Question about probability riddle: doesn't make sense

[stinkynuts]

I guess where the crucial point lies is in this peculiarity, which intuitively makes no sense:

IF I see the frog that croaked, we can all agree that the odds are 50/50 on the two frog side, since we can eliminate the frog that croaked, and the other has a 50/50 chance of being male or female.

However, if we DON"T see the frog that croaked, then it seems that the consensus is that there is a 2/3 in chance that you will have a female among the pair.


My question is, what possible difference does seeing the frog that croaked make? We already know that one of them is male because it croaked. So the other still has a 50/50 chance of being male or female.

 

[fuji]

The difference is whether you are assigning one and sampling one, or sampling two and then adding information.

For the record, I think the frog case is not sampling two, but only sampling one. It's not two random picks. It's a random ordering of a known quantity and a random pick.

But in the coin example, it's two random picks and the 2/3 odds will be real--and make for a profitable drinking game with your buddies next time you hit the pub.

 

[TeeJay]

That's intuitive, but wrong. There are in fact twice as many ways to pick a M-F combination as a M-M combination.

There's a great Wikipedia article about how the exact wording of the question determines whether the answer is 2/3 or 1/2.

https://en.m.wikipedia.org/wiki/Boy_or_Girl_paradox#Bayesian_analysis

No because in either scenario (m-f or f-m) result is same
The second example linked to has an extra bit of info (which is the older child)
In frog example it is irrelevant

That video took a classic math problem, tried to explain it in laymans terms, and failed

 

[TeeJay]

This is similar to the "Let's make a deal" problem

At the end of the show, there was one contestant, and three curtains. Behind one, there was the big prize, between the other two were much lesser prizes.

So the contestant is asked to choose one curtain.

AFTER the contestant chooses, the host then unveils one curtain, (because he knows where the big prize is), which is NOT the big prize. The contestant then gets to switch curtains if he wishes.

Should he switch?

Not even close
This APPEARS to be random chance, but the problem is game is of course rigged

 

[fuji]

No because in either scenario (m-f or f-m) result is same
The second example linked to has an extra bit of info (which is the older child)
In frog example it is irrelevant

That video took a classic math problem, tried to explain it in laymans terms, and failed

You may think the M-F and F-M combinations are the same, but there are two ways to come up with that combination and only one way to get M-M, so it's twice as likely. Since F-F has been eliminated that leaves you with two chances at M-F and one chance at M-M.

 

[stinkynuts]

This video says that the original video is wrong, and that in fact it is 1/2, not 2/3. For the reason Fuji mentioned.


https://www.youtube.com/watch?v=go3xtDdsNQM

 

[johnhenrygalt]

on the other hand......

This guy doesn't get it. Switching gives you a 2/3 chance of winning. For those who have trouble following logic and probability in the abstract, you can run an experiment. Find a friend and actually play the game 200 times - 100 times with always switching and 100 times with never switching, then check the results.

Another way to think of the problem is pretend that there are not 3 but 50 doors: one door with a car and 49 with goats. After you pick one door, the host reveals 48 goats and leaves you with 2 unopened doors - the one you picked and one other. Here it should be obvious that one would always switch to win the car (this increases your chances from 1-in-50 to 49-in-50).

 

[johnhenrygalt]

Someone go run this experiment:

Get two friends to flip coins in another room. Friend A flips one coin and Friend B flips two. If neither of the coins Friend B flips are heads, both friends re-flip their coins. If Friend B gets at least one heads, he calls out to you and says so. Now, you win a dream vacation if you pick a friend who has flipped a tails. According to what the video in this thread has told you, it makes the most sense to pick Friend B every time. Now run this experiment 1.0x10^20 times. If the video is right, you should win ~6.7*10^19 dream vacations.

But don't be surprised when you actually won roughly 5.0x10^19 vacations. The reason is because there are actually two distinct and equally possible combinations of HH [call it H1H2 and H2H1] (you lose in both cases), and one each of HT and TH (you win both times).

Wrong. Try it yourself. You'll see that with friend B you'd win 2/3 of the time. - H1H2 and H2H1 aren't two distinct possibilities. You don't even need a friend. Flip 2 coins simultaneously - reflip if you get 2 tails. Then record how many times you got one head, one tail and how many times you got 2 heads. By the time you've recorded 100 events you should see the 2/3 - 1/3 pattern develop.

 

[johnhenrygalt]

You can try this yourself in a pub.

Get a friend to flip two coins behind a screen. You get to pick heads or tails. If he has not got the coin you guessed it's a bye, and you play again.

