>> If you throw 1000 dices, is it possible to get all one? Yes. Is it likely? Not at all.
> That's literally as likely as any other possible outcome.
???
If you want any outcome, they're equally likely.
But the prev post chose a particular outcome, and any particular outcome is rare.
There's no contradiction.
So what's the insight?
This distinction is popularly represented by the "Monty Hall problem": should you take the offer of the other door.
The problem involves 3 doors with a prize behind only one, where you choose 1 of the three, then Monty shows you what's behind 1 of the remaining 2, which is not the prize, then asks you if you would like to switch to the remaining door.
You might think that your odds won't change because nothing behind the doors has changed, or might get worse because the offer is a second chance to pick the dud.
But instead of 3 doors, imagine 1000 doors. You pick 1. Monty shows you what's behind 998 that aren't the prize and asks you if you want to switch.
> But the prev post chose a particular outcome, and any particular outcome is rare.
No, we first observed a particular outcome (the giant ring). This would be like running coin flips for long enough, spotting some interesting sequence that wasn’t decided beforehand, then deciding it must not be random because that sequence should have been incredibly rare.
Sure, that sequence was rare but it was just as likely as all the other sequences which we didn’t end up seeing.
> But instead of 3 doors, imagine 1000 doors. You pick 1. Monty shows you what's behind 998 that aren't the prize and asks you if you want to switch. By switching, your 1-of-1000 odds become 1-of-2.
No they should become 999 out of 1000. If your door is 1 in 1000 then the other door must have all other possibilities.
Also, the Monty haul problem is counter intuitive because it depends on the exact rules under which he operates. Suppose the classic 1 in 3 odds of a win, but an evil Monty haul where he only gives the option if you would win, now swapping is a guaranteed loss. Mathematically the answer is obvious when all the rules are guaranteed, but people’s internal heuristics don’t automatically trust rules as stated.
> By switching, your 1-of-1000 odds become 1-of-2.
It's not 50/50. That means you had a 50% chance to get the door correct on the first guess out of 1000. By showing the non-winning doors, the odds collapse into the remaining door. You had a 1/1000 chance of getting it right the first time, after the reveal all 998 are now assigned to the remaining door.
> That's literally as likely as any other possible outcome.
???
If you want any outcome, they're equally likely.
But the prev post chose a particular outcome, and any particular outcome is rare.
There's no contradiction.
So what's the insight?
This distinction is popularly represented by the "Monty Hall problem": should you take the offer of the other door.
The problem involves 3 doors with a prize behind only one, where you choose 1 of the three, then Monty shows you what's behind 1 of the remaining 2, which is not the prize, then asks you if you would like to switch to the remaining door.
You might think that your odds won't change because nothing behind the doors has changed, or might get worse because the offer is a second chance to pick the dud.
But instead of 3 doors, imagine 1000 doors. You pick 1. Monty shows you what's behind 998 that aren't the prize and asks you if you want to switch.
By switching, your 1-of-1000 odds become 1-of-2.
The particulars matter.