:fa_bar_chart_o: Probability is confusing.
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@Grunnen said:
Let's change the assumption of the 100% chance that a male frog will croak.
There was no assumption about how frequent they croak. It was simply that one did croak. You could alternatively talk about how frequently they sit on stumps vs hang out in clearings. It wouldn't change the probabilities at all, because we came upon a situation where one was on the stump and two were in the clearing.
You could say something about the a priori chances of that happening, but it wouldn't change the probabilities in the problem as presented.
If you arrive at a result that is only valid under the assumption that a male frog will croak with 100% certainty, you must have used that assumption somehow/somewhere.
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If you arrive at a result that is only valid under the assumption that a male frog will croak with 100% certainty, you must have used that assumption somehow/somewhere.
What result would that be?
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The 2/3-probability.
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If you repeat an experiment 1000 times with you blindfolded, so that you only hear a frog croak (completely omitting the experimental tries where neither of the frogs croaked), you'll find that one of the two frogs is female around 2/3 of the time. We agree on that point.
Now, if there was another observer who wasn't blindfolded, so he was able to see which frog croaked, the results would still be that one of the two frogs was female around 2/3 of the time. That can't change. His reality can't be different than yours. So neither can the probability, because it must match what actually happened. If it doesn't, there's no use to calculating it.
The problem here is that we're keeping or discarding experimental runs based on whether the frogs croak.
The blindfolded observer buckets all tries where a frog croaked together, and 2/3 of the time when a frog croaks, one of the frogs turns out to be female.
The unblindfolded observer has two separate buckets for the frog on the left and the frog on the right (assuming for the sake of the argument that the frogs stay put for the duration of the experiment). Half the time when LeftFrog croaks, a frog turns out to be female. Half the time when RightFrog croaks, a frog turns out to be female.
But LeftFrog croaks less than half of the time - half of the time that any frog croaked, LeftFrog croaked. And likewise RightFrog. I have to go catch a train but I'll put some numbers on it and if they don't come out to 2/3 of the time total for the unblindfolded observer, then I'll eat my hat for lunch tomorrow.I think I confused myself there and I have frankly no idea what numbers to put on anything - in my defence, I was in a hurry. But the point is that we're talking about a different phase space in each case.
When we talk about running the experiment with a blindfolded and a sighted observer, the person seeing what happens doesn't change anything, because we're still considering the same phase space as the person in the woods who heard a frog croak but didn't see it. We're considering all the times when LeftFrog croaked and the times when RightFrog croaked.
But the observer in the woods who saw one of the frogs croak is calculating their probability based on the phase space of all the times when LeftFrog croaked or all the times when RightFrog croaked.
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It can be shown even simpler, directly.
Let x be the chance that a male frog croaks. Then, if a croak is observed:
MM: P(croak) = x + (1-x)*x
MF: P(croak) = x
FM: P(croak) = x
FFP(at least a female) = (MF + FM) / (MM + MF + FM) = 2x / (2x + x + (1 - x) * x) = 2 / (4 - x)
Now let P(at least a female) = 2/3
Then: 2/3 = 2/(4-x), thus x = 1
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That's one (frequentist) definition.
Try the standard definition.
Nevertheless, your intuition about the different observers is misleading you.
How so? In the scenario, if the observers calculate the probability the way that you say is correct, then one will find that the probability he calculated doesn't match reality; therefore, the probability that he calculated was wrong.
Probability is not relative frequency.
A calculated probability is. An observed probability is not.
If you run an experiment, say, 10 times, and based on your observation of some result 7 times out of 10 you calculate that the probability was 70%, then no... that is not relative frequency, because you didn't run the experiment enough times for the results to converge.
However, if you look at the parameters of the experiment and you calculate that the result should occur 2/3 of the time, then that is relative frequency. And, if you can accurately model the parameters of the situation and then simulate it, your results over a great many number of simulations has to converge to the probability you calculated; otherwise either your simulation was modeled incorrectly or you calculated the probability incorrectly.
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An Example: Tossing a coin.
If you create a computer model (or physical experiment) with a coin toss with every single initial condition known (the force imparted to the coin, the winds in the area, etc etc), then you run it 1 billion times in a row with the same initial conditions, YOU WILL GET THE SAME RESULT EVERY TIME (paper)
It is only if you vary the initial conditions of tossing the coin in some "random" (unknown) way, by changing the force imparted to the coin slightly, adding updrafts, etc, will you get a different result. Therefore you cannot "repeat" the event - every one is unique!
