Well?

OP you faggot.
Make up some new and original questions. That are not purposely made ambiguous or to have flaws that allow for retarded correct answers. If you can't then kill yourself you faggot.

Oh, I see, I made the assumption that a card with a letter would always have a number on the other side. Ayy

A K and 7 then

A
>make sure it fits the rule
7
>make sure it's not a vowel on the other side

why not:
K
>it's not a vowel, nobody gives a shit
2
>either it's a vowel and it fits the rule or isn't a vowel and it's irrelevant

say

pretty close, you just need to change one thing there

Read

all except k

>The population of these frogs is split evenly between males and females
>There are only 3 frogs

>The population is only the three frogs you saw in this particular moment
Report to the nearest McDonalds immediately.

lick the frogs behind you right

the one in front of you is 50/50 obviously

the one behind you theres at least one male so

MM
FM
MF

2/3 chance on a female

A must have an even number on the other side. Turn it to find out.

K must not have a vowel on the other side. Turn it to find out.

2 can have whatever on the other side. You don't have to turn it.

7 must not have a vowel on the other side. Turn it to find out.

A, K, 7.

But what if K has a vowel on the other side? WHAT THEN?

>lick the frogs behind you right
Nope.

>the one in front of you is 50/50 obviously
Nope. Try again.

>the one behind you theres at least one male so
>MM
>FM
>MF
>2/3 chance on a female
You are making the classic mistake of assuming that all distinct states are equally likely. They are not in this case. Think about why.

If the population is evenly split, there are only 8 possible combinations of frogs:
MM M
MM F
MF M
MF F
FM M
FM F
FF M
FF F
Out of these, the last two are not possible for the situation described. That leaves 6 different options. Below I have written the chances of survival if you go to the two frogs/one frog:
MM M 0/0
MM F 0/1
MF M 1/0
MF F 1/1
FM M 1/0
FM F 1/1
You can easily see that if you go to the singular frog you have a 3/6 or 1/2 chance to survive, while if you go to the two frogs you have a 4/6 or 2/3 chance of survival. Of course the probability of you hearing a frog behind you croak is only 6/8 or 3/4. So in the case where a frog doesn't croak behind you (25% of the time) you should still go to to the two frogs because your chances of survival become 75% whereas the single frog remains at 50%.

You made the same mistake of assuming that every possible combination is equally likely. There is something you're missing.

Alright help me out here then

Very simple : All of them.

In order to determine that one card does not break the rule, you must check all cards to ensure all of them follow it.

I'll give you a hint: the answer is dependent on the chance of males croaking in the time you were listening.

which you never gave us...

Actually; let me rephrase that

A and 7 must be tipped over.

It does not explicitly state that consonants mean odd numbers. Consonants can be even or odd. But vowels can only be even. So therefor, even can be vowel or consonant, odd can only be consonant.

A must have a even number on the other side
K does not matter, it can have either and still follow the rule
2 does not matter, it can be vowel or consonant.
7 does matter, it can only be a consonant.

[spoiler]I didn't see that some are not mutually exclusive. pls no bully

well I guess if its MM for the frogs hearing a croak is more likely

but then again no clue how to put that in numbers

This was easy.. if you would have put a little mind, you will solve it yourself

So it's a variable.

If K has a vowel on the other side then the rule is broken. Jesus how dumb are you people?

You can just as easily say that not hearing a croak from the singular frog could highly increase the chances that that frog is female. Also, if you're going to expect information that you don't provide, you might as well add that male frogs only croak around a female frogs. With the information given to us, makes the most sense.

its kinda implicit that one side has a character and the other one has a number

but whatever

>what is algebra

>You can just as easily say that not hearing a croak from the singular frog could highly increase the chances that that frog is female.
You could say that, but then you would just be making shit up that isn't in the problem, and you would be answering a different problem.

>Also, if you're going to expect information that you don't provide, you might as well add that male frogs only croak around a female frogs.
I didn't "expect information" that I didn't provide. You should be capable of answering a question with an unknown variable.

>With the information given to us, makes the most sense.
No, that makes no sense at all. I'll give you another hint. In order for a frog which we didn't hear croak to have an equal chance of being male or female, males would have to be as likely to not croak as females. But we know this is not true because we already heard a male croak and we know females don't croak. At this point I'm spoonfeeding you the answer.

Where is it implied?

in the video this image is from

The question being asked is not in the video, it's in the image in the original post.

Very

A & 2

alright spoonfeed me more senpai

Nope. Figure it out.

no u

Shut up fag. I know you won't tell anyone if they're right so that you can post this shit again with out people instantly spewing out the correct answer.

If you could actually get it right I would tell you.

STFU liar! I saw a thread a while back with this exact question (albeit with slightly less ambiguity) and you just said; 'so far, so and so many people got it right' without so much as giving a clue to whom was right.

A and 7. K could have an even or odd on the other side, it doesn't matter which, and the rule is still valid. 2 could have a vowel or consonant on the other side and it doesn't change the rule. A must have an even number and 7 can't have a vowel, or else the rule isn't true.

