Three Doors with one prize

One other proof.
“Run the experiment in Reverse” :

Start off with 2 shut doors.
Your chances are 1-in-2 of picking the right door.

Now the host shows you a third OPEN empty door.
Does that change the 1-in-2 initial chance?

Of course not.

You could always lash the problem into CafeOBJ
It's a little similar to Portia's Casket ... dunno know ?????????????


-- in fopl.mod
set include FOPL-CLAUSE on

mod! PROBLEM{
protecting(BOOL)
-- the basic facts.
pred gc :
pred sc :
pred g :
pred s :

ax [F0] : gc <-> ~(sc) .
ax [F1] : g <-> ~(gc) .
ax [F2] : s <-> (s <-> ~(g)) .

}

open PROBLEM
option reset .
goal(gc) .
flag(auto, on) .
flag(very-verbose,on) .
flag (universal-symmetry,on)
param(max-proofs, 1) .
resolve .
close

eof

** EMPTY CLAUSE__________________________

19:[hyper:18,5,4]

** PROOF ________________________________

4:[unit-del:7] g
5:[] ~(s) | ~(g)
6:[] s | ~(g)
7:[] ~(gc)
18:[hyper:4,6] s
19:[hyper:18,5,4]

** ______________________________________

Look at what all this constant rain has reduced you all to :lol:
This is sooooo off topic....

They appear to have soggy membrane pain.

You have two strategies: A) Always stay with your original choice; Always change.

A) You have 1/3 chance of winning the prize - trivial.

If you pick the box with the prize initially (1/3) you will obviously lose it when you swap, so to win you must pick an empty box first off. Also, whichever empty box you pick, the host must show you the other empty one which allows you to swap to the box with the prize in it. Ergo, you definitely win if you pick either of the empty boxes, so your chance of winning is 2/3 with this strategy.

I think.

Anyway, there is another radio station which has a competition where callers pick numbers from 1 to 10, one of which has the prize (which to keep it on topic is a week's holiday at an observatory of the winner's choice anywhere in the world, with full unrestricted use of the telescopes and equipment/assistants etc.). If the first caller gets it wrong the chance passes to the next caller (who has obviously heard the wrong guess) and so on. What is the optimal position in the queue to have the best chance of winning the prize (I'm sure there's a tautology in there somewhere )?

Gus --

Well done! That's the first analysis of Monty's Paradox that didn't make my brain hurt. (And it even arrived at the right answer.)

-- Jeff.