Dice puzzle – redux
June 13, 2008 – 11:03 pmI’d blogged previously about rolling a die and being given whatever you roll in dollars. You get given two more chances if you don’t like what you get the first time, or the second time, for a max of 3 rolls. What’s your strategy? You can see the answer in my previous post.
A couple of people have asked – what if there’s a pair of dice?
It’s the same principle. You could write all 36 permutations (but you’d live in fear you’ll be asked for a 3 dice solution). You know you can only roll a sum from 2 to a 12, and that you can get these sums from more than one combination of numbers on the dice.
You’ll see in the table below we again start by looking at the last throw. The expected payoff on the third throw is 7 (as you may have figured given the expected payoff for 1 die is 3.5). Following the same strategy as for the 1 die problem, for the second roll we pretend anything less than 7 is 7 for the purposes of figuring the expected payoff, which for roll 2 is 7.97. And so on for roll 1 which has an expected payoff of 8.54 (this is what you’d expect to win on average from this game).
So the strategy to maximize the expected payoff: on roll 1, stick with 8 or higher. On roll 2, stick with 7 or higher.
2 Responses to “Dice puzzle – redux”
for all those non-theory people out there (not necessarily including myself, but I tend to be heathily skeptical) i tried the dice problem in excel to get a brute force solution to the expected payoff of a single dice. 2.8? can’t be right… turns out there is a bug in excel 2003! the random number generator sometimes gives negative numbers, throwing off the uniform randomness of this pseudo random number generator. It only happens when the number of trials gets sufficiently large, around 5000. for smaller number of trials (or after a hotfix) we get on average 3.5.
By benkster on Jun 14, 2008
check out a cool way to make that picture spreadsheet an actual working embedded spreadsheet
By benkster on Jun 14, 2008