Mathematical Conundrum or Not? Number Six

Srap Tasmaner

So you're saying that before I look I can say that there's a 50% chance that my envelope is envelope X but after looking I can't? — Michael

Suppose X = 1.

You pick an envelope. On opening it, you find $2. You have chosen the 2X envelope but you don't know it.

Does that mean there is a 1/2 chance that X = 2?

No. X = 1. It's just not true that half the time 2 = 1 and half the time it doesn't.

Your not knowing whether you have the X or the 2X envelope doesn't change anything.

Michael

↪Srap Tasmaner

This doesn't address my question.

After having picked an envelope, but before looking, would I be right in saying that there's a 50% chance that my envelope is envelope X?

After having picked an envelope, and after looking, would I be right in saying that there's a 50% chance that my envelope is envelope X?

I would say "yes" to both. You seem to want to say "no" to the second, but what about the first?

Michael

Suppose X = 1.

You pick an envelope. On opening it, you find $2. You have chosen the 2X envelope but you don't know it.

Does that mean there is a 1/2 chance that X = 2?

No. X = 1. It's just not true that half the time 2 = 1 and half the time it doesn't.

Your not knowing whether you have the X or the 2X envelope doesn't change anything. — Srap Tasmaner

This is misleading reasoning.

Compare:

1. I flip a coin and hide the result. I ask you for the probability that it is heads. You say 50%.

2. I flip a coin and hide the result. Suppose it landed tails. I ask you for the probability that it is heads. You say 50%.

In this second case, are you saying that half the times heads = tails? Of course not.

Even if it landed tails (unknown to you) you're still right to say that there's a 50% chance that it landed heads, and even if I picked the 2X envelope (unknown to me) I'm still right to say that there's a 50% chance that I picked the X envelope.

Michael

If you could prove that always switching is the best strategy over the long term, doesn't that amount to proving that you are more likely to have chosen the smaller envelope? Why doesn't that bother you? — Srap Tasmaner

We're not saying that you're more likely to have chosen the smaller envelope. We're saying that the amount you can win by switching is greater than the amount you can lose. If I have £10 in my envelope then I will either lose £5 or gain £10. There's twice as much to gain as there is to lose. And given that there's a 50% chance of picking the smaller envelope, there's an even chance of winning. That's a 2:1 payout.

Michael

That's a 2:1 payout. — Michael

Regarding this, and the example of repeated games, consider betting twice on a coin toss with a 2:1 payout, but in the first game you bet £10 and in the second game you bet £20.

If we win one and lose one then we break even. This is exactly what happens if we play two Two Envelope games with a £10 and £20 envelope and always switch (assuming we get one in the first game and the other in the second game).

So I think the example of repeating the game is misleading. What we want to know is, given this single game, is it better to switch or not to switch? And with the coin toss, is it better to bet or not to bet?

I'd say it's better to bet on a coin toss with a 2:1 payout.

BlueBanana

1. We pick an envelope at random
2. There's a 50% chance that my envelope is the X envelope and a 50% chance that my envelope is the 2X envelope.
3. I open my envelope and see £10
4. From 2 and 3, there's a 50% chance that my £10 envelope is the X envelope and a 50% chance that my £10 envelope is the 2X envelope. — Michael

False, learning the amount changes the odds. There is no way to choose those two amounts of money in such a way that any amount is as likely to be X or 2X.

Snakes Alive

I'm just doing this:

1. We pick an envelope at random
2. There's a 50% chance that my envelope is the X envelope and a 50% chance that my envelope is the 2X envelope.
3. I open my envelope and see £10
4. From 2 and 3, there's a 50% chance that my £10 envelope is the X envelope and a 50% chance that my £10 envelope is the 2X envelope.
5. From 4, there's a 50% chance that the other envelope contains £20 and a 50% chance that the other envelope contains £5. — Michael

Check my previous post – I explained why this reasoning is fallacious.

Michael

Check my previous post – I explained why this reasoning is fallacious. — Snakes Alive

Which step does it show to be wrong? Presumably it must be 2), 4), or 5)?

Snakes Alive

↪Michael

There is no step that's wrong – what's wrong is that in order to use this reasoning to provide an average based on a ratio of the amount drawn from the first envelope, you have to average a single variable ("how much I drew" = Y) across two distinct situations, as if this value would be the same across them as an independent variable (so that I average 1.25 Y by switching). But it isn't. How much you drew is definable only in terms of X, which switches in the two scenarios: in one scenario, it's 5, and in the other it's 10. So it's only coherent to calculate the average based on the odds in terms of X, since Y (how much you drew) is a variable dependent on X. If you do this, you'll see that the average payout is 1.5X, whether you switch or not.