If you actually read the thread you would see that I congratulated several people who got the right answer, you rabid ignoramus.

Okay so from what I understand, with the back two frogs, you KNOW at least 1 is male, but with the front one you don't know if it's male or not. So these are the possibilities for the back two:
MM
MF
FM
There's a 2/6, or 1/3 chance you will get a female. With the front, here are the possibilities:
M
F
Obviously you have a 1/2 chance it will be female. 1/2 > 1/3 so your best bet is to lick the front one.

Did you read my replies to those posts? You assumed incorrectly that those possibilities are all equally likely. They are not. The line frog can't have equal chance of being male or female because that would imply males are as likely to not croak as females. But we know they croak and females don't.

What line frog are you talking about?

It's a, because it's not a question about which card to look at, it's an if then statement about the a card

There's more than one card, try again.

>Being this retarded.
Stay classy Veeky Forums

Wow you're retarded. If it isn't specified then it's safe to assume that there's an equal chance of a male frog croaking regardless of the circumstance. You're putting an unknown condition on the unknown variable only after we solved for the variable with the specified conditions.

It's 7 too. If that card has a vowel the rule is broken.

>I don't want to admit that I'm wrong so I'll just call him a retard.
Okay.

Lone frog

bump

A & 7 you retard

'no'

For the frogs...

I'll use this notation: P(m|dc) = probability that a frog is male given that it didn't croak in the time period we have been standing there.

So we have:
P(m) = P(f) = 0.5
P(dc|f) = 1
P(dc) = P(dc|m)P(m) + P(dc|f)P(f) = 0.5*P(dc|m) + 0.5

For the frogs behind us we look at the probability that they’re both males given that one croaked and one did not. (n = intersect, u = union)

P(m n m | (c n dc) u (dc n c))
= P(m n m | c n dc) + P(m n m | dc n c) --- independent events
= 2*P(m n m | c n dc) --- by symmetry
= 2[P(m|c)P(m|dc)]
= 2[1*P(m|dc)]
= 2*P(m|dc)

So it’s twice as likely that the 2 frogs behind us are males as the one in front is male. Therefore lick the one in front to double survival probability.

Survival probability is:
P(f|dc) = P(dc|f)P(f)/P(dc) --- by Bayes
= 1/(P(dc|m)+1)
...which is >0.5 since P(dc|m)

50% guys, this is easy
It either happens or it doesn't

A, K, and 7.
"If a card has a vowel on one side, it must have an even number on the other side."
A is flipped, because if it has a odd number or letter, the statement is false.
K is flipped over, because if it has a vowel the statement is false.
2 is not flipped, because the converse is not stated.
7 is flipped over, because if it has a vowel the statement is false. (Same reason K is flipped over)

It's A and 7.
Isn't this type of logic overviewed in baby proofs, logic101, discrete math, mathematical reasoning, and probability?

Ive seen this same fucking question in terms of a police line up, icecream sales, and that fucking hats shit. Generic IF THEN

Nice try, but you made a mistake.

Agreed, that's the conclusion I came to as well.

what if the other side of K is a vowel user

>Isn't this type of logic overviewed in baby proofs, logic101, discrete math, mathematical reasoning, and probability?

That's exactly why seeing so many people get it wrong is hilarious. People don't read it, they just assume they know the answer because they've encountered a different variation of the problem before and assume the same answer applies.

F=cure=not(croak)
P(M)=P(F)
M=croak
1frog{?}1frog{?}+1frog{M}
Is the answer: Pick either one because they both contain one frog that has a 1/2 probability of being female.

Nope.

>Turning around you see two frogs of the species where the croak came from.
>You only have enough time to run to the frog in front of you and lick it or to the two frogs behind you.
Is this a trick question? Because there no longer are two frogs behind you because you turned around and now they are in front of you?

Is the whole point of this thing that the croaking frog behind you (which is male) and the one beside it are both more likely to be male and wanting to attract a mate?

All of them except the 2. If any of them contain vowels the rule is not upheld. If the 2 contains anything the rule is upheld so ignore it.

As the problem is stated here, we have two things to verify:
>if vowel then otherside even number
>if odd number then other side not vowel
Therefore we must at least turn over all vowels and odd numbers, A and 7. However, we must also verify this for the opposite faces of the cards. Hence we also turn over K, because K might have a vowel on the other side. We do not need to turn over 2: if there is a vowel on the other side, all conditions are valid; likewise if the other side shows a consonant or odd number.
Again, A,K,7 must be turned.

This problem is often stated, however, with the assumption that every card has a number on one side and a consonant on the other. In this case, we turn over A because the other side might be odd and we turn over 7 because the other side might be a vowel. However, we do not need to turn over K because all numbers are valid on the other side. 2, again, also does not need turning.
Then the anser is A,7.

Would it matter if you had a choice of between licking 8999 male frogs with one frog of unknown gender and licking one frog of unknown gender?

So what is it fag? I'm sick of your question(and these types of questions on Veeky Forums) being reposted.

