Mathematical Conundrum or Not? Number Five

BlueBanana

No, your calculations with Kolmogorov's definition give an answer that contradicts mine. They do not directly address or disprove that 75/225=1/3.

As for the calculations themselves, they show the odds throwing heads and it being Monday happening, not the odds of it being the current situation at the moment the question is asked. There's also 1/2 chances of throwing tails, then it being Monday, and then it being Tuesday, happening.

Basically the thing I question on them is the usage of the term "heads" and claiming its odds to be 1/2. Do you mean the odds of throwing heads or the odds of them having been thrown when the question is asked? The former leads to what I described above; right answer to the wrong question. The latter is circular reasoning.

Michael

No, your calculations with Kolmogorov's definition give an answer that contradicts mine. They do not directly address or disprove that 75/225=1/3. — BlueBanana

It's not supposed to show that 75/225 != 1/3. It's supposed to show that P(Heads) != 75/225.

Basically the thing I question on them is the usage of the term "heads" and claiming its odds to be 1/2. Do you mean the odds of throwing heads or the odds of them having been thrown when the question is asked? The former leads to what I described above; right answer to the wrong question. The latter is circular reasoning. — BlueBanana

I address the case of being asked here:

$P(Heads|Awake) = \frac{P(Heads ∩ Awake)}{P(Awake)}$

The answer depends on whether P(Awake) = 1 or 3/4. The halfers have to say 1, the thirders 3/4.

I say P(Awake) = 1 because Heads ∩ Tuesday isn't part of the sample space.

This is more clear if we don't consider days at all and just ask her once if it's heads and twice if it's tails (with amnesia in between). The sample space is H₁, T₁, and T₂ – and in every case she's awake.

Srap Tasmaner

↪Michael

Here's a slightly different argument.

If, when awakened, Beauty knew it was Monday, she would answer 50%; if she knew it was Tuesday, she would answer 0%. There's a 100% chance of being interviewed on Monday, but only a 50% chance of being interviewed on Tuesday, based on the fair coin toss. That means it is twice as likely that the interview is being conducted on Monday.

Thus 50% has a 2/3 chance of being the right answer (because the interview is being conducted on Monday) and 0% has a 1/3 chance of being correct (because the interview is being conducted on Tuesday).

From that we can calculate a value for my conditional expectation of heads, given that I am being interviewed:

$\small E(H \mid A) = \cfrac{2}{3}(50\%) + \cfrac{1}{3}(0\%)$

$\small \phantom{E(H \mid A)} \approx 33\%$

The wagering argument confirms that this is the proper degree of belief.

Michael

If, when awakened, Beauty knew it was Monday, she would answer 50% — Srap Tasmaner

$P(Heads|Monday) = \frac{P(Heads ∩ Monday)}{P(Monday)}$

$P(Heads ∩ Monday) = P(Heads) * P(Monday|Heads)$

$P(Heads ∩ Monday) = 0.5 * 1 = 0.5$

$P(Tails ∩ Monday) = P(Tails) * P(Monday|Tails)$

$P(Tails ∩ Monday) = 0.5 * 0.5 = 0.25$

$P(Monday) = P(Tails ∩ Monday) + P(Heads ∩ Monday) = 0.75$

$P(Heads|Monday) = \frac{0.5}{0.75} = \frac{2}{3}$

$\small E(H \mid A) = 0.75(\frac{2}{3}) + 0.25(0)$

$\small \phantom{E(H \mid A)} = 50\%$

Michael

The wagering argument confirms that this is the proper degree of belief — Srap Tasmaner

Again, that's just a matter of a greater payout. There are two wins with every tails and only one with heads. That doesn't mean tails is more likely. It only means that if you're a winner then you're twice as likely to have won with a guess of tails.

BlueBanana

↪Michael

With the common sense, how does that sound rational to you? Heads and tails being thrown are each as likely, so how does the knowledge that it's Monday change that?

Michael

With the common sense, how does that sound rational to you? Heads and tails being thrown are each as likely, so how does the knowledge that it's Monday change that? — BlueBanana

I suppose I could ask you the same question. Heads and tails being throw are each as likely, so how does knowing that you'll wake up twice if it's tails change that?

I suppose both our answers are that common sense isn't a good measure of probability, e.g. the Monty Hall problem.

BlueBanana

There are two wins with every tails — Michael

With each tails thrown, not with each tails guessed. The guesses are separate. You get to guess more times with tails, so naturally it makes sense to guess tails was thrown.