If he has the coin you guessed, he slides it over to you, keeping the remaining coin secret. Now you have to guess that coin and if you guess right he buys the next round. If you guess wrong, you buy the round.

If you bet the second coin is the opposite of the first one, you win the bet 2/3rds of the time and wind up paying for only a third of the drinks.

It's really going to work.

This is correct. You don't even need the pub or a friend; you can do it yourself.

 

[stinkynuts]

Now that I think about it, I don't think the remaining combinations are:

MF
FM
MM

This is because MF an FM are the same thing. This is not a permutation, but a combination problem.

 

[TeeJay]

You may think the M-F and F-M combinations are the same,

They are the same, since in this flawed puzzle either combination has same result
The author of this video thought he was being smart but the reverse was true

 

[RemyMartin]

From the sample space, he should also eliminate one of the (male-female), since (male-Female) is same as (Female-male).

by doing that you only have two choices left , which is still 50%.

thats from post #8

 

[fuji]

They are the same, since in this flawed puzzle either combination has same result
The author of this video thought he was being smart but the reverse was true

They are not the same. If I flip two coins H-T and T-H are not the same outcome, they are two different outcomes. If you have two frogs, M-F and F-M are not the same outcomes. The left frog being male is different than the right frog being male.

There's a problem with the video, but there's no problem with the coin toss example. You can't arbitrarily collapse the two different ways of getting a male and a female into a single case.

 

[TeeJay]

They are not the same. If I flip two coins H-T and T-H are not the same outcome, they are two different outcomes. If you have two frogs, M-F and F-M are not the same outcomes. The left frog being male is different than the right frog being male.

They are exactly the same and have same result

The whole premise was to lick the correct frog to save your life
If you lick the M then F or the F then M in both situations you are still alive

Go back and watch original video if you can't grasp the logic

 

[fuji]

They are exactly the same and have same result

No, they aren't the same. Male on the left is not the same as male on the right. There are three different ways to get at least one male, and they are different ways: both male, one male on the left, one male on the right.

You really shouldn't post on this thread when you have this little understanding of basic probability. I believe they teach this stuff in highschool, but maybe you didn't get that far in mathematics.

 

[TeeJay]

No, they aren't the same. Male on the left is not the same as male on the right. There are three different ways to get at least one male, and they are different ways: both male, one male on the left, one male on the right.

You really shouldn't post on this thread when you have this little understanding of basic probability. I believe they teach this stuff in highschool, but maybe you didn't get that far in mathematics.

Pot & Kettle huh

Ok genius then explain for the masses
If I am poisoned and need to lick one frog
How is M-F any different than F-M?

 

[escortsxxx]

While this is the current theory of probability I believe it the case of the Zeno paradox, or the case of the missing dollar, that is the theory being applied incorrectly. Look up the Zeno paradox, where a faster runner could not catch a slower running no hard he tries . . .
But to convert the frog to SP terms, if there was a 50/50 chance for any sp to give good service and we had 2 sps in one place and 1 sp in the other, and you tried one sp at place A which was not to your taste which sp should you try next?

https://en.wikipedia.org/wiki/Monty_Hall_problem

 

[fuji]

Pot & Kettle huh

Ok genius then explain for the masses
If I am poisoned and need to lick one frog
How is M-F any different than F-M?

It's not pot and kettle. I understand it. You don't.

M-F is different from F-M because in one case the male frog is on the left, and in the other the male frog is on the right. Left versus right are different.

You may not CARE about the difference, but nevertheless it's a difference.

The point being there are four different possibilities, all different from one another: MM, MF, FM, and FF. The croak only eliminated FF leaving three remaining possibilities.

I think there's a valid discussion about whether hearing the croak changed the sample space--there's additional information that one actually croaked--but there is no validity in claiming MF is the same as FM. It's not the same.

 

[stinkynuts]

I still can't for some reason can't wrap my head around the following logic, that I perceive to be correct:

There are two frogs
One croaked, so it is male
We don't know the sex of the second one
The second one can be either male or female
So the probability that there will be a female among the pair of frogs is 50%


In this case I don't think that it matters if the frog on the left or right croaked
The probability that the other frog is female is still 50%

If the frog on the right croaked, then the probability that the frog on the left is female is 50%
If the frog on the left croaked, then the probability that the frog on the right is female is 50%

No matter which case, it is still 50%.

 

[escortsxxx]

However computer models have shown this is not the case - or at least math wizards claim this to be so.