Michael

You're really confusing me now. Perhaps you could explain what you meant by this objection:

No, it doesn't. If you toss a coin once and then again if it's tails then there's a 0.75 chance that you'll finish with a heads. It doesn't matter if ...

I have a new original scenario. Here you said "Completely different situation. There's no 'eliminated' outcome." as if to explain why the probability ...

Where has Tuesday come from? It wasn't mentioned at all in the experiment I described here. Then you might as well say that we've eliminated the trivi...

What possibility have we eliminated?

But Beauty is aware of how the experiment is to be conducted. If I know that you flipped a coin and that if it's heads you'll give me £1 and if it's t...

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 d...

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

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 t...

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. I...

What eliminated outcome?

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 on...

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 he...

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

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...

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 tai...

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. ...

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 ? Monda...

It's not supposed to show that 75/225 != 1/3. It's supposed to show that P(Heads) != 75/225. I address the case of being asked here: P(Heads|Awake) = ...

I’ve showed the reasoning multiple times. It’s the Kolmogorov definition of conditional probability.

All you're saying is: Iff 100 people asked then it was tails 100 winners Therefore, it was tails But that's obvious, and not relevant.

No. There are more winners because you asked more people. This doesn't make any sense.

For any given person there's a 50% chance that they're right, so it doesn't matter if they pick heads or tails. It's just that if it's tails and they ...

She's given it twice: once on Monday and once on Tuesday. You don't even have to go through the hassle of putting her to sleep. Just tell her that if ...

I've gone over this: If I offer you one free lottery ticket if you correctly guess heads and two free lottery tickets if you correctly guess tails the...

OK, I think I've found the disagreement between halfers and thirders: P(Heads|Awake) = \frac{P(Heads ? Awake)}{P(Awake)} To get a result of \frac{1}{3...

The classical approach only works if each outcome is equally likely. As I would argue each outcome isn't equally likely it would be wrong to apply the...

P(Monday|Heads) means "the probability that it's Monday given the fact that it's heads". If I know that today is heads then the probability that today...

Disappointment? What do you think I want?

That's the P(Monday|Heads) = 1. So what's the condition? P(Heads|Awake)? Well, let's apply the Kolmogorov definition again: P(Heads|Awake) = \frac{P(H...

He isn't. He was looking into (among other things) campaign finance violations, and then passed it onto the Southern District of New York.

But it doesn't follow from that that each of the other three outcomes are equally likely. How do you get from P(Tuesday?Heads) = 0 to P(Monday?Heads) ...

I've done so multiple times: Kolmogorov definition: P(A|B) = \frac{P(A ? B)}{P(B)} P(A ? B) = P(A|B) * P(B) P(Monday ? Heads) = P(Monday|Heads) * P(He...

I think the mistake is you're going from P(A will happen) = P(B and C will happen) to P(A is now) = P(B is now) = P(C is now) But I don't see how that...

Why? If the flip had a 50% chance of being heads and if heads guarantees Monday then there's a 50% chance that today is heads and Monday. You seem to ...

There's a 50% chance that both B and D will happen, but there's a 25% chance that any specific day is B.

Yes, it's the original Kolmogorov definition: P(A|B) = \frac{P(A ? B)}{P(B)} P(Monday|Heads) = \frac{P(Heads ? Monday)}{P(Heads)} P(Monday|Heads) = \f...

See my edit.

If you tell me that you flipped a coin ten minutes ago I'm going to say that there's a 50% chance that it landed heads. And if you tell me that I'll g...

The topic is about credence. In my example of the free lottery tickets, my credence is that heads and tails are equally likely, despite knowing that t...

If I offer you one free lottery ticket if you correctly guess heads and two free lottery tickets if you correctly guess tails then tails is the better...

But applying the probability rule she knows that there's only a 25% chance that it's Tuesday, whereas a 75% chance that it's Monday. It stills works o...

It wouldn't be an equally safe bet, but that's irrelevant. If I offer you one free lottery ticket if you correctly guess heads and two free lottery ti...

I don't think there's any freedom. We just apply the Kolmogorov definition: P(A|B) = \frac{P(A ? B)}{P(B)} P(A ? B) = P(A|B) * P(B) P(Monday ? Heads) ...

Then there's 6 states, not 7. You're counting the tails state twice, which you shouldn't do. The two tails days need to share the probability that it'...

Or a weighted coin that’s 5/6 chance of heads.

That’s not right. Consider instead a dice with 5 red faces and 1 blue face. The three waking states are red and Monday, blue and Monday, or blue and T...

I’ll admit I was too quick to think it relevant when I thought it agreed with me. But after testing it I realised the problem. In this case, it’s not ...