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/sci/ - Science & Math

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9469658 No.9469658 [Reply] [Original]

Let's see if /sci/ can solve this.
You have two coins. One coin has A A on their sides and other coin has A B on their sides.
You pick a coin at random and flip it. The upper side is A, what is the probability that bottom side is also A?

>> No.9469683

Two thirds

>> No.9469719

1/2

>> No.9469761

2/3

>> No.9469766

>>9469683
>>9469761
brainlets

>>9469719
correct answer

>> No.9469775
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9469775

>>9469658
2/3

>> No.9469781
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9469781

>>9469766
Do you even probability

>> No.9469795

>>9469781
Brainlet

>> No.9469797
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9469797

It's 50/50, either it happens or it doesn't

>> No.9469802

1/2

>> No.9469807

>>9469775
God damn Liar Game was a good manga.

>> No.9469824

>>9469658

Probability of both sides being A = 1/2
Probability of A on one side = 3/4

Therefore probability of bottom being A given top is A is (1/2)/(3/4) = 2/3

>> No.9469825

>>9469719
>>9469781
>>9469797
>>9469802

Too many brainlets

>> No.9469829

>>9469658
3/4

>> No.9469830

>>9469829
I mean, 1/2.

>> No.9469848

>>9469658
we had a thread just like this a few days ago except it was with balls. Answer is 2/3.
it's obvious if you think about actually running an experiment where you flip the coins a bunch of times. you will see 1-A1, 1-A2, 2A, and 2B, in equal proportion. which means there is a 50% chance that you will have an A and see an A on the other side, a 25% chance that you will see an A and have a B on the other side, and a 25% chance of having a B with an A on the other side.
Since we eliminate seeing the B initially, there is left the 50% chance of having AflipA and the 25 % chance of having the AflipB, which is a 50/75 proportion or 2/3.

>> No.9469861

>>9469658
Godfuckingdammit, can you stop baiting these "1/2" retards?

https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

>> No.9469880
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9469880

>>9469861
>i read it on wikipedia so this must be right

>> No.9470048

>>9469861
We only have 2 coins here anon.

>> No.9470080

>>9469848
Any of these problems can be split in two parts: before and after picking the item.

Let's say, what is the probability of you being born? Pretty low if we calculate it. But it doesn't matter, because you have been born already.
Should we sum your probability of being born with any of your other probabilistic actions?

The A coin is already has been born. The choice was made.
Now there is lying a coin with A face, and the is another coin in the box. That's the only things you need to consider. Now I'm asking you: one of these coins has AA sides, and other coin has AB sides, what is the probability that the coin you see is AA coin?

>> No.9470083

-1/12

>> No.9470085

>>9470080
Wait, we can't see the other coin? That changes a lot.

>> No.9470088

>>9469658
If you're going to dio the same problem again, you should have made it more obvious this time.

>You have two, 20-sided die. One is a standard 20- sided die, the other has 20 on all of its faces. You pick a site at random and roll it. It shows 20. What is the probability that you rolled the die all 20 die?

>> No.9470096

>>9469830
It either happens it it doesn't, amiright.

>> No.9470108

>>9470085
Actually, you can't see either coin

>> No.9470110

>>9470108
Okay, well that changes a lot. I change my answer to 3/4

>> No.9470113

>>9470110
Sorry, 2/3*

>> No.9471231

>>9470083
Close, but the question says we have finite coins.