Did you read the post? I can do another one working through your example specifically.

Michael

There is no step that's wrong — Snakes Alive

So the premises are true and the inferences are valid but the argument isn't sound? That doesn't work.

Snakes Alive

↪Michael

No, your conclusion doesn't establish what you want it to (that switching is more profitable).

Michael

No, your conclusion doesn't establish what you want it to (that switching is more profitable). — Snakes Alive

What do you mean by switching being more profitable? I'm certainly not saying that if you switch then you will earn more money, because I accept that one could lose £5.

I'm also not saying that you're more likely to make a profit, because I accept that the chance of winning is equal to the chance of losing.

All I am saying is that this single game has a 2:1 payout with an even chance of winning.

Snakes Alive

↪Michael

What I am saying is that as a strategy, switching does not increase your chances of earning more money, regardless of how many times the game is played. Your average earnings are the same regardless of whether you switch or not, and regardless of how many times the game is played.

If as a player my goal is to get as much money as possible, the two strategies are indifferently effective.

Michael

What I am saying is that as a strategy, switching does not increase your chances of earning more money regardless of how many times the game is played. — Snakes Alive

Neither am I.

Your average earnings are the same regardless of whether you switch or not, and regardless of how many times the game is played.

And this is misleading. Again, consider the case of betting on a coin toss with a 2:1 payout, but half the bets are £10 and half the bets are £20. On average you will break even.

But take one game in isolation? You have a 2:1 payout with an even chance of winning. That's a bet worth making.

So in our case, I have £10 and the other envelope contains either £5 or £20. There's an even chance of winning, but the payout is 2:1. Obviously like the coin toss if we play again and our envelope has £20 then we stand to break even, but like the coin toss the best (switch) in a single game is worth making.

Snakes Alive

But take one game in isolation? You have a 2:1payout with an even chance of winning. That's a bet worth making.

So in our case, I have £10 and the other envelope contains either £5 or £20. There's an even chance of winning, but the payout is 2:1. — Michael

This reasoning is fallacious. Did you read the post?

Michael

This reasoning is fallacious. Did you read the post? — Snakes Alive

Yes, and you haven't explained how the reasoning is fallacious. In fact you agreed that my premises were true and the inferences are valid, the final inference being "there's a 50% chance that the other envelope contains £20 and a 50% chance that the other envelope contains £5".

If there's a 50% chance that the other envelope contains £20 and a 50% chance that the other envelope contains £5, then switching my £10 envelope gives me a 2:1 payout with an even chance of winning.

Snakes Alive

↪Michael

The post explains why this reasoning is fallacious.

The fallacy is that there is some value, X, determined in each case. You are acting as if there is an independent variable Y, viz. what you drew first, and that switching will get you either .5Y or 2Y, and so on average the "bet" is worth taking, since your average payout is then 1.25Y.

This is fallacious, because there is no such variable: Y is defined in terms of X (either it is X, or 2X), and across the two scenarios you're averaging, the value of Y changes. Hence, there is no single value of Y across the two situations, and the value 1.25Y is a chimera.

If you define Y in terms of X, this illusion disappears. Thus, as you said, you either drew X or 2X. Therefore, there's a 50% chance that switching gets you X, and a 50% chance that it gets you 2X. Likewise if you stay. The average payout for either is 1.5X. Switching does not offer favorable odds.

Michael

The fallacy is that there is some value, X, determined in each case. You are acting as if there is an independent variable Y, viz. what you drew first, and that switching will get you either .5Y or 2Y, and so on average the "bet" is worth taking, since your average payout is then 1.25Y.

This is fallacious, because there is no such variable: Y is defined in terms of X (either it is X, or 2X), and across the two scenarios you're averaging, the value of Y changes. Hence, there is no single value of Y across the two situations, and the value 1.25Y is a chimera. — Snakes Alive

I'm not talking about averages at all. I explained why this is misleading with the example of betting on a coin toss with a 2:1 payout, but where you bet £10 half the time and £20 half the time. On average you will break even, but if we just consider a single game then a 2:1 payout with an even chance of winning is a bet worth making.

So, again, my only conclusion is this:

If there's a 50% chance that the other envelope contains £20 and a 50% chance that the other envelope contains £5, then switching my £10 envelope gives me a 2:1 payout with an even chance of winning. This is a bet worth making.