BlueBanana

I suppose I could ask you the same question. Heads and tails being throw are each as likely, so how does knowing that you'll wake up twice if it's tails change that? — Michael

You get to guess more when tails are thrown, so if you're guessing it's likelier tails were thrown.

Michael

You get to guess more when tails are thrown, so if you're guessing it's likelier tails were thrown. — BlueBanana

The probability that you get a guess is 1 in either case. The above reasoning only works if you say that if it's heads there's a 50% chance of getting to guess and if it's tails then there's a 100% chance of getting to guess. Only then can you say that tails is more likely.

BlueBanana

↪Michael

How about this: I toss a coin and ask you what I got, but if it's heads I toss it again and if it's tails that round isn't played. What are you guessing when I ask you what I got?

The above reasoning only works if --- — Michael

It works now. It explains why guessing tails is profitable.

Srap Tasmaner

My calculation is that expecting that profit is the same as if there were a single $1 bet at even odds that the result of a coin toss will be tails and the coin had a 3/4 probability of coming up tails. To see this, note that in that case the expected profit is

$1 x 3/4 + (-$1) x 1/4 = $0.50 — andrewk

I think I see why this is happening, even though the odds are 2-1 against heads.

By betting tails, I get to double what I risk only when my profit is guaranteed to double. — Srap Tasmaner

The $1 payoff matrix for a fair coin , betting at even money, is just

          Bet
          H    T
Toss  H   1   -1
      T  -1    1

If the coin is biased 3:1 tails:heads, we can multiply to get

           Bet
           H     T
Toss  H   .25  -.25
      T  -.75   .75

Betting tails you win $0.50 on average; betting heads you lose $0.50 on average.

Here's the $1 payoff matrix for our Sleeping Beauty, at even money:

           Bet
           H    T
Toss  H    1   -1
      T   -1    2

It's a fair coin, so multiplying gives you:

           Bet
           H    T
Toss  H   .5   -.5
      T  -.5    1

So betting tails again gives you a profit of $0.50 on average; betting heads you break even.

So it is because your tails profit is disproportional to your tails risk.

Michael

It explains why guessing tails is profitable. — BlueBanana

Being more profitable isn't the same as being more likely.

BlueBanana

↪Michael

It is in this context. With each guess there's the same bet, same amount of money, same chances of winning. So if guessing tails doesn't win more times, why is it profitable?

Michael

We flip a coin. If it's heads then the result stands. If it's tails then we flip again and the new result stands.

What is the probability that it's heads? 2/3, because two of the three outcomes are heads? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)?

This seems to be the crux of the disagreement. I say the latter. You seem to be saying the former.

Michael

So if guessing tails doesn't win more times, why is it profitable? — BlueBanana

It does win more times, but not because tails is more likely. It wins more times because there are two guesses for each flip of a tails compared to one for each flip of heads.

BlueBanana

We flip a coin. If it's heads then the result stands. If it's tails then we flip again and the new result stands.

What is the probability that it's heads? 2/3, because two of the three outcomes are heads? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)?

This seems to be the crux of the disagreement. — Michael

Completely different situation. There's no "eliminated" outcome. There's no difference between an outside observer and the subject.

BlueBanana

It wins more times — Michael

This is just another way of saying it's more likely to win.

there are two guesses for each flip of a tails — Michael

Because of which each guess is more likely to have been caused by tails.

Michael

Completely different situation. There's no "eliminated" outcome. — BlueBanana

What eliminated outcome?

BlueBanana

↪Michael

Heads and Tuesday.

Michael

Heads and Tuesday. — BlueBanana

That's irrelevant and doesn't have anything to do with the probability. We can change the scenario slightly to:

If it's heads then we wake her once. If it's tails then we wake her twice.

What's her credence that it's heads?

BlueBanana

That's irrelevant and doesn't have anything to do with the probability. We can change the scenario slightly to:

If it's heads then we wake her once. If it's tails then we wake her twice.

What's her credence that it's heads? — Michael

Then it's likelier the waking up was caused by flipping tails. As I said,

there are two guesses for each flip of a tails
— Michael

Because of which each guess is more likely to have been caused by tails. — BlueBanana

Michael

Then it's likelier the waking up was caused by flipping tails. — BlueBanana

So we're back to this and can ignore your "there's no 'eliminated' outcome" objection:

We flip a coin. If it's heads then the result stands. If it's tails then we flip again and the new result stands.

What is the probability that it's heads? 2/3, because two of the three outcomes are heads? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)?

