You are playing a game for money. There are two envelopes on a table.
You know that one contains $X and the other $2X, [but you do not
know which envelope is which or what the number X is]. Initially you
are allowed to pick one of the envelopes, to open it, and see that it
contains $Y . You then have a choice: walk away with the $Y or return
the envelope to the table and walk away with whatever is in the other
envelope. What should you do?
You could have X or 2X. If you have X and you switch then you get 2X but lose X so you gain X; so you get a +1 X. However, if you have 2X and switch then you gain X and lose 2X; so you get a -1 X. — Jeremiah
You just said the same exact thing i did — Jeremiah
It is the same thing, the math comes out exactly the same. — Jeremiah
I'm saying that if I have £10 then I either lose £5 by switching or gain £10. — Michael
My reasoning is:
I have £10. If X is 5 then I lose £5 by switching. If X is 10 then I gain £10 by switching.
So it's either -5 or +10. This is not the same. — Michael
Case one X = 5
If you have 2X and switch then you are left with 5 bucks. You had 10 and now you have 5. You got a -X.
Case two X = 10
If you have X and switch then you are left with 20 bucks. You had 10 and now you have 20. You got a +X.
Exactly what I said before. — Jeremiah
Case one x = 5. You lose 5 bucks.
You had 10 and now you have 5. Means you lost 5 and in that case X = 5. Therefore you got a -X.
Case two x = 10. You gain 10 bucks.
You had 10 and now you have 20. Means you gained 10 and in that case X=10. There you got a +X
The same outcome as you. — Jeremiah
<?php $switch = $no_switch = 0; for ($i = 1; $i <= 1000000; ++$i) { // Our envelope contains £10 $chosen = 10; // The other envelope contains either £5 or £20 $other = random_int(0, 1) ? 5 : 20; // If we switch $switch += $other; // If we don't switch $no_switch += $chosen; } echo 'Switch: £' . number_format($switch) . PHP_EOL; echo 'No Switch: £' . number_format($no_switch);
Two envelopes are e.g. 5 (X) and 10 (2X) - Pick 10(2X) and switch get X - But if you've already picked 2X you cannot get 2(2X). There is no 4X (20). That's no longer a possibility. Once you pick once, you eliminate either a double or a halving. Your array presumes three possibilities 5, 10, and 20. That contradicts the description in the OP. Right? — Baden
Case one X = 5
If you have 2X and switch then you are left with 5 bucks. You had 10 and now you have 5. You got a -X.
Case two X = 10
If you have X and switch then you are left with 20 bucks. You had 10 and now you have 20. You got a +X. — Jeremiah
In the event of L if you switch then you gain -X, which in this case is a lose of 5 bucks. If you don't switch then you gain 0.
In the event of K if you switch then you gain +X, which is gain of 10. If you don't switch then you gain 0. — Jeremiah
So switching gives me a 50% chance of gaining £10 and a 50% chance of losing £5. There's more to gain by switching than there is to lose. — Michael
Therefore picking 10 is incompatible with both a possibility of 5 and of 20. — Baden
No, you never get to that position. Switching gives you a 100% chance of gaining £10 iff you are in scenario two and 100% of losing £5 iff you are in scenario one. You can never be in both. That's impossible by the time you get to seeing the £10. The fact that you don't know which scenario you are in needs to be separated out from the actual possibilities available to you at that time which are dependent on the reality of the scenario that applies at that time.
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