Math question

Is this how you people have fun

Bayes theorem
Are you still in school op?

>>76204156
1/2 if you disagree your mother is a cheap streetwhore

>>76204156
This is a trick question and the answer is, IT DEPENDS WHAT CARD YOU START WITH AS FIRST DRAW AND SECOND, so you can get either 1/2 or another 1/3. Enjoy.

>>76204312
I'm 23.

If you draw cards at random:
1/3 of the time, you'll grab the red card.
1/3 of the time, you'll grab the blue card.
1/6 of the time, you'll grab the mixed card and see its red side.
1/6 of the time, you'll grab the mixed card and see its blue side.

You're discarding any occurrence where you see a blue side. This leaves you with:
1/3 of the time, you'll grab the red card.
1/6 of the time, you'll grab the mixed card and see its red side.

You're picking the red card twice as much, so this gets normalized to 2/3 chances you get the red card, 1/3 chances you pick the mixed card. So the odds the other side is blue are 1/3.

>You're picking the red card twice as much, so this gets normalized to 2/3 chances

What

>>76204694
The sum of all probabilities must yield 1=100%. But since you're discarding half of them (because you saw a blue side), you need to "normalize" the shit by multiplying everything by two, so the sum is 100% again.

>>76204363
Hmm, what is your answer?

>>76204156
The graphic is deceiving:

It makes it look like you always pick the cards with that side up like in the picture, in the actual question however it is clear that when you pick a card either side can be up.

So, there are not 3 possibilities when picking a random card, but actually 6. (Each card can have 2 sides up). Of these 6 possibilities 3 make end up with you looking at a red card. Of these 3 possibilities only 1 has a blue side on the other side.

So the answer is 1/3, not 1/2 like it looks like in the picture.

Great addition to my /v/ bait folder. Thank you Nippon-kun.

>>76205312
>/v/ bait folder.
And people still wonder why /v/ is /small b/ in Spanish...

>>76204156

This sounds Monty Hall-ish, but I think there's something in the telling that may change it.

There are three red sides in play. The blue card can just be discarded at this point. You've just locked in one red, so now 2 of 3 remaining possibilities are also red.

2/3?

>>76206642
sorry, just woke up.

1/3 that it's blue.

>>76206682
good job pajeet.

now please use that sheet of paper to wipe your ass

File: IMG_20170618_182305.jpg (142KB, 1231x832px) Image search: [Google]

142KB, 1231x832px

Answer is 1/3

>>76206814
right-o

>>76206751
No. I have Bidet. It was just a useless page lying on by desk.

>>76204156

P(rr)=1/3
P(bb)=1/3
P(rb)=1/3

P(r)=0.5
P(b)=0.5

P(b|r)=p(rb)/p(r)=1/6

Where is my cookie?

Bertrand's Box Parado
/thread

>>76207087 (You) #
>P(b|r)=p(rb)/p(r)=(1/3)/0.5

Whoops its (1/3)/0.5=2/3 not 1/6

P(b|r)=2/3

lmao just open the box and see what card you want nerds

>>76206814
You missed the alternative route too.

I'm not sure if this is advanced trolling or if people legit have no idea how vhance works.
The box could have 1000 of the two blue sides card and it wouldn't matter because we are calculating a chance with a precondition.
When calculating with a precondition you divide the chance of the outcome with the chance of the precondition. In this case the chance of the precondition is 2/3 card and every card has an outcome chance of 1/3, so obviously you have a 1/2 chance to select any card that's red on the front. From there it's obvious that the other side being blue is also 1/2.