From the blog post:
for everything that happens— every physical thing that happens does so under very specific, unique circumstances. Thus, nothing can have a limiting relative frequency; there are no reference classes.
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Let x be the chance that a male frog croaks. Then, if a croak is observed:
Yes, when a croak was observed the probability that one of the frogs croaked is 1. That's my point. But it doesn't say anything about the prior probability of a male croaking. That's my point.
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Okay, I've been thinking a bit more.
The actual outcome is that, whether or not you hear a croak, you will choose to go into the clearing with the two frogs. And a-priori you will have a chance of 75% to live if you do that.
The whole rest of the story is indeed a trade-off between your knowledge about the situation being more or less specific, and the possible situations within that knowledge being more or less likely.
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How so? In the scenario, if the observers calculate the probability the way that you say is correct, then one will find that the probability he calculated doesn't match reality; therefore, the probability that he calculated was wrong.
You're not making sense here. How do you think their respective probability estimates "do not reflect reality?" What reality? Their estimates were based on their knowledge. The outcome was previously determined.
But let's suppose the outcome was not determined. Instead of a frog's sex, we were talking about a future roll of a die (and maybe the unknown was the number of faces or something). Then they would still be correct to have different estimates if they had different information. That the outcome went one way or the other says absolutely nothing about their estimates. Low probability things happen, too.
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The actual outcome is that, whether or not you hear a croak, you will choose to go into the clearing with the two frogs. And a-priori you will have a chance of 75% to live if you do that.
If you heard no croak, then yes, survival chance is 75%.
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@Grunnen said:
The actual outcome is that, whether or not you hear a croak, you will choose to go into the clearing with the two frogs. And a-priori you will have a chance of 75% to live if you do that.
If you heard no croak, then yes, survival chance is 75%.
No, if you heard no croak, then the survival chance is dependent on the probability of a male croaking (e.g. 100% survival chance if a male croaks with 100% probability).But a-priori your survival chance is 75%.
With hearing or not hearing a croak, you can narrow your situation down to a situation being more likely or less likely and with a higher or with a lower survival chance, but if you multiply both factors and add up all possibilities, you'll end up with 75% again. And that's why this riddle doesn't violate the law of large numbers.
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What reality?
The reality that "when (a / any / at least one) frog croaks in the clearing, 2/3 of the time, one of them is female".
Both of the observers were measuring that. One observer had the additional information of which frog croaked, but his results still have to match what the other observer tallied, because ultimately they're measuring the same thing.
The only way that their results could be different would be if one of the observers was counting or ignoring situations that the other observer wasn't. But both observers are supposed to be counting the number of times there was a croak, and of those times, how many times one of the frogs was female.
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No, if you heard no croak, then the survival chance is dependent on the probability of a male croaking (e.g. 100% survival chance if a male croaks with 100% probability).
Yes, if you knew about the probability of hearing a male croak, you could further condition the probability.
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Both of the observers were measuring that. One observer had the additional information of which frog croaked, but his results still have to match what the other observer tallied, because ultimately they're measuring the same thing.
No, they're measuring different things. One has additional information (namely, which frog croaked). Let's look at it this way:
Given frogs X and Y, A only heard, B saw that X croaked, C didn't hear anything because he had headphones on and D heard a croak but also saw Y laying eggs earlier in the day:
A's possibilities (for X/Y)
M/M 1/3 F/F 0 M/F 1/3 F/M 1/3B's possibilities:
M/M 1/2 F/F 0 M/F 1/2 F/M 0C's possibilities:
M/M 1/4 F/F 1/4 M/F 1/4 F/M 1/4D's possibilities:
M/M 0 F/F 0 M/F 1 F/M 0So A believes that he has a 2/3 chance of surviving, but B believes he has only a 50% chance of surviving. C believes 1/4, and D knows which is male and which is female. All are reasoning about the same situation, but with different knowledge, and each comes to a different and correct conclusion.
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Let's say, you have 100 blind observers and 100 seeing observers and conduct 100 experiments with one of each participating together but independently. And let's assume that a male frog croaks every time.
25 times, the blind observer will not hear croaks and live, and the seeing observer will not hear a croak and live
25 times, the blind observer will hear croaks and the seeing observer will see the left frog croak. They will both live.
25 times, the blind observer will hear croaks and the seeing observer will see the right frog croak. They will both live.
25 times, the blind observer will hear croaks and the seeing observer will see both frogs croak. They will both die.75 blind observers hear croaks, 25 of them will die, 2/3 chance of survival.
50 seeing observers will see the left frog croak, 25 of them will die, 1/2 chance of survival.
