Well?

File: 1444931083848.jpg (20KB, 306x306px) Image search: [Google]

20KB, 306x306px

>>8209156
>this bait again

turn K and 7

>>8209156

If A has an odd number on the other side it fails

If the K has an even number on the other side the property is not broken, it is wovels that must have even numbers, not even numbers that must have wovels

If the 2 has a wovel on the other side the property is upheld. If the 2 has a consonant the property is upheld because consonants can have even and odd numbers.

If the 7 has a wovel on the other side the property is not uphold.

Thus A and 7 must be turned.

>>8209163
>>8209166
No

>>8209168
It's actually A and 7, the video said so too

>>8209156
You are a biologist travelling in the rainforest. You are bitten by a deadly venomous snake. You know that the antidote for the venom is secreted by the female of a certain species of frog found in this rainforest. The population of these frogs is split evenly between males and females, and they are visually indistinguishable from each other. You also know that the males have a distinctive croak and the females don't croak. You see a frog of the species in front of you. At the same time, you hear a male croak behind you. 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 and lick them both before you pass out from the venom. Which choice maximizes your chance of survival and what is the probability of survival?

>>8209175
Nope. Try again.

>>8209179
You need to flip K too to make sure there is no vowel on the other side. But the original problem states that each card has a letter on one side and a number on the other so in that case it's A and 7.

>>8209179
Here's your (You)

>>8209156
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.

>>8209168

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

>>8209188
say

>>8209200
pretty close, you just need to change one thing there

>>8209203
Read >>8209204

all except k

>>8209178
>The population of these frogs is split evenly between males and females
>There are only 3 frogs

>>8209303
>The population is only the three frogs you saw in this particular moment
Report to the nearest McDonalds immediately.

>>8209178
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.

>>8209285
But what if K has a vowel on the other side? WHAT THEN?

>>8209343
>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.

>>8209178
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%.

>>8209360
You made the same mistake of assuming that every possible combination is equally likely. There is something you're missing.

>>8209379
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.

>>8209389
I'll give you a hint: the answer is dependent on the chance of males croaking in the time you were listening.

>>8209404
which you never gave us...

>>8209398
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

>>8209406
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

>>8209156
This was easy.. if you would have put a little mind, you will solve it yourself

>>8209406
So it's a variable.

>>8209412
If K has a vowel on the other side then the rule is broken. Jesus how dumb are you people?

>>8209414
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, >>8209360 makes the most sense.

>>8209429
its kinda implicit that one side has a character and the other one has a number

but whatever

>>8209414
>what is algebra

>>8209430
>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, >>8209360 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.

>>8209435
Where is it implied?

>>8209449
in the video this image is from

>>8209454
The question being asked is not in the video, it's in the image in the original post.

>>8209429
Very

>>8209156
A & 2

>>8209445
alright spoonfeed me more senpai

>>8209527
Nope. Figure it out.

>>8209533
no u

>>8209178
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.

>>8209551
If you could actually get it right I would tell you.

>>8209559
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.

>>8209156
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.

>>8209565
If you actually read the thread you would see that I congratulated several people who got the right answer, you rabid ignoramus.

>>8209178
>>8209343
>>8209360
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.

>>8209642
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.

>>8209669
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

>>8209680
There's more than one card, try again.

>>8209682
>Being this retarded.
Stay classy /sci/

>>8209430
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.

>>8209690
It's 7 too. If that card has a vowel the rule is broken.

>>8209690
>I don't want to admit that I'm wrong so I'll just call him a retard.
Okay.

>>8209676
Lone frog

A & 7 you retard

>>8209642
'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)<1

This doesn't take into account the likelihood that a male frog near a male frog may be more/less likely to croak than if it's near a female. I assume that they can tell each other apart with sense other than sight so this should really be another variable.

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

>>8209156
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)

>>8209156
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

>>8211534
Nice try, but you made a mistake.

>>8209346
Agreed, that's the conclusion I came to as well.

>>8211567
what if the other side of K is a vowel anon

>>8211567
>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.

>>8209178
F=cure=not(croak)
P(M)=P(F)
M=croak
1frog{?}<choice>1frog{?}+1frog{M}
Is the answer: Pick either one because they both contain one frog that has a 1/2 probability of being female.

>>8212245
Nope.

