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.
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.