Because MATH, that's why.

[theweaselking]

You roll two fair 6-sided dice.

One of them is a 6.

What are the odds that the other is also a six?

[Poll #1770498]

Comments

[tsunami-ryuu.livejournal.com]

Ahhh, tricksy question. I answered incorrectly because I didn't think that one through. It's all in the wording.

[theweaselking.livejournal.com]

Seriously, people?

[dianavilliers.livejournal.com]

1/11.

[silmaril.livejournal.com]

1/11.

(This is how tired I am: I had to actually write out the complete set, delete everything that didn't have a six in it, and count. I was left with {(6,1),(6,2),(6,3),(6,4),(6,5),(1,6),(2,6),(3,6),(4,6),(5,6),(6,6)}, which I rewrote here and recounted just to make sure my count was right. I think if you were to pull out the flaying knives now I might be able to do the formal derivation, but not otherwise. Tiiiiiiired.)

[a1057soul.livejournal.com]

depending on how you read the question... there are 11/36 possible combinations where a 6 is present: 1-5,6 and 6,1-6.

The question asks if one of the die is a 6 and what is the probability that the other is ALSO a six... this to me indicates BOTH sixes, which has a probability of 1/36 based on each having 6 outcomes on two dies... assuming of course 'fair sided' means numbering from 1-6 (this isn't stated).

The probability of rolling a 2nd six after identifying the first is 1/6, but the probability of rolling 2 sixes is 1/36.

But then I generally do my math like this (http://www.giantitp.com/comics/oots0034.html).

:o.
Dan

[crazy-alexy.livejournal.com]

This shit is why I hate math. -.-

[ipslore.livejournal.com]

Okay, there's four possibilities:
1) *The one that's a six is a six, and the other one is a six
2) *The one that's a six is a six, and the other one is not a six
3) *The one that's a six is not a six, and the other one is a six
4) *The one that's a six is not a six, and the other one is not a six.

Clearly, we can eliminate 3) and 4). That leaves us with a 1 in 6 probability that the die other than the die that's a six is also a six.

(alternate, smart-ass answer: the odds are 0, because 'one of them' means 'exactly one of them'.)

[kafziel.livejournal.com]

Without reading other comments: is it 1/11?

[dreamshade.livejournal.com]

Bah, I understood the general idea because I'd already heard the "one of my children is a boy, what are the chances that the other is" puzzle, but then I went and read the comments before doing the exact math and spoiled it for myself.

[dscotton.livejournal.com]

I got 1/11. 11/36 die combinations include a six. 1 of those consists of two sixes.

[krinndnz.livejournal.com]

Yes, questions are harder to answer when they're worded badly. Every time you post this (e.g. the raptors) I think you're an asshole.

[pappy-legba.livejournal.com]

Livejournaler confuses poor wording for cleverness, film at 11.

[ali-in-london.livejournal.com]

[x] Some initial answer, but then reading the comments convinced me I was wrong.

[tempter.livejournal.com]

I'm curious what you think the answer is.

[fridgepunk.livejournal.com]

1/6 because the die that came up six is loaded and always comes up 6.

Problem?

[heavenscalyx.livejournal.com]

More evidence that I still have PTSD and memory gaps from my horrible required statistics course 20 years ago.

[thornae.livejournal.com]

If I'm rolling them? 1.


(Actually, I made the same table as dianavilliers in my head, and got the same answer. But I've been watching a lot of The Mentalist lately...)

[cavalorn.livejournal.com]

Kicking myself now for misreading the question.

I'd seen it as an issue of 'two dice rolled under separate cups - you look at one of them, see it is a six, and therefore the die under the other cup has a one in six chance of being a six.'

But that's not the question you were asking at all, I see.

In fact, it's much more like this:

Your GM rolls 2d6 behind his GM's screen. He looks at them and says 'Well, one of these is a 6.' You then have to figure out the odds of the other one being a 6 too.

[littlebus.livejournal.com]

The question is too ambiguously worded to be a useful probability question, but is a good semantics riddle.