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# [Poll] Can YOU solve the frog riddle?

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Here we may reign secure, and in my choice, To reign is worth ambition though in HellRegistered User regular
edited February 2022
My personal guess is that I don't have hearing good enough to pinpoint a frog. That is to say, just because the sound came from the direction of the clearing doesn't mean that I know it came from the clearing, that's not something I can reliably map. Meaning that I lack the certainty that one of the two frogs in the clearing is male, which puts me at much better odds than just licking a single frog

Edit: Partially what I mean by this is that unlike the Monty Hall problem, a clear inspiration here, this situation doesn't feel counterintuitive to me. Like, without thinking about probability I would gravitate towards licking two frogs as a clear better choice, y'know?

Straightzi on
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No face Registered User regular
They’re both 50%. Flip a coin, lol
NO MATH

ONLY

FROGS

My frog videos of choice will not be bottom paged

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Registered User regular
Larlar
3cl1ps3 wrote: »
The Boy or Girl paradox surrounds a set of questions in probability theory, which are also known as The Two Child Problem,[1] Mr. Smith's Children[2] and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner featured it in his October 1959 "Mathematical Games column" in Scientific American. He titled it The Two Children Problem, and phrased the paradox as follows:

Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

Gardner initially gave the answers 1/2 and 1/3, respectively, but later acknowledged that the second question was ambiguous.[1] Its answer could be 1/2, depending on the procedure by which the information "at least one of them is a boy" was obtained. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Maya Bar-Hillel and Ruma Falk,[3] and Raymond S. Nickerson.[4]...The paradox has stimulated a great deal of controversy.[4] The paradox stems from whether the problem setup is similar for the two questions.[2][7] The intuitive answer is 1/2.[2] This answer is intuitive if the question leads the reader to believe that there are two equally likely possibilities for the sex of the second child (i.e., boy and girl),[2][8] and that the probability of these outcomes is absolute, not conditional.[9]

There's no way to embed this one without fucking the formatting, but the actual answer is we're both right depending on the framing and resultant model. Sweet.

Eh.
I'd say that the problem telling us that at least one coin is Heads will always tell us that at least one coin is Heads for all scenarios where that is true.
And that is what should always be assumed of the math problem.

Which leads to 1/3 chance of HH.

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Registered User regular
Larlar
That is, math isn't ever going to tell us that one of the frogs in the clearing is female.

Math wants us to die.

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Registered User regular
If you count [TH] and [HT] as two separate outcomes, one with the mystery frog on the left and one with it on the right, then you have to count [HH] twice also. You have a [HH] with the mystery frog on the left and a [HH] with the mystery frog on the right. That leaves you with [TH] [HT] [HH] [HH] [TT] [TT], and you're back at 50% chance of getting the good frog.

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Registered User regular
I guess I got nerdsniped on this one since I spent time drawing pictures of the ambiguity, or at least one representation, that would result in the differing 1/2 vs. 2/3 (or 1/3) judgements:

1/2 case:

1/3 or 2/3 case:

For my perspective, my immediate thought processes ended up with the 2/3 case first. But I can definitely see the natural processes that produce a 1/2 result, and I think I'm on board with that interpretation.

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Registered User regular
Larlar
The 1/2 case diagram is incorrect

Because when I look at TH and pick coin 1, I don't then report the result of 'at least one is tails'.
I instead go and pick coin 2.
Which is H.
And then report 'at least one is Heads'

And if both are tails, I squash the frogs.

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Registered User regular
So for 2 frogs where you know 1 is male but don't know which, the probability of there being 1 female frog is 2/3 because the possibilities are MM, FM, and MF?

Then for 3 frogs where you know 2 are male but don't know which, the probability of there being 1 female frog is 3/4 because the possibilities are MMM, FMM, MFM, and MMF?

Then for 9 frogs yada yada probability of 1 female is 9/10?