50 seeing observers will see the right frog croak, 25 of them will die, 1/2 chance of survival.
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50 seeing observers will see the left frog croak, 25 of them will die, 1/2 chance of survival.
50 seeing observers will see the right frog croak, 25 of them will die, 1/2 chance of survival.Yes, but if you combine those two groups, 75 of those observers see one or both frogs croak, and of those 75, 2/3 of them live.
The only way that a seeing observer gets more information is if he sees both frogs croak, and in that case, he should go to the frog on the stump.
But that's different from the original scenario, because we were assuming that only one or the other frog croaked (assuming they don't croak at exactly the same time); as soon as one croaked, we stopped waiting and made our decision.
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@Grunnen said:
50 seeing observers will see the left frog croak, 25 of them will die, 1/2 chance of survival.
50 seeing observers will see the right frog croak, 25 of them will die, 1/2 chance of survival.Yes, but if you combine those two groups, 75 of those observers see one or both frogs croak, and of those 75, 2/3 of them live.
Exactly.By the way, it gets interesting if you assume that a male will croak 100% of the time, except that it will shut up if another male has croaked first.
In 25% of the cases, there will be two females, no croaks, the observer will live
In 25% of the cases, the left frog will be male and croak, the observer will live
In 25% of the cases, the right frog will be male and croak, the observer will live.
In 12,5% of the cases, both frogs will be male, the left one will croak, the observer will die.
In 12,5% of the cases, both frogs will be male, the right one will croak, the observer will die.Chance of survival after seeing which one of the frogs croaked? Not 1/2, but 2/3 again!
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You can get the same result if you stop the observation as soon as one frog croaks, which was actually the scenario in the original: it was assumed that we don't wait to find out if both frogs croak before we make the decision whether to head toward the stump or toward the clearing.
If neither the seeing nor the non-seeing individual waits long enough to hear both frogs croak, then both of them have a 2/3 chance of survival.
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What about transfrogs? Are we accounting for transfrogs?
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Yes.
But wait a minute, there is indeed the stump too. With the frog that didn't croak, but which could still be a male, with a certain probability.
So if we assume that x is the chance of a male frog croaking (and x>0, because we heard a croak!)
chance of survival at the stump: 0,5x / (0,5 + 0,5x) = x/(x+1) < 0,5 if x>0
chance of survival at the clearing: 2/(4-x) > 0,5 if x>0Ok, I'm happy now!
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What about transfrogs? Are we accounting for transfrogs?
What did you think I meant by stump frog?
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Have. To. Solve. This.
Let A be the frog on the stump and B and C be the frogs in the clearing. Let x be the probability that a male frog wants to croak.
If we wait for a short amount of time but certainly not longer than that any one of the three(!) frogs croaks, the probabilities of the frog at the tree stump also aren't independent. Now let's write the whole thing out:
Gender Wants to croak Croaks first Survival ABC A B C MMM 1/8 no no no (1-x)^3 dead end no no yes (1-x)^2*x C 1 no no yes no (1-x)^2*x B 1 no no yes yes x^2*(1-x) B 1/2 no C 1/2 no yes no no (1-x)^2*x dead end yes no yes x^2*(1-x) A 1/2 dead end C 1/2 no yes yes no x^2*(1-x) A 1/2 dead end B 1/2 no yes yes yes x^3 A 1/3 dead end B 1/3 no C 1/3 no MMF 1/8 no no (1-x)^2 dead end yes no x(1-x) dead end no yes x(1-x) B 1 clearing yes yes x^2 A 1/2 dead end B 1/2 clearing MFM 1/8 no no (1-x)^2 dead end no yes x(1-x) C 1 clearing yes no x(1-x) dead end yes yes x^2 A 1/2 dead end C 1/2 clearing MFF 1/8 dead end FMM 1/8 no no (1-x)^2 dead end yes no x(1-x) B 1 stump no yes x(1-x) C 1 stump yes yes x^2 B 1/2 stump C 1/2 stump FMF 1/8 no 1-x dead end yes x B 1 stump and clearing FFM 1/8 no 1-x dead end yes x C 1 stump and clearing FFF 1/8 dead end'Dead end' marks impossible cases, namely those cases where no croak from the clearing is heard.
And now, here it comes, if I am not mistaken:
Survival only at clearing: 1/8 * ( 2x(1-x) + x^2 )
Survival only at stump: 1/8 * ( 2x(1-x) + x^2 )It even makes sense intuitively. If there is one male frog in the clearing, the chances for both other frogs to be female are the same.