>>8212267
>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?

>>8212245
>>8212278
>>8209178

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?

>>8209156
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.

>>8209156
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.

>>8212323
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?

>>8209178
>>8212353
So what is it fag? I'm sick of your question(and these types of questions on /sci/) being reposted.

>>8212353
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.

>>8209156

all of them?

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

>>8212413
That's my first thought but surely that's too simple.

File: image.jpg (72KB, 380x468px) Image search: [Google]

72KB, 380x468px

>>8209178
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.

>>8212412
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.

>>8212419
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.

>>8209156
All except the card with a '2' on it.

>>8212278
No.

>>8212323
No.

>>8212353
No.

>>8212416
Yes. But do you know what the probability is?

>>8212421
>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.

>>8212424
1 in 2.

>>8212424
>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).

>>8211559
Hm. You're right. I guess it is A K and 7. I was convinced it was only A and 7.

>>8212440
>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?

>>8212424
> 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?

>>8212482
There's only one amount of time that passed. You don't have enough time to wait. Try a different variable.

>>8212488
>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.
>>8212482
>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?

>>8212496
>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.

>>8212503
Eh. Fuck you and your question. I hope you stay awake at night thinking about how much time and resources you've wasted.

>>8212508
Yeah, take your ball and go home. Take a probability course before pretending to know what you're talking about.

>>8212518
Go back to sitting in your chair at your dead-end job pretending you can still contribute anything significant to humanity with your life.

>>8211534
>>8212215
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?

Pr(Male croaks in the time interval) = n
Pr(Male) = Pr(Female) = 1/2
Pr(Survival if you choose Lone Frog) = Pr(Female|Not Croak) = Pr(Not Croak|Female) / (Pr(Not Croak|Female) + Pr(Not Croak|Male))
= 1 / (1 + 1 - n)
= 1 / (2 - n)
Pr(Survival if you choose Two Frogs) = same shit

So you can pick either, with survival chance of 1/(2-n) where n is the probability a male croaks in the time interval you were listening.

>>8213018
Can you explain why it'd be the same for two frogs?

>>8213044
Since we heard a croak, there is one frog that croaked and one that didn't. Therefore:

Pr(Survival if you choose Two Frogs) = Pr(Female|Croak) + Pr(Female|Not Croak)
= 0 + Pr(Female|Not Croak)
= 1 / (2 - n) as shown in the case of the lone frog

>>8212834
Yeah I don't really understand your notation but you shouldn't just be adding conditional probabilities like that. You have to multiply them by the relative chance of a conditions occurring, which is 0.5.

>>8213107
Of course, thanks!

So what's the answer?

MM M
MM F
MF M
MF F
FM M
FM F

6 possibilities. The top one is less likely, as we know there are an equal amount of males and females. The other 5 options are of equal likelihood. Since we do not know the total population of frogs on the island, we can't say how less likely this top option is, so we'll have to use a range. Lets first assume there is an essentially infinite amount of frogs on the island, making the top option just as likely as the others, leaving us with 2/3 chance live if you go for the two frogs and 1/2 chance live if single frog.
Now the lowest possibility, 4 frogs on the island. If this is the case, the top option is literally impossible, leaving us with 4/5 if you go with the 2 frogs and 3/5 if you go with the single.
So you should go to the 2 frogs. There is somewhere between a 66.6rpt% and 80% chance of survival there where as the single frog has somewhere between a 50% and 60% chance of survival.

>>8214703
Nope. Try again.

>>8214729
Is this correct?

>>8213018
>>8213048

>>8214735
Yes.

>>8214703
^ I'm this guy

I'll preemptively bring up that I expect you to mention I didn't take into account that males aren't as likely not to croak as females and I'd like to address that but stating it doesn't apply. Since I've already broken up the chances to address that we dont know which of the two frogs is male, we dont need to 'double down' and also break it up by which of the two frogs croaked, as the single frog is just as likely not to croak as the frog that may or may not be male of the duo. If there was more information given as to the odds a male frog croaks in a given time frame for example, or there were more frogs with the single frog and they also didn't croak, then this would change, but as of now, there is one frog on each side that has not croaked, so our percentages will not be affected by this.

>>8214703
Has anyone else addressed that we dont know the population of frogs on the island? I feel like this is a factor that affects the answer.