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Registered User regular
Larlar
Yes
But then the problem gets more and more side-eye.
Like did someone select 100 random frogs and then at least 99 were male?
Or did someone select 99 male frogs and one random frog?

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Registered User regular
discrider wrote: »
The 1/2 case diagram is incorrect

Because when I look at TH and pick coin 1, I don't then report the result of 'at least one is tails'.
I instead go and pick coin 2.
Which is H.
And then report 'at least one is Heads'

And if both are tails, I squash the frogs.
I think you're misunderstanding how the observation is generated here. If I had a TH set and picked coin 1, then I actually would have reported "at least one tails"! For that first chart, generation of the signal is in a completely forward direction:
• Secretly generate item A as heads or tails with equal probability.
• Secretly generate item B as heads or tails with equal probability.
• Randomly select A or B to report, again with equal chance.
• Report the selected item's state as public information.
In this form of generation, there's no sense of 'redrawing' what to report to force a particular outcome. We just report what we got. Sometimes, that'll be 'tails', and sometimes that'll be 'heads'. So maybe it just happens to be that the report we got was "heads". In arguing the 1/2 answer, it ends up that the HH case is twice as likely to be an originator of a 'heads' report compared to TH and HT.

Having an additional force in there to redraw any possible 'tails' results into 'heads' is a different scenario from that diagram. That bias towards always reporting 'heads' where possible allows for the 1/3 or 2/3 answer to come about, since it provides no particular advantage to the HH case. But that bias does create a question of what actually gets reported when nature generates a TT set, and that asymmetry can feel a bit unnatural.

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passed out on the floor nowRegistered User regular
edited February 2022
The pair of frogs is 75% likely to save your ass
If there are three frogs but you know at least 2 are male:

MMM
MMF
MFM
MFF
FMM
FMF
FFM
FFF

(you might notice there are 2^n combos)

3/4 of the now possible outcomes contain at least one F. It goes up from there!

4 Frogs, >=3 Male:

MMMM
MMMF
MMFM
MMFF
MFMM
MFMF
MFFM
MFFF

FMMM
FMMF
FMFM

FMFF
FFMM
FFMF
FFFM
FFFF

4/5!

The likelihood that you would choose a random sample of n frogs with no females in falls the more frogs you sample, no matter how many of the sampled frogs you later find out are male, unless you find out they are all male.

I blame Tom Stoppard for all this confusion, and yes I would like to be buried stuffed into a locker, preferrably one with a bunch of NIN and MSI stickers stuck on it.

edit: anyway this is the story of how I got this page ranked when people are doing porn searches

MrMonroe on
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Registered User regular
edited February 2022
Larlar
MrBlarney wrote: »
discrider wrote: »
The 1/2 case diagram is incorrect

Because when I look at TH and pick coin 1, I don't then report the result of 'at least one is tails'.
I instead go and pick coin 2.
Which is H.
And then report 'at least one is Heads'

And if both are tails, I squash the frogs.
I think you're misunderstanding how the observation is generated here. If I had a TH set and picked coin 1, then I actually would have reported "at least one tails"! For that first chart, generation of the signal is in a completely forward direction:
• Secretly generate item A as heads or tails with equal probability.
• Secretly generate item B as heads or tails with equal probability.
• Randomly select A or B to report, again with equal chance.
• Report the selected item's state as public information.
In this form of generation, there's no sense of 'redrawing' what to report to force a particular outcome. We just report what we got. Sometimes, that'll be 'tails', and sometimes that'll be 'heads'. So maybe it just happens to be that the report we got was "heads". In arguing the 1/2 answer, it ends up that the HH case is twice as likely to be an originator of a 'heads' report compared to TH and HT.

Having an additional force in there to redraw any possible 'tails' results into 'heads' is a different scenario from that diagram. That bias towards always reporting 'heads' where possible allows for the 1/3 or 2/3 answer to come about, since it provides no particular advantage to the HH case. But that bias does create a question of what actually gets reported when nature generates a TT set, and that asymmetry can feel a bit unnatural.

