose Olive or Yellow, respectively. Isn’t probability wonderful!

And if you still want more, it turns out that if you roll the Grime dice in pairs, the order of the word length chain reverses!