>>8214743
>Since I've already broken up the chances to address that we dont know which of the two frogs is male, we dont need to 'double down' and also break it up by which of the two frogs croaked, as the single frog is just as likely not to croak as the frog that may or may not be male of the duo.
The pair of frogs is indeed reducible to the single frog, but then why did you write that they give different probabilities of survival? And that doesn't address the fact that the single frog does cannot have a 1/2 chance of being male because that could only be true if males are as likely to not croak as females.

>>8214745
Then let me make the problem less ambiguous: Instead of the population being split evenly, let us say that the chance a random frog is male is 1/2.

>>8214753

The pair of frogs is indeed reducible to the single frog, but then why did you write that they give different probabilities of survival?

"we dont need to 'double down'"

>>that doesn't address the fact that the single frog does cannot have a 1/2 chance of being male because that could only be true if males are as likely to not croak as females.

This statement is true, I should have taken that into account. I know it will make the options with 2 Fs more likely, but I'm unsure how I would do the math for that.

>>8214756
This is not the same as having a split population, but if that is what you intended, yes, that would be the better way to state it. I'll get back to you with a new answer.

>>8214756
>>8214785
Wait, no, actually, you can't have it both ways like that. If the chance that a random frog is male is 1/2 then knowing one of the 3 frogs is male wouldn't affect anything. If I flip 3 coins and know one of them landed on heads that doesnt change the odds that the other two are heads or tails. The way you are saying the problem you *do* want it to be based on population. That *does* need to be addressed here.

Card 1 has non-even on other side: false

Card 1 has even on other side: true

Card 2 has vowel on other side: false

Card 2 has non-vowel on other side: true

Card 3 has vowel on other side: true

Card 3 has non-vowel on other side: true

Card 4 has vowel on other side: false

Card 4 has non-vowel on other side: true

So you need to turn over 1,2,4

>>8209156
Just turn over the cards A and 2. How the fuck is this even hard? It's literally 2 seconds of thought.

>>8214964
lol you're wrong - have to turn over 7 because there could be a vowel on the other side

>>8214964
and you don't have to turn over 2 lmfao you're retarded

>>8214964
lmfao dummy and you have to turn over k because there could be a vowel on the other side

>>8214964
so in short ur retarded lol

>>8209156
A,K,7.

There could be a vowel on the other side of the K card.

>>8209156
??? Just A, right? Wtf?

>>8215311
Isn't it this? Will someone else come along and say something. New to /sci/

>>8215908
No.

>>8209346
^This is the correct answer

>>8214897
I'm not trying to have it both ways, I'm saying that the chance a random frog is male is 1/2.

>If the chance that a random frog is male is 1/2 then knowing one of the 3 frogs is male wouldn't affect anything.
It doesn't.

>If I flip 3 coins and know one of them landed on heads that doesnt change the odds that the other two are heads or tails. The way you are saying the problem you *do* want it to be based on population.
No I'm not. Where do you see this?

If a card has a vowel on one side, it must have an even number on the other side.

If Vowel, then Even.

V -> E
~(V -> E) <-> ~E

therefore, check all cards ~E

>>8209346
>>8216493

But it says "if a card has a vowel on one side, it must have an even number on the other side".

The opposite may or may not be true. A card showing an even number may not have a vowel on the other side. It never said that non-vowels must have an odd number, nor did it say that if an odd number is present on a side then a non-vlwe must be on the other side.

As such, all that's necessary to do is to flip the card that is showering a vowel.

Isn't this true?

>>8216921

read
>>8216911

>>8216926
Not a mathfag, tell me what it means?

Based on the information provided you would have to flip over every single card except the 2. I am arguing under the assumption that any symbol could be opposite any card. For example the A could have another A on the other side, or a house, or a dog, etc..
The logical trick being played is the concept that an original statement and it's converse are not equal.


Thus saying, 'If a vowel is on one side, then there must be an even number on the opposite side." That is not equal to the converse "If there is an even number on one side then there must be a vowel on the other side"
Thus the original statement is only true in the case where a vowel is opposite anything other than a vowel. So we need to make sure that A has an even number opposite and K, 7 have anything other than a vowel opposite.

For the 2, if it is a vowel, then great it supports our statement, and if it isn't then the statement doesn't apply to the card.