Yeah.
Not having a forcing agent is a forcing agent by itself though.
And the forcing agent is an assumption built into the math problem.

You either can have 1/3, 1/2 or 1/1 chance of having HH given 'at least one head' depending on how the person posing the question labels the sets.

And I'd say repicking after encountering a tails is the best assumption to make here.
It's the set of all outcomes where the statement is true, not the set of all outcomes where the statement is true but I give up measuring halfway through checking half the sets.

discrider on
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RIESLING OCEANRegistered User regular
Larlar
Perrsun wrote: »

God, this jam rips so hard.

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Sleepy Registered User regular
edited February 2022
Larlar
There's a 3/4 chance of there being a male frog in the pair. There is a 1/2 chance of having a male and female frog. Therefore there is a 2/3 chance that there is a female frog given that there is a male frog, because P(A|B)=P(A+B)/P(B)=(1/2)/(3/4)=2/3. (The writeup linked on the first page overcomplicates this a lot... though it can be made even more pointlessly complicated by considering that the probability of hearing a croak is clearly a function of time

Polaritie on
Steam: Polaritie
3DS: 0473-8507-2652
Switch: SW-5185-4991-5118
PSN: AbEntropy
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might be real Registered User regular
The pair of frogs is 75% likely to save your ass
i didn't read the poll options closely enough

but i agree with 67% if the question was set up the way he intended to set it up

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boom Registered User regular
the primary reason to become a statistician is to construct scenarios that when you reveal the math to people makes them go "hey fuck you"

i'm fairly confident i could empty quote this post every few pages in this thread and get more agrees each time at this point

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Registered User, Moderator mod
They’re both 50%. Flip a coin, lol
Polaritie wrote: »
There's a 3/4 chance of there being a male frog in the pair. There is a 1/2 chance of having a male and female frog. Therefore there is a 2/3 chance that there is a female frog given that there is a male frog, because P(A|B)=P(A+B)/P(B)=(1/2)/(3/4)=2/3. (The writeup linked on the first page overcomplicates this a lot... though it can be made even more pointlessly complicated by considering that the probability of hearing a croak is clearly a function of time

there's a 100% chance of having a male frog in the pair as the problem is described though?

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No face Registered User regular
They’re both 50%. Flip a coin, lol
TOO MUCH MATH TALK

NOT ENOUGH FROG VIDEOS

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Sleepy Registered User regular
Larlar
Polaritie wrote: »
There's a 3/4 chance of there being a male frog in the pair. There is a 1/2 chance of having a male and female frog. Therefore there is a 2/3 chance that there is a female frog given that there is a male frog, because P(A|B)=P(A+B)/P(B)=(1/2)/(3/4)=2/3. (The writeup linked on the first page overcomplicates this a lot... though it can be made even more pointlessly complicated by considering that the probability of hearing a croak is clearly a function of time

there's a 100% chance of having a male frog in the pair as the problem is described though?

Yes, which is why we're calculating P(A|B).

Steam: Polaritie
3DS: 0473-8507-2652
Switch: SW-5185-4991-5118
PSN: AbEntropy
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Registered User regular
This is easy, I simply ask the frog on the stump which of the frogs in the clearing it would lick, then I lick the other one.

My friend is working on a roguelike game you can play if you want to. (It has free demo)
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Registered User regular
Let's solve this empirically by checking a million random pairs of frogs from which we hear a croak and seeing how many contain a female.

The code:
import random

# constants
num_frogs = 2
total_samples = 1000000

# initialise variables
sample = 0

while (sample < total_samples):
# Generate frogs
frogs = []
for i in range(num_frogs):
frogs.append(random.choice("MF"))

# Throw out sample and try again if both are female since we heard a croak
if (all(frog == "F" for frog in frogs)):
continue

# Did we find any lady frogs?
if (any(frog == "F" for frog in frogs)):

sample += 1



The results:
Found a lady frog 666418 times
Approximate probability of finding a lady frog = 0.666418


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Registered User regular
Alternatively...

