I really liked the article. So allow me to miss the forest for a moment; I want to chop down this tree:
Let’s solve the green box problem:
Try zero coins: EV: 100 coins.
Try one coin, give up if no payout: 45% of 180.2 + 55% of 99= c. 135.5 (I hope.)
(I think this is right, but welcome corrections; 90%x50%x178, +.2 for first coin winning (EV of that 2 not 1.8), + keeper coins. I definitely got this wrong the first time I wrote it out, so I’m less confident I got it right this time. Edit before posting: Not just once.)
Try two coins, give up if no payout:
45% of 180.2 (pays off first time)
4.5% of 178.2 (second time)
50.5% of 98. Total: c.138.6
I used to be quite good at things like this. I also used to watch Hill Street Blues. I make the third round very close:
45% of 180.2
4.5% of 178.2
.45% of 176.2
50.05% of 97
Or c. 138.45.
So, I pick two as the answer.
Quibble with the sportsball graph:
You have little confidence, for sure, but chance of winning doesn’t follow that graph, and there’s just no reason it should. If the Piggers are playing the Oatmeals, and you know nothing about them, I’d guess at the junior high level the curve would be fairly flat, but not that flat. If they are professional sportsballers of the Elite Sportsballers League, the curve is going to have a higher peak at 50; the Hyperboles are not going to be 100% to lose or win to the Breakfast Cerealers in higher level play. At the junior high level, there will be some c. 100%ers, but I think the flatline is unlikely, and I think the impression that it should be a flat line is mistaken.
Once again, I liked the article. It was engaging and interesting. (And I hope I got the problem right.)
I also get “stop after two losses,” although my numbers come out slightly differently. However, I suck at this sort of problem, so it’s quite likely I’ve got it wrong.
My temptation would be to solve it numerically (by brute force), i.e. code up a simulation and run it a million times and get the answer by seeing which strategy does best. Often that’s the right approach. However, sometimes you can’t simulate, and an analytical (exact, a priori) answer is better.
I think you are right about the sportsball case! I’ve updated my meta-meta-probability curve accordingly :-)
Can you think of a better example, in which the curve ought to be dead flat?
Jaynes uses “the probability that there was once life on Mars” in his discussion of this. I’m not sure that’s such a great example either.
I think you are right about the sportsball case! I’ve updated my meta-meta-probability curve accordingly :-)
The wikipedia article on the Beta distribution has a good discussion of possible priors to use. The Jeffreys prior is probably the one I’d use for Sportsball, but the Bayes-Laplace prior is generally acceptable as a representation of ignorance.
The example I like to give is the uncertain digital coin- I generate some double p between 0 and 1 using a random number generator, and then write a function “flip” which generates another double, and compares it to p. This is analogous to your blue box, and if you’re confident in the RNG means you have a tight meta-meta-probability curve, which justifies the uniform prior.
Jaynes uses “the probability that there was once life on Mars” in his discussion of this. I’m not sure that’s such a great example either.
Yeah, that seems like a good candidate for the Haldane prior to me.
I really liked the article. So allow me to miss the forest for a moment; I want to chop down this tree:
Let’s solve the green box problem:
Try zero coins: EV: 100 coins.
Try one coin, give up if no payout: 45% of 180.2 + 55% of 99= c. 135.5 (I hope.)
(I think this is right, but welcome corrections; 90%x50%x178, +.2 for first coin winning (EV of that 2 not 1.8), + keeper coins. I definitely got this wrong the first time I wrote it out, so I’m less confident I got it right this time. Edit before posting: Not just once.)
Try two coins, give up if no payout:
45% of 180.2 (pays off first time) 4.5% of 178.2 (second time)
50.5% of 98. Total: c.138.6
I used to be quite good at things like this. I also used to watch Hill Street Blues. I make the third round very close:
45% of 180.2 4.5% of 178.2 .45% of 176.2
50.05% of 97
Or c. 138.45.
So, I pick two as the answer.
Quibble with the sportsball graph:
You have little confidence, for sure, but chance of winning doesn’t follow that graph, and there’s just no reason it should. If the Piggers are playing the Oatmeals, and you know nothing about them, I’d guess at the junior high level the curve would be fairly flat, but not that flat. If they are professional sportsballers of the Elite Sportsballers League, the curve is going to have a higher peak at 50; the Hyperboles are not going to be 100% to lose or win to the Breakfast Cerealers in higher level play. At the junior high level, there will be some c. 100%ers, but I think the flatline is unlikely, and I think the impression that it should be a flat line is mistaken.
Once again, I liked the article. It was engaging and interesting. (And I hope I got the problem right.)
Glad you liked it!
I also get “stop after two losses,” although my numbers come out slightly differently. However, I suck at this sort of problem, so it’s quite likely I’ve got it wrong.
My temptation would be to solve it numerically (by brute force), i.e. code up a simulation and run it a million times and get the answer by seeing which strategy does best. Often that’s the right approach. However, sometimes you can’t simulate, and an analytical (exact, a priori) answer is better.
I think you are right about the sportsball case! I’ve updated my meta-meta-probability curve accordingly :-)
Can you think of a better example, in which the curve ought to be dead flat?
Jaynes uses “the probability that there was once life on Mars” in his discussion of this. I’m not sure that’s such a great example either.
The wikipedia article on the Beta distribution has a good discussion of possible priors to use. The Jeffreys prior is probably the one I’d use for Sportsball, but the Bayes-Laplace prior is generally acceptable as a representation of ignorance.
The example I like to give is the uncertain digital coin- I generate some double p between 0 and 1 using a random number generator, and then write a function “flip” which generates another double, and compares it to p. This is analogous to your blue box, and if you’re confident in the RNG means you have a tight meta-meta-probability curve, which justifies the uniform prior.
Yeah, that seems like a good candidate for the Haldane prior to me.
178.2 should be 178.4 (180.2 − 1.8) and 176.2 should be 176.6 (178.4 − 1.8)
This doesn’t change the result, though:
After 2 failed tries, even if you do have the good box, the most your net gain relative to standing pat can be is 98 additional coins.
But, the odds ratio of good box to bad box after 2 failed coins is 1:100 or less than 1% probability of good box.
So your expected gain from entering the third coin is upper bounded by (98 x 0.01) - (1 x 0.99) which is less than 0.
The answer I got also was to give up after putting in two coins and losing both times (assuming risk neutrality), if you get a green box.