>>8215908


You would be correct if the original statement was instead "If the card has a vowel showing on THIS AND ONLY THIS side, it must have an even number on the other side"

>>8216940
It means that you should learn more math.

>>8216940
If then statements are always true except when the if is true and the then is false.

So we have to look for cards where there could be or is a vowel but not an even.

In this case, we look at a,k,7.
If a does not have an even, the proposition is false.
If k does have a vowel, the proposition is false
If 7 does have a vowel, the proposition is false.

In general, if then statements are not reflexive. If a then b does not imply if b then a.

This is why we don't worry about the 2 because regardless of what is on the other side, the proposition is always true.

If it was stated that each card had a letter and a number, we wouldn't have to check the k for the same reason.

>>8216954
But that doesn't make sense.

A is the only vowel here, and it is the simplest way to determine whether or not the rule has been followed. If yes, the rule MAY BE followed. If not, the rule is broken (conclusive).

However, the statement doesn't say anything about odd numbers or non-vowels, so flipping them over achieves nothing. What is on the other side of the card K and 7 doesn't matter, as it does not contradict nor prove the statement.

As for 2, there's no need to flip it either, because the statement must also imply the converse -that having an even number on one side MUST mean that it has a vowel on the other side- before you justify flipping the card. Since it doesn't, it may have whatever written on the other side. It doesn't matter as it wouldn't flout the rule no matter what is written on the other side.

>>8216960
What is on the other side of k,7 can contradict the statement, which is why we look.

See my post above yours

>>8216960
If there is a vowel on the opposite side from 7, then the rule is broken. Thus, you need to check.

>>8216963
I see now. Thank you.

Btw, are you the same person who posted >>8216911 ?

>>8216968
Nope

but if then statements are written mathematically with the -> symbol which means implies.
The ~ means negation
So you can parse the statement and it comes out the same.

>>8209178
Monty Hall: Rainforest edition

>>8209156
A&7

if A is odd then it's false
if 7 is vowel then it's false
and vise versa

>>8216940

In Propositional Logic(1), statements(2) can be replaced with statement variables(3), so the statement from the puzzle, "If a card has a vowel on one side, it must have an even number on the other side" can be represented using the following interpretation(4).

Statement:

"If a card has a vowel on one side, it must have an even number on the other side"

Symbolization key:

V = The card has a vowel on one side
E = The card has an even number on the other side

Symbolization:

V → E

The puzzle then asks us, "Which card(s) must be turned over to determine whether or not the rule has been followed"

This is another way of saying, "Which cards will make the statement V → E false?"

In order to know whether V → E is false, we only need to know when the statement of the form A → B can be false. Since we're using propositional logic, each statement can either be true, or false. Since there are two variables, V and E, there can only be 4 unique pairs of truth values; (1,1), (1,0), (0,1), and (0, 0). In Propositional Logic, there are five connectives, and the connective we are using is the conditional, "→". Between any two statements, simple or compound, a conditional statement is only(5) considered false if the consequent (the statement on the right-side of the arrow) is false. This means we only need to look for a card which is not even, since V → E can only be false if E is false(6).
(1) also called "Sentential Logic" or "Sentential Calculus", meaning literally "Logic of Sentences" or "Calculus of Sentences"

(2) also called 'sentences' or 'propositions'

(3) also called 'sentence variables' or 'propositional variables'

(4) interpretation keys allow us to assign, or, equate sentences from one language to another, not always natural languages such as English, but also formal languages as well

(5) "~(V -> E) <-> ~E"

(6) E is false can be represented by "~E"

There's a lot more I missed but I only get 2000 characters sorry I hope this was helpful enough.

>>8216911
Your logical equivalency doesn't hold.

V->E would yield ~E -> ~V.

You can't switch to biconditional unless you prove both conditions.

>>8216999
You're almost there, but the truth table is only used to demonstrate how you would treat the overall sentential connective in a conditional.

But there is an objectively demonstrable answer to this one, provided you are using the same system of logic:

Let V = a card has a vowel on one side;
Let E = (that same card) has an even number on the other side.

V -> E
~E -> ~V

Are the logical equivalencies that we can draw from this.

So to check V -> E you have to look at the A and see if it has an even number.

