Probably Overthinking It

Here’s a probability puzzle to ruin your week.

In the Royal Game of Ur, players advance tokens along a track with 14 spaces. To determine how many spaces to advance, a player rolls 4 dice with 4 sides. Two corners on each die are marked; the other two are not. The total number of marked corners — which is 0, 1, 2, 3, or 4 — is the number of spaces to advance.

For example, if the total on your first roll is 2, you could advance a token to space 2. If you roll a 3 on the next roll, you could advance the same token to space 5.

Suppose you have a token on space 13. How many rolls did it take to get there?

Hint: you might want to start by computing the distribution of k given n, where k is the number of the space and n is the number of rolls.  Then think about the prior distribution of n.

I’ll post a solution later this week, but I have to confess: I believe my solution is correct, but there is still part of it I am not satisfied with.

[UPDATE November 1, 2018]

Here’s the thread on Twitter where a few people discuss this problem.

And here’s my solution.  As you will see there are still some unresolved questions.

Here’s another solution from Austin Rochford, which estimates the posterior distribution by simulation.

And here’s a solution from vlad, also based on simulation, using WebPPL:

Here are a series of problems I posed in my Bayesian statistics class:

1) Suppose you meet an adult resident of the U.S. who is 170 cm tall. What is the probability that they are male?

2) Suppose I choose two U.S. residents at random and A is taller than B.  How tall is A?

3) In a room of 10 randomly chosen U.S. residents, A is the second tallest.  How tall is A?  And what is the probability that A is male?

As background: For adult male residents of the US, the mean and standard deviation of height are 178 cm and 7.7 cm. For adult female residents the corresponding stats are 163 cm and 7.3 cm.  And 51% of the adult population is female.

If you solve the problems in order, you can reuse code from the first two to solve the third.

Here’s my solution, using a grid algorithm and the libraries from Think Bayes:

When I tweeted about this problem, I heard from Colin Carroll, who wrote a solution using PyMC:

And vlad posted a this solution using WebPPL, a browser-based environment for probablistic programming:

You can run that solution at WebPPL.

Last week I posed this problem in my Bayesian Statistics class:

Suppose there are 10 people in my Dungeons and Dragons club; on any game day, each of them has a 70% chance of showing up.

Each player has one character and each character has 6 attributes, each of which is generated by rolling and adding up 3 6-sided dice.

At the beginning of the game, I ask whose character has the lowest attribute. The wizard says, “My constitution is 5; does anyone have a lower attribute?”, and no one does.

The warrior says “My strength is 16; does anyone have a higher attribute?”, and no one does.

How many characters are in the party?

My solution is in this Jupyter notebook: