Well?

Well?

2/3 brainlet

...

You are a zoologist traveling through a rainforest when you are bitten by a venomous snake. You know the venom will quickly kill you unless you can find the antidote secreted on the back of the female of a certain species of frog. You also know that males and females of this species are in equal proportion and are indistinguishable except for the fact that the males have a distibctive croak. You see one such frog in front of you, and at the same time hear a single distinctive croak behind you. You turn around to see two frogs of the species next to each other. You only have enough time to run to the lone frog and lick it's back or to the pair of frogs and lick their backs. Which do you choose and what is your chance of survival?

A man hires three prostitutes and wants to have sex with all three of them. They all might have different sexually transmitted diseases and they all want to use condoms. Unluckily, they have only two condoms. Plus, they are in the forest and can’t buy new condoms. Can the man have sex with all three of the women without danger to any of the four?

What a stupid fucking scenario -- I'll humor this question for the fun of it.

Chance of survival:

50 50
first it's a coin and second it either happens or it doesn't

50/50
It's a trick question

Makes no difference. 50/50 either way.

What you're insinuating here is that she is more likely to take the trick coin than the normal coin from her pocket. Those odds are 50/50 therefore if she flips heads there's still 50/50 chance of it being the trick coin

No, cant fuck two women with one condom cuz might spread one females std to the other

Does the man potentially have any STDs? If not
>put on condom
>fuck #1
>change condom, but keep old one
>fuck #2
>invert condom #1 over condom #2, fuck #3

If the guy potentially has STDs I'm lost though.

No, she is more likely to have taken the trick coin if what she flipped was heads.

If she flipped tails, it would be a 0% chance. If she flips heads, there are 3 possibilities: side 1 of trick coin, side 2 of trick coin, heads of normal coin. So given she flips heads, there's a 2/3 chance it was the trick coin.

2/3 of course.

Ika's coin: {Heads (A), Heads (B)}
Normal coin: {Heads (C), Tails (D))

Now, when she flips "Heads", you either have case A, B, or C. Thus, you have 2/3 probability to have either A or B.

This is what I learned probability theory for babyyy. Let T be trick (as in the probability she picks the trick coin) and H be head (as in the probability she lands heads.

[math] P(T | H) = \frac{P(T \cap H)}{P(H)} [/math]

[math] P(H) = \frac{3}{4} [/math] and [math] P(T \cap H) = \frac{2}{4} [/math] so the answer is [math] \frac{2}{3} [/math]

Nice answer, shame you can't explain it.

It's 2/3. There are three "heads" in her pocket, 2 of which are on the trick coin.

>you're insinuating here is that she is more likely to take the trick coin
Nope, half the non-trick-coin cases are eliminated by the statement "if Eiko lands a head", while none of the trick coin cases are eliminated.

>Btfo of The Monty Hall Problem

The post.

Fuck two prostitutes with the two condoms, turn one inside out and place it over the other and fuck the third. Then marry three prostitute and raise your child together.

It's not 50/50. Try again.

How is flipping a coin going to help her solve the problem?

Kek
Maybe it was something like "will this ticket win the lottery?" question where there are 2 choices, yes or no, so the chance is 50-50. So if she flips a coin and it lands on what she bet, it gets added or subtracted to the 50, making it 100 or 0 respectively. Of course, if it was the double coin, the probability would be 100, added to 50 is 150, creating a stack overflow and returning the chance back to 0.
Man, maths are so cool

Why did you dig up this old meme image to post here

You are all dumb.
There are two coins in her pocket.
Naturally the probability of drawing either of the coins is 1/2.

2/3

Consider the case where one coin had two heads and the other had two tails. Then consider suicide.

Let D = double headed coin and H = got a head

P(D|H) = P(H|D)P(D)/(P(H|D)P(D) + P(H|~D)P(~D)) = .5/((1*.5)+(.5^2)) = 2/3