The code:
import random

# constants
num_frogs = 2
total_samples = 1000000

# initialise variables
sample = 0

while (sample < total_samples):
# Generate frogs
frogs = []
for i in range(num_frogs):
frogs.append(random.choice("MF"))

# Have one of them croak
croak = random.choice(frogs)

# Throw out sample if the croak was female since we heard a male
if (croak == 'F'):
continue

# Are there any lady frogs?
if (any(frog == "F" for frog in frogs)):

sample += 1



The results:
Found a lady frog 499756 times
Approximate probability of finding a lady frog = 0.499756


It's all about how you frame the generating mechanism underlying the scenario.

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Registered User regular
Let's enumerate the possible universes of the frogs in the clearing by intersecting which frog is the male who croaked and the sex of the silent frog.

The left frog is the male who croaked. The right frog is also male.
The left frog is the male who croaked. The right frog is female.
The right frog is the male who croaked. The left frog is also male.
The right frog is the male who croaked. The left frog is female.

In two of these universes, both frogs are male. In two of these universes, the other frog is female. The exercise evidently allows us to survive if either frog is female, so there is apparently a 50-50 chance at the clearing, same as the stump.

They're asking "what are the odds that at least one member of the set has the required attribute, given that an unknown member of the set does not possess that attribute?" Which is an interesting problem, but I think it actually works out to just be the odds of finding at least one member of a set sized n-1 with the target attribute.

My favorite musical instrument is the air-raid siren.
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Registered User, Moderator mod
They’re both 50%. Flip a coin, lol
at the very least this illustrates an interesting ambiguity in the question if you can write code to solve the problem two ways that comes up with different answers

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Yeah ZestRegistered User regular
They’re both 50%. Flip a coin, lol
This shit is just the Oxford comma for statisticians and nothing anybody says can convince me otherwise.

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Registered User regular
Let's find this frog and force him to tell us his secrets

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Registered User regular
What is a frog's preferred programming language?

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might be real Registered User regular
The pair of frogs is 75% likely to save your ass
ChicoBlue wrote: »
What is a frog's preferred programming language?

javascript

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Registered User, Moderator mod
They’re both 50%. Flip a coin, lol
anything but Python

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Registered User, Moderator mod
They’re both 50%. Flip a coin, lol
https://esolangs.org/wiki/I_like_frog

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Registered User regular
They’re both 50%. Flip a coin, lol
Frogs are good with any programming language, but they’re more skilled with debugging than development.

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Registered User, Moderator mod
They’re both 50%. Flip a coin, lol
ah good point, they probably prefer a language with as many bugs as possible

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passed out on the floor nowRegistered User regular
The pair of frogs is 75% likely to save your ass
Pinfeldorf wrote: »
This shit is just the Oxford comma for statisticians and nothing anybody says can convince me otherwise.

Allow me to introduce you to my friends, base rate errors and the conjunction fallacy.

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Registered User regular
Larlar
So after reading the information what I get is that taking a sample from a population level has a different probability than if you are looking at a particular pair and observe information about it. I think the original scenario is supposed to make you think it's the former, but I think the latter interpretation fits better.

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In the scuppers with the staggers and jagsRegistered User regular

Almost all math arguments on the internet are the result of ambiguity in the question's starting conditions.

Rodeo frog will return next week with more upsetting facts.

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Registered User regular
Larlar
The question is, how do you formulate the croaking question to include the possibility that 'at least one frog is male' is the author making you feel like you have a shot with the two guaranteed males in the clearing, as otherwise they would have told you about the female there.

