Yes - you have to calculate again - but only to get the same result.
With your first choice you had a 1:2 chance of having the right box – BUT you don’t know if you do (because you got a $100). So you now have to choose again (not “get to choose” but “have to choose”).
Your first choice did nothing other than to let you know that you had not yet lost.
Your next choice is the same as the first but will end the game – 50% chance of choosing the $100 over the $1.
You had 2 tries for the prize.
If you had asked of the probabilities from the beginning of getting a $100 both times - the results would be different –
First choice - 3:4 for getting a $100
Second choice (assuming that you got one already) - 1:2
So – .75 * .5 = 0.375 or 37.5% or 3/8
Obsrvr, okay, so you still maintain that having selected the first bill, and seeing that it’s $100, you have been given no new information (even probabilistic, incomplete information) about what box you’re likely to have chosen.
I’m not going to go into my experimental proof, or bayesian proof, yet, because first I think I want to try to persuade you that the first bill you pull out SHOULD be seen as information about what box you’re likely to have chosen. I have a more extreme scenario:
Instead of two boxes with two bills each, imagine I had two bags with 50 coloured balls each. One of the bags has 50 blue balls. The other bag has 1 blue ball, and 49 red balls. I present you the two bags, you don’t know which one is which, so you choose one bag at random. Then you stick your hand in (don’t peek!) and pick out one ball at random. The one ball you picked out is blue.
Does the fact that you selected a blue ball give you any possible information about what bag you selected? Are you more likely to have selected the bag full of blue balls, once you see that you’ve selected a blue ball? Or are you just as likely to have selected the bag with 49 red balls?
Like I said before, the boxes are not what is being calculated. What is being calculated is the probability of picking a certain denomination bill out of a total number of bills.
You have a 60% chance of picking a $100 bill out of 3 $100’s and 2 $1 bills, which is 5 total bills. The boxes mean NOTHING!
“Initially drawn” just means “a previous unrelated scenario.”
1st scenario - 4 $100 and 2 $1, for a total of 6 bills.
2nd (unrelated) scenario - 3 $100 and 2 $1, for a total of 5 bills.
Every time a bill is picked the game is over. If you want to start a new game from the remaining bills then the probability starts over, with new calculations.
You seem to be missing that your “second choice” isn’t actually a choice - you are already stuck with only the box your first choice yielded. So there are not two choices being made.
Right, so Obsrvr, I’m glad you see that in that scenario, the color of the first ball you chose is information (though imperfect) about what bag you chose.
So what we have is a continuum of scenarios.
50 blue vs 1 blue 49 red.
49 blue vs 1 blue 48 red.
48 blue vs 1 blue 47 red.
…
and so on, decreasing the number of balls until we have
3 blue vs 1 blue 2 red.
2 blue vs 1 blue 1 red.
Now in your opinion, selecting a blue ball in the top scenario gives you information about what bag you selected.
And I’m willing to bet that that opinion continues onto the second scenario where each bag ahas 49 balls, and I’m willing it applies to the 48 scenario as well.
BUT, it stops before the bottom scenario, where each bag has 2 balls only. In the bottom scenario, the fact that you’ve chosen a blue ball first gives you NO information about what bag you selected, in your opinion. So… where does it stop? Is there a specific point where the selection of the blue ball STOPS giving you information about what bag you selected?
Motor, it seems like you have some misunderstandings about the thought experiment. The question isn’t “what’s the probability of choosing any particular ball?”
The question is, you’ve already chosen a bag, and you’ve already chosen a ball, and it was blue, so what’s likely to be left in the bag you’ve chosen?
I thought I pre-answered that question for you – “BECAUSE it wasn’t 50% blue:red.”
It stops when there is no distinction to be seen.
Information IS distinction – no distinction (50/50) = no information.
Now you switched the scenario to a new scenario, of which there is a known box of $1 and $100, of which you picked the $1 out first. There is a 0% chance of picking a $1 bill out, and 100% chance picking a $100 bill.
There is only 1 possibility, because there is only 1 denomination left.
Can you translate your balls/bags reasoning to the boxes/bills scenario and ask where the line is drawn that drawing an X is no longer (or shows was never) informative?