To check ~E -> ~V you have to look at the 7 to see that it *doesn't* have a vowel on it.

The "trick" here is that we also have to check the K, because it qualifies as ~E, being not an even number, and we have to make sure it doesn't have a vowel on it either.

We're so trained to see letters and numbers as two utterly different categories that we forget that with this puzzle.

I wonder if Hebrew students would have an easier time of it? Would the puzzle even work in Hebrew?

>>8209156
Oh my god this thread again.
All except 2 must be turned.
Stop posting this crap.

>>8212234
>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.
It's every card except 2 you imbecile. Obviously you draw just as premature conclusions as anons you criticize.

>>8218164
>It's every card except 2 you imbecile.

I know, I was making fun of the guy I was responding to for getting it wrong.

Each card until you encounter an error, or have turned over all cards.

>>8218914
Never mind. It doesn't have to not have an even number.

All cards except K.

>>8218922
Wrong. You have to turn over k.

Why bother shitposting if you put no effort into it?

>>8219078
You don't have to turn over K. The system is simply that a vowel must be paired with an even number, not that an even number must be paired with a vowel.

>>8219214
>If a card has a vowel on one side, it must have an even number on the other side.
You need to flip K to ensure that there isn't a vowel on the other side which would break the rule.

>All cards except K
Flipping 2 is actually unnecessary since the rule's only requiring that cards with vowels must have an even number on it's other side. So it doesn't matter what's behind 2. As for A and 7, those indeed need to be flipped over.

All cards except 2 should be flipped.

>>8219227
>You need to flip K to ensure that there isn't a vowel on the other side which would break the rule.
True. I had assumed it was always a letter number pair, but this need not be the case.

File: congress_plastic_playing_cards.jpg (79KB, 450x350px) Image search: [Google]

79KB, 450x350px

>>8209156
None of them. The rule has not been followed. Have you tards never seen the back of a playing card before?

>>8219214
i'm probably being trolled but

imagine the K card was K/A. then it it has a vowel on one side but no even number on the other side. However if it was K/5, for example, there's no issue. So you have to check it.

>>8219253
Show me playing cards that resemble ops pic

>>8219257
You're not being trolled. The thread has shown people are actually that dumb.

>>8219257
I've addressed this.
>>8219228

File: imp.png (3KB, 322x96px) Image search: [Google]

3KB, 322x96px

>>8217409
>>8217443

I replied to these but then fucked it up again. Oh well

Meant to write

¬(V→E) ↔ ( V ∧ (¬E))

which does hold and here is a nice truth table to accompany it.

>>8209156
A, K and 7 with the specified rules.

A and 7 if there's a number on one side and letter on the other side.

File: unnamed (6).gif (1MB, 256x192px) Image search: [Google]

1MB, 256x192px

>>8209156
all of them
these are like pattern questions
every answer is right in pattern questions

>>8209156
From my point of view, the rule doesn't specify that the card saying 2 necessarily has to have a vowel on the other side, it might be a complete coincident having an even number assigned to some other letter. The same goes for 'K' and '7' ( vice versa, obviously).

The rule only specifies that for every vowel, there is an even number assigned on the other side of the card. So turning around 'A' might give you the answer whether or not the rule is valid or not.

>>8212424
>Yes. But do you know what the probability is?
Why should I care? The question was what I should do, and the answer is, it doesn't matter which frog(s) I lick.

Trying to shift the goalposts on us?

>>8219718
>Trying to shift the goalposts on us?

Exactly my problem with this. He admits the problem is 50/50, but wants us to mode it in the specific way he happens to want us to for whatever reason. Like I said before, the issue I take is that there are plenty of different things you could choose to factor in. I could model the average time between when a given frog croaks and when it is next likely to croak, but I don't know if the scientist can estimate with this information. I assume he wants us to perform Bayesian reasoning along with the assumption that we can model the chance of a given (male) frog croaking as an independent event.

So we can do things like, say 'A' is the chance of a frog croaking per period. Given the chances of a frog being male is independently 1/2, we can say that the chances of two frogs both being male, given we hear one croak during a period, is (1 - A)/(2 - A).

>>8209156
Oh, I have seen this video during my HATE PSYCHOLOGY days.

https://www.youtube.com/watch?v=t7NE7apn-PA