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Scritch scratch scritch scratch scritch scratch scritch scratch scritch scratch scritch scratch scritch scratch scritch scratch scritch scratch scritch scratch scritch scratch scritch scratchRegistered User regular
discrider wrote: »
3cl1ps3 wrote: »
discrider wrote: »
3cl1ps3 wrote: »
My post explains how and why that way of modeling it is wrong. The model that results in 67% assumes either coin can be tails. We already know that one of the coins can not tails, therefore that assumption has to be discarded. You (and the video) are doing the exact same thing that gets people to the "don't switch" i.e. wrong answer i. Monty Hall :P

I'm not gonna keep arguing it though, I said what I had to say on it and if you don't buy it then no amount of me trying to explain it will convince you.

Where is my math wrong then?

Unless me stating the result to you has somehow increased the chance of me flipping two Heads after I've flipped them (probability of croak), then the chance that I've flipped two heads is after I've told you I've flipped at least one is:
Original chance of HH (25%) / Chance of having flipped at least one H (100%-TT= 75%)

Whereas your matrix should look like
[H H] [T H]
[H T] [T T]

And I don't see where you justify doubling the chance of HH at all.

These two situations:

-I have flipped two coins, one is heads. What are the odds the second is tails?

and

-I have flipped a coin. It is heads. I will flip it again. What are the odds it will be tails?

are the same. The answer in both cases is 50% because the second flip is its own independent event. Sequential independent events have to be modeled differently from aggregate events.

Since we know the results of one flip already, we have to discard the aggregate model of the whole set and move to individual modeling of the next partial set (this is also the step that gets you to 50/50 in Monty Hall).

Whether the problem is "you flipped two coins, one is heads" or "you flipped a coin twice and got heads once" the math doesn't change. "You don't know which coin" doesn't matter because the coins are both already flipped, so you have to discard any solution in which the coin that came up heads could be tails, because it can't - the flip already happened and we know it was heads. The framing screws up the model.

You don't know which of the coins came up heads though.
So you can't discard all solutions in which the coin that came up heads came up tails.

So I flip a coin, it comes up heads, what is the chance that it comes up tails next?
Looks like:
[H H] [H T] [T H] [T T]

But I flip a coin twice, it came up heads at least once looks like
[H H] [H T] [T H] [T T]
Because we can't eliminate either T H or H T.

The framing of
-I have flipped two coins, one is heads. What are the odds the second is tails?
Tricks the reader into the first 50% scenario by putting an ordering on the results with the word 'second'.
Yes it you flip two coins and the first comes up H then the second has a 50% of tails, but if you can't distinguish between the two coins then it is twice as likely that you have one tail and one head than it is if you have two heads.

On your second example I think you have to have a second HH

If coin 1 is heads then the possibilities are

HH HT

If coin 2 is heads the possibilities are

TH HH

So in total there's 4 outcomes and two of them have a tails.

I don't THINK that's double counting the HH possibility? Either coin could be the one that already came up heads, so we have to account for both results of both coins.

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boom Registered User regular
at the very least this illustrates an interesting ambiguity in the question if you can write code to solve the problem two ways that comes up with different answers

the source of all these style of internet arguments is mischievous ambiguity

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Registered User regular
MrBlarney wrote: »
Alternatively...

The code:
import random

# constants
num_frogs = 2
total_samples = 1000000

# initialise variables
sample = 0

while (sample < total_samples):
# Generate frogs
frogs = []
for i in range(num_frogs):
frogs.append(random.choice("MF"))

# Have one of them croak
croak = random.choice(frogs)

# Throw out sample if the croak was female since we heard a male
if (croak == 'F'):
continue

# Are there any lady frogs?
if (any(frog == "F" for frog in frogs)):

sample += 1



The results:
Found a lady frog 499756 times
Approximate probability of finding a lady frog = 0.499756


It's all about how you frame the generating mechanism underlying the scenario.

In your version you know which frog croaked, which affects the answer.

I didn't watch the video but my understanding was that you don't know which frog croaked, only that one of them did, and that added unknown affects the probability.