Wait, so you're actually trying to restate an 'equivalent' to the Monty Hall problem here? I marvel at your stupidity if that's what you actually think you've done then. No it would not make a difference under your conditions.

all of them?

Just because one card obeys the rule we have can't assume the others do

That's my first thought but surely that's too simple.

The frogs in the back and the frog in the front both have an equal probability of containing a female you dipshits. It doesn't make a difference.

No, you retard. If you actually read the question; you'd know that It was clearly stated that you had enough time to lick both frogs if the two frogs were chosen.

Which would make no difference since at least one of them is male, right? Are you implying that there might be another, unseen frog that is actually the male then, given you haven't confirmed by sight that it was one of the two you saw? Then the problem won't have a probability as an answer unless further information is given. Otherwise it's 1/2 for each non male independently.

All except the card with a '2' on it.

No.

No.

No.

Yes. But do you know what the probability is?

>Then the problem won't have a probability as an answer unless further information is given.
This isn't a homework question. It's not about calculating a number, it's about getting there. There is not enough information to come to a specific probability, but there is only one right answer involving a variable.

>Otherwise it's 1/2 for each non male independently.
As I have said many times, it cannot be 1/2 because that would imply that males are as likely to not croak as females.

Jesus, sure is summer in here.

Turn them all over because fuck you.

1 in 2.

>it's about getting there
Then we need more fucking information. How likely is a male frog to not croak in a given situation? I'm not the scientist so I have no bloody clue. If we have one frog on either side then the choice is either "frog A" or "frog B" and whatever the probability they have of being female is, it's the same on both sides given we know at least one of the frogs on the two frog side is male. If you answer something like "well, the frog croaking only tells up that there aren't two female frogs and hence the chances of there being two male frogs are only 1/3 out of the possible options", then you need to start thinking about actual data and not contrived problems that ignore how information actually affects statistical outcomes. If a frog is unlikely to croak in a given instance, then hearing one frog out of two croak lends significant data to the probability of both of them both being male (as two male frogs are more likely to croak once than a single male frog).

Hm. You're right. I guess it is A K and 7. I was convinced it was only A and 7.

>Then we need more fucking information. How likely is a male frog to not croak in a given situation?
No, you need to use a variable. What is this, kindergarten?

>If we have one frog on either side then the choice is either "frog A" or "frog B" and whatever the probability they have of being female is, it's the same on both sides given we know at least one of the frogs on the two frog side is male.
This is the correct answer for the wrong reason. As an example, if you flip two coins and at least one is heads, then the chance that the two coins contain a tails is not the same as the chance of getting tails from flipping a single coin. But the frogs is not an analogous problem. The whole point of this problem is for you to figure out how its different. But you didn't even get that far because you don't understand basic probability.

>If a frog is unlikely to croak in a given instance, then hearing one frog out of two croak lends significant data to the probability of both of them both being male (as two male frogs are more likely to croak once than a single male frog).
This is better reasoning, but not getting at anything useful. If the chance of a male frog croaking while you were listening is greater than 1/2, i.e. male frogs are likely to croak, then it is not true that a pair of frogs is more likely to give a single croak than a male frog and a female frog. Why do you think that is?

> it cannot be 1/2 because that would imply that males are as likely to not croak as females.
So... Would the probability of survival be something similar to:
(T + 1) / (T + 2)
T = amount of time that has passed
Since only a female with a 0 probability of croaking would yield an infinite amount of time without croaking. So as time tends towards infinity, you can still just assume that the probability of a male croaking is infinitesimal so why bother?

There's only one amount of time that passed. You don't have enough time to wait. Try a different variable.

>There's only one amount of time that passed. You don't have enough time to wait.
But I don't know the amount of time passed. I could have been lazy and stood there and it could have been my way of running.
>Try a different variable.
Can I switch that 'T' out with C (the probability that that a male has of croaking) and subtracted 1 from the numerator and denominator. Then it'd be all good?

This is rather simple.

It makes no difference which option you choose. Survival is 50% for either choice.

Anyone saying different is either a total retard, a troll, a cunt, or a fuckwit.

Given that there is exactly a 25% chance that anyone saying different is one of the above, what are the chances that they are also all of the above, and enjoy regular anal sex their mother as well?

>But I don't know the amount of time passed.
You don't need to.

>Can I switch that 'T' out with C (the probability that that a male has of croaking)
that a male has of croaking *while you were listening*

>and subtracted 1 from the numerator and denominator. Then it'd be all good?
No, that would just be nonsense. I just spoonfed you the answer.

Eh. Fuck you and your question. I hope you stay awake at night thinking about how much time and resources you've wasted.

Yeah, take your ball and go home. Take a probability course before pretending to know what you're talking about.

Go back to sitting in your chair at your dead-end job pretending you can still contribute anything significant to humanity with your life.

Ok since you say that the probabilities are equal, I'm guessing the mistake is in one of these 2 lines for the two frogs behind:

P(m n m | (c n dc) u (dc n c))
= P(m n m | c n dc) + P(m n m | dc n c) --- independent events

I just couldn't see why this reasoning was wrong so I stuck with this. Why is it wrong?