Jeremiah

I am gonna argue that the outcome of the coin flip and when Beauty is awakened are two completely different events.

The coin flip with pull a subset of one from the sample space.

Our sample space contains only two possible outcomes:

H - Monday/Heads
T - Monday/Tails and Tuesday/Tails

So our event will have a 1 to 1 odds of H or T. That is a 50% chance each. That part we should all be able to agree on.

Now the second event is when Beauty determines her credence, and I am arguing that this is actually independent of the coin flip, because Beauty does not know the results of the coin flip. She can only make her determination based on what she knows about the experiment, which would mean the event from the coin flip is out of play here. It does not matter if it landed on H or T, as Beauty does not have that information.

The question is what is her credence, or what is her belief, that is independent of the actual coin flip and completely in the hands of how Beauty decides to interpret the experiment.

BlueBanana

and can ignore your "there's no 'eliminated" outcome" objection — Michael

No, we can't. Eliminating that possibility diminishes the chances of heads being the correct guess.

We flip a coin. If it's heads then the result stands. If it's tails then we flip again and the new result stands.

What is the probability that it's heads? 2/3, because two of the three outcomes are heads? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)? — Michael

3/4 but the situation is completely different. No difference between observers. Outcomes lead to same amounts of inquiries.

Now this example right here:

I toss a coin and ask you what I got, but if it's heads I toss it again and if it's tails that round isn't played. What are you guessing when I ask you what I got? — BlueBanana

This one's cool because it too can be calculated quite simply. 1/2 chances of tails and (1/2)*(1/2)=1/4 chances of both no question and heads.

Now what happens if you repeat the test? The Sleeping Beauty problem.

Michael

No, we can't. Eliminating that possibility diminishes the chances of heads being the correct guess. — BlueBanana

Eliminate what possibility? If it's heads we wake her once. If it's tails we wake her twice. There's nothing to eliminate.

andrewk

↪Michael

In line 4 it assumes that P(Heads) is 0.5, That would be reasonable for the probability space of an independent observer. But this probability space is the epistemological probability space of Beauty, and we have no axioms that justify our assuming that the probabilities will be the same in that probability space as in the independent observer's one.

Under Kolmogorov's framework, an event that occurs in more than one probability space can have different probabilities in different spaces. Indeed, that freedom is central to the mathematics of derivative pricing in finance, where alternative probability spaces are employed that use what are called 'risk-neutral probabilities'.

BlueBanana

↪Michael

Not from there. That's where you get when you first take the trivial case where she's woken twice either way.

That's like saying you can't eliminate 1 from the set {1} to get { } because there's nothing in { } to eliminate.

Michael

Not from there. That's where you get when you first take the trivial case where she's woken twice either way. — BlueBanana

I have no idea what you're saying, so let's start from scratch:

Mary volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice during the experiment, Mary will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening.

A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Mary will be awakened and interviewed once only. If the coin comes up tails, she will be awakened and interviewed twice. In either case, the experiment ends after the final interview.

Any time Mary is awakened and interviewed, she is asked, 'What is your belief now for the proposition that the coin landed heads?"

There are three possibilities; waking for the only time if heads, waking for the first time if tails, and waking for the second time if tails.

There's nothing to eliminate. This is it. So back to this:

We flip a coin. If it's heads then the result stands. If it's tails then we flip again and the new result stands.

What is the probability that it's heads? 2/3, because two of the three outcomes are heads? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)? — Michael

3/4 — BlueBanana

Good. So let's change it up slightly:

We flip a coin. If it's heads then we wake Mary once. If it's tails then we wake her twice.

What is the probability that it's her first awakening? 2/3, because two of the three outcomes are first awakenings? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)?

andrewk

Nevertheless, if it's Tuesday only tails will save her. — Benkei

We can assume without loss of generality that she decides before the experiment begins what she is going to guess when awoken.

Say she chose heads. Then when awoken she can know with certainty that it is not Tuesday because if it were, the coin must have landed Tails, so she would be dead, having already guessed Heads on Monday.

So if she chose Heads she will never get to Tuesday.

A variant that one might take in seeking to avoid this objection is where the execution occurs on Wednesday, so she will still be woken and asked to guess on Tuesday even if she guessed heads on Monday. But on examination we see that this doesn't change the preferences because she knows on waking that if it is Tuesday and her chosen strategy was Heads then she is already condemned to death so there is nothing at stake in today's answer.

Mathematical Conundrum or Not? Number Five

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