/sci/ - Science & Math

[Bayesian Conspirator !t8Lgs/NqoU No.2073741]

Its time for yet another troll probability thread.

Your friend rolls three dice and doesn't let you look. You ask him if he rolled at least one four. What is the probability that he rolled the same number for all three dice, given that a) he answers 'no' to your question and b) he answers 'yes'.

Pic unrelated.

[Anonymous No.2073777]

a) 1/125
b) 1/(216-125) = 1/91

[Anonymous No.2073782]

>>2073777

Oops, I meant 5/125, so 1/25.

[Anonymous No.2073805]

I flipped two coins. At least one of them is a head. What's the probability they're both heads?

> mfw 100 replies later, people still don't understand why it's 1/3

[Anonymous No.2075556]

Let A denote the event that at least one die is 4.
Let B denote the event that all three dice are equal valued.
case no:
P(B|~A) = P(~A|B)*P(B)/P(~A)
1/2 * 1/36 / 1/8 = 1/9
case yes:
P(B|A) = P(A|B)*P(B)/P(A)
1/2 * 1/36 / 7/8 = 1/63

[Anonymous No.2075561]

>>2075556
ah crap, (1/6)^3 is 216, not 36.....
1/9 -> 1/54
1/63 -> 1/378

[Anonymous No.2075563]

>>2073777
>>2073782
This.

[Anonymous No.2075587]

>did you roll at least one four?
No:
So, the dice are all 1-3 or 5-6.
125 outcomes, of which 5 are triples.
1 in 25.
Yes:
So at least one dice is a 4.
216 outcomes. 125 of which are not viable.
6 possible trips, 5 of which are not viable.
1 in 91.

>flip two coins. At least one is heads. What's the probability of both being heads?
4 outcomes, 1 of which is eliminated by the condition.
3 viable outcomes
one is double heads
1 in 3.

>>2075556
>>2075561
Dice aren't ordered.

[Anonymous No.2075589]

>>2075561
ahhh, more crap
I need to learn how to read.
>at least one four
>at least one die with at least one four

[Anonymous No.2075595]

>he says no

1/25

>he says yes

1/36

[Anonymous No.2075603]

A: Rolled a 4
B: Rolled trips

>P(B|~A) = P(~A|B)*P(B)/P(~A)
Probability of "not A" given B is 5/6.
Probability of "not A" is 125/216.
Probability of B is 1/36.
1/25. So yes, the formula works. You just fail at something.

>P(B|A) = P(A|B)*P(B)/P(A)
Probability of A given B is 1/6
Probability of B is 1/36
Probability of A is 91/216.
Making 1/91. Formula still works, and you once again did something bad.

[Anonymous No.2075617]

>>2075603
yes, I mentioned, I thought it was "one diceat least 4", instead of "at least one dice 4"

[Anonymous No.2076269]

>>2073741
In the absence of any additional information, the answers are:
a) 1/6
b) 1/6

To get a different answer, you would need to know the probability of a particular answer being given for each combination. E.g. if he always answers "yes" if he rolls three sixes and always answers "no" for any other combination, then the answers would be a) 0 and b) 1.

Adding additional "facts" in meaningless without the associated conditional probability matrix. That's why questions such as these are essentially trolling; If you posted them to a probability forum, your post would just get deleted with no comment beyond "read the FAQ".

[Anonymous No.2076500]

>>2076269
I don't see issue, other than taking into the account that he might be flat out lying?
Like that, you would never be able to have a word problem, since you would need to start assuming strange things about the dies (maybe they only had 3 sides? maybe statement in the question itself is a lie!) and

Besides, it wouldn't be 1/6 either way (but 1/36).