Snakes Alive

On average you will break even, but if we just consider a single game then a 2:1 payout with an even chance of winning is a bet worth making. — Michael

This cannot be.

Michael

If you define Y in terms of X, this illusion disappears. Thus, as you said, you either drew X or 2X. Therefore, there's a 50% chance that switching gets you X, and a 50% chance that it gets you 2X. Likewise if you stay. The average payout for either is 1.5X. Switching does not offer favorable odds.

Like others you're conflating different values of X. Given that I know that I have £10:

Thus, as you said, you either drew X or 2X. Therefore, there's a 50% chance that switching gets you X (£5), and a 50% chance that it gets you 2X (£20). Likewise if you stay. The average payout for either is (£12.5).

Snakes Alive

↪Michael

The value of X is fixed prior to the decision. Your reasoning switches the value of X across the two scenarios.

Michael

This cannot be. — Snakes Alive

Yes it can. We bet twice on a coin toss with a 2:1 payout. Half the time we bet £10, half the time we bet £20. We'd expect to break even. We switch envelopes with a 2:1 payout. Half the time we have the £10 envelope and half the time we have the £20 envelope. We'd expect to break even.

Snakes Alive

↪Michael

No.

If there is a 2:1 payout on a coin toss, then the bet is always worth taking, regardless of the amount bet, and one will not break even on average.

But this is getting far away from the main topic anyway.

What remains true is that there is no benefit to switching, either in the one case or across cases.

Srap Tasmaner

↪Michael

I modified your simulation.

<?php

  $switch = $no_switch = 0;

  for ($i = 1; $i <= 1000000; ++$i)
  {

    // set X
    $X = random_int(1, 100);

    // randomly select whether you get the X or the 2X
    $choice = random_int(0, 1) ? 1 : 2;
    
    // if you swap, you get the other one, duh
    $swap = 1 + $choice % 2;
    
    // If we switch
    $switch += $X * $swap;

    // If we don't switch
    $no_switch += $X * $choice;

  }

  echo 'Switch: £' . number_format($switch) . PHP_EOL;
  echo 'No Switch: £' . number_format($no_switch);

Run it here.

Heh - the $choice line is funny. I've never written PHP before.

Michael

Your reasoning switches the value of X across the two scenarios. — Snakes Alive

So does yours. Here was your argument:

There's a 50% chance that switching gets you X, and a 50% chance that it gets you 2X. Likewise if you stay. The average payout for either is 1.5X.

But you're forgetting that I know that I have £10. If there's a 50% chance that switching gets me X then my £10 is 2X, and so "getting X" is getting £5. And if there's a 50% chance that switching gets me 2X then my £10 is X, and so "getting 2X" is getting £20. The X in "getting 2X" has a different value to the X in "getting X", and so you can't add them together this way.

Jeremiah

↪Michael

The very definition of probability is the frequency of occurrences over repeated random events.

The first time is no different than the 10,000th time. We could set this up 10,000 different times, bring in 10,000 different people so that each event is a "first" time and it still would not reflect your model.

Michael

The very definition of probability is the frequency of occurrences over repeated random events. — Jeremiah

That's the frequentist's interpretation.

Although I don't understand what this has to do with my argument. I'm not saying that you're more likely to win if you switch. I'm saying that switching offers a 2:1 payout with an even chance of winning, and so is a good bet to make.

Michael

↪Srap Tasmaner

That doesn't model the case we're considering, which is that we know we have £10. This goes back to what I was saying here:

So in your example you've considered repeating the game using the same value envelopes and say that half the time $10 is picked and half the time $20 is picked whereas in my example I've considered repeating the game using the same starting envelope and say that half the time the other envelope contains $5 and half the time the other envelope contains $20.

Is there some rule of statistics that says that one or the other is the proper way to assess the best strategy for a single instance of the game (where you know that there's $10 in your envelope)?

I would have thought that if we want to know the best strategy given the information we have then the repeated games we consider require us to have that same information, and the variations are in the possible unknowns – which is what my example does.

Snakes Alive

↪Michael

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.

The shoe is on the other foot; you are assuming that the amount drawn, say 10, is constant in terms of X across both considered scenarios. This is not the case.

Snakes Alive

That doesn't model the case we're considering, which is that we know we have £10. — Michael

The value that you get from the first envelope is irrelevant.

Mathematical Conundrum or Not? Number Six

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