Suppose u think you’re 80% likely to have left a power adapter somewhere inside a case with 4 otherwise-identical compartments. You check 3 compartments without finding your adapter. What’s the probability that the adapter is inside the remaining compartment?
I think the simplest way to compute this in full rigor is via the odds formula of Bayes Rule (the regular version works as well but is too complicated to do in your head):
Prior odds for [Adapter is in any compartment]: (4:1)
Relative chances of observed event [I didn’t find the adapter] given that it’s in any compartment vs. not: (25%: 100%) = (1:4)
Posterior odds [Adapter is in any compartment]: (4:1) ⋅ (1:4) = (4:4) = (1:1) = 0.5
Alternative way via intuition: treat “not there” as a fifth compartment. Probability mass for adapter being in compartments #1-#5 evolves as follows upon seeing empty compartments #1, #2, #3:
I think the people who mention Monty Hall in the Twitter comments have misunderstood why the answer there is 13 and falsely believe it’s 0.8 in this case. (That was the most commonly chosen answer.)
Monty Hall depends on how the moderator chooses the door they open. If they choose the door randomly (which is not the normal version), then probability evolves like so:
(13,13,13)→(12,0,12)
So Eliezer’s compartment problem is analogous to the non-standard version of Monty Hall. In the standard version where the moderator deliberately opens [the door among {#2, #3} with the goat], the probability mass of the opened door flows exclusively into the third door, i.e.,
You’re assuming the adapter is as likely to be in any compartment as any other. (If they aren’t, and I have more information and choose to open the three most likely compartments, then p<20%, where p=”the probability that the adapter is inside the remaining compartment”.)
I think the people who mention Monty Hall in the Twitter comments have misunderstood why the answer there is 13 and falsely believe it’s 0.8 in this case. (That was the most commonly chosen answer.)
They’re handling it like the probability it’s in the case is 100%. And thus, it must certainly be in the case in the fourth compartment. This works with 1:0 in favor of it being in the case, but doesn’t for any non-zero value on the right of 1:0.
In order for it to be 80% after the three tries, they’d have to do this intentionally/adversarially choosing.
The obvious fix is play a game. (Physically.) With the associated probabilities. And keep score.
Most people are really bad at probability.
I think the simplest way to compute this in full rigor is via the odds formula of Bayes Rule (the regular version works as well but is too complicated to do in your head):
Prior odds for [Adapter is in any compartment]: (4:1)
Relative chances of observed event [I didn’t find the adapter] given that it’s in any compartment vs. not: (25%: 100%) = (1:4)
Posterior odds [Adapter is in any compartment]: (4:1) ⋅ (1:4) = (4:4) = (1:1) = 0.5
Alternative way via intuition: treat “not there” as a fifth compartment. Probability mass for adapter being in compartments #1-#5 evolves as follows upon seeing empty compartments #1, #2, #3:
(15,15,15,15,15)→(0,14,14,14,14)→(0,0,13,13,13)→(0,0,0,12,12)
I think the people who mention Monty Hall in the Twitter comments have misunderstood why the answer there is 13 and falsely believe it’s 0.8 in this case. (That was the most commonly chosen answer.)
Monty Hall depends on how the moderator chooses the door they open. If they choose the door randomly (which is not the normal version), then probability evolves like so:
(13,13,13)→(12,0,12)
So Eliezer’s compartment problem is analogous to the non-standard version of Monty Hall. In the standard version where the moderator deliberately opens [the door among {#2, #3} with the goat], the probability mass of the opened door flows exclusively into the third door, i.e.,
(13,13,13)→(13,0,23)
You’re assuming the adapter is as likely to be in any compartment as any other. (If they aren’t, and I have more information and choose to open the three most likely compartments, then p<20%, where p=”the probability that the adapter is inside the remaining compartment”.)
They’re handling it like the probability it’s in the case is 100%. And thus, it must certainly be in the case in the fourth compartment. This works with 1:0 in favor of it being in the case, but doesn’t for any non-zero value on the right of 1:0.
In order for it to be 80% after the three tries, they’d have to do this intentionally/adversarially choosing.
The obvious fix is play a game. (Physically.) With the associated probabilities. And keep score.
https://www.gwern.net/docs/statistics/bayes/1994-falk