No.
The value of X is fixed. The two possibilities you consider must treat X as the same. You don't know what X is; but you do know that whatever it is, it does not change across the two scenarios you are considering. Your fallacious reasoning assumes that it does. — Snakes Alive
The amount in my envelope is also fixed. So as the amount in my envelope is fixed to £10 and if X is fixed to 10 then it would be false to say that there's a 50% chance of switching into X and a 50% chance of switching into 2X. We can only say that there's a 100% chance of switching into 2X (or if X is 5 then a 100% chance of switching into X). — Michael
3. There's a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20. — Michael
While you know how much you have drawn, you don't know whether this is X or 2X.
Hence, there is a 50% chance of your having drawn each, and a 50% chance of switching into the other, if you decide to switch. — Snakes Alive
This is the same opportunity that not switching gives you.
In terms of probability, sure, but not earnings. The average of the above is £12.50, but the average of not switching is £10.00. — Michael
There's a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20. — Michael
This conclusion does not show what you want it to. From this, it does not follow that switching is advantageous. — Snakes Alive
Yes it does. It shows that there's a 2:1 payout with an even chance of winning. How is that not advantageous? — Michael
The payout is on average 1.5X. This is the same payout that not switching affords. — Snakes Alive
That doesn't model the case we're considering, which is that we know we have £10. — Michael
It models it precisely. Your knowing what's in one envelope does not change what's in the other envelope.
Here's an example.
Suppose for a given trial, X = 12. There's an envelope with 12 in it and another with 24.
Suppose you pick the one with 12 and look at it. Y = 12. We'll call the amount in the other envelope Z. You reason, quite correctly, that
(1) Y = 12 → . Z = 6 ⊻ Z = 24
This is absolutely true because Z = 24.
Suppose you picked the 24 envelope. Y = 24. You reason, quite correctly, that
(2) Y = 24 → . Z = 12 ⊻ Z = 48
This is absolutely true because Z = 12.
What you have to accept is that for each trial there are only 2 values in the space. In the example above, Z = 6 = X/2 is not actually a possibility; Z = 48 = 4X is not actually a possibility. You can choose between 12 and 24 as many times as you like and you will never, ever get either 6 or 48. Your thinking they are possibilities does not make them so. — Srap Tasmaner
All switching does is change which envelope is picked; the switcher is thus committed both to thinking that switching is advantageous and indifferent. — Snakes Alive
The problem is that if you do this, then you are changing the value of X across the two situations, and treating things as if there is one epistemic possibility, say that X = 5, and another, say that X = 10. While this is on one sense so (and is reflected already in saying it the way I've put it: that you may have either drawn X or 2X), the problem comes when you try to use this changed value to average the two situations.
You can't do this, since it presumes there are two epistemic possibilities, each of which has a distinct value for X. This is not so: we know a priori that whatever X is, it is fixed. — Snakes Alive
Multiple trials show the two strategies converging.
<?php $switch = $no_switch = $max_2x = 0; for ($i = 1; $i <= 10000; ++$i) { // Ensure that all envelopes are even and not double an odd number $x = random_int(1, 100) * 4; $envelopes = [$x, $x * 2]; $choice = $envelopes[$key = random_int(0, 1)]; $other = $envelopes[(int) !$key]; $no_switch += $choice; if ( !$max_2x || $choice <= $max_2x / 2 ) { $switch += $other; } else { $switch += $choice; } $max_2x = max($other, $choice, $max_2x); } echo 'Gain: ' . (($switch / $no_switch) - 1);
That's because although it's a 2:1 payout with an even chance of winning the bet when you lose is twice as big as the bet when you win. Obviously you're going to break even. It would be like flipping a coin and hiding the result (heads) and having one person bet £10 on heads and another £20 on tails and offering them both a 2:1 payout. It's an advantageous bet for both people (given that they don't know the result or about each other's offer), despite the fact that the average payout is £0. — Michael
Also, a 2:1 payout would be getting $30 on a win and nothing on a loss, after betting $10. — Snakes Alive
It works because you limited the selection of numbers to below 400, and so funneled the choice to switching only when lower numbers relative to that range appear. This is not a feature of the original example, and if there were no a priori range from which the values were chosen, the strategy would have no benefit. — Snakes Alive
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