{"id":80,"date":"2018-10-21T16:56:08","date_gmt":"2018-10-21T16:56:08","guid":{"rendered":"https:\/\/www.allendowney.com\/blog\/?p=80"},"modified":"2018-11-01T14:13:50","modified_gmt":"2018-11-01T14:13:50","slug":"the-game-of-ur-problem","status":"publish","type":"post","link":"https:\/\/www.allendowney.com\/blog\/2018\/10\/21\/the-game-of-ur-problem\/","title":{"rendered":"The Game of Ur problem"},"content":{"rendered":"

Here’s a probability puzzle to ruin your week.<\/p>\n

In the Royal Game of Ur<\/a>, 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.<\/a><\/p>\n

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.<\/p>\n

Suppose you have a token on space 13. How many rolls did it take to get there?<\/p>\n

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.\u00a0 Then think about the prior distribution of n.<\/p>\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.<\/p>\n

[UPDATE November 1, 2018]<\/p>\n

Here’s the thread on Twitter<\/a> where a few people discuss this problem.<\/p>\n

And\u00a0here’s my solution<\/a>.\u00a0 As you will see there are still some unresolved questions.<\/p>\n

Here’s another solution<\/a> from Austin Rochford<\/a>, which estimates the posterior distribution by simulation.<\/p>\n

And here’s a solution from vlad<\/a>, also based on simulation, using WebPPL:<\/p>\n