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1/11

We can solve this problem by writing out every single outcome of two dice rolls, but we can simplify the sample space because it's given that at least one dice is a 6. Therefore our remaining options are:
1,6
2,6
3,6
4,6
5,6
6,1
6,2
6,3
6,4
6,5
6,6
out of these 11 total possible outcomes, only 1 outcome has 2 sixes. Therefore the probability is 1/11.

1 in 6

Either it will or it won't

clearly 1/6

But what if the person decided beforehand to tell you "at least one of the dice is {whatever the 1st dice rolled}", and in this case it just so happened that the 1st dice rolled a 6?

1/6
One dice will show a six, so the chance for 1 six to occur is 1. The chance for the second six to occur is 1 * 1/6 since a dice has six sides.

A red and a green die are rolled. The red comes up 6. What is the probability that the green is a 6?

verry true, but these partitions are not of same size

testing atm, so far 100% chance

will report again when more data collected

The difference from this and the doors variant is that you always adjust one of the dices to a 6 (the one you show and remove). It's different.

You're basically knocking its probability out of the game and its no different than a paper with a "6" on it with no other values.

1:6

Then that changes the sample space because the dice are distinguishable. Let's write out ever possibility in the form of a table.
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
For the outcomes in the first row, he would have said "at least one of the dice is a 1" The question still remains what is the probability the remaining die is a 6? in this case the answer is 1/6. Similarly in the second row and so on, the answer is 1/6. Each row has a 1/6 chance of occurring, so (1/6)*(1/6) is the chance of any one situation happening, so 6*(1/6)*(1/6) is the total chance of the remaining dice showing a 6. When the dice are distinguishable, the answer is usually what intuition tells us (1/6). When the dice are indistinguishable they answer is counterintuitive

Looks like a spin off of the Monty Hall problem or the coin flipping problem.

I like this solution. I was thinking of the problem in terms of directly calculating the binomial distributions and going from there. Although the way we go about getting to the solution isn't really all that different.

[eqn]Pr(n, k, p) = \binom n k p^k(1-p)^{n-k}[/eqn]

Where the pmf would look like:

[eqn]Pr(k) = \binom 2 k (1/6)^k(1-1/6)^{2-k}[/eqn]

[eqn]Pr(0) = 25/36[/eqn]
[eqn]Pr(1) = 10/36[/eqn]
[eqn]Pr(2) = 1/36[/eqn]

Exclude the [math]k = 0[/math] outcome because at least one die is a 6. The renormalization factor is:

[eqn]a*(Pr(1)+Pr(2))=1[/eqn]
[eqn]a=\frac{1}{Pr(1)+Pr(2)}[/eqn]
[eqn]a=\frac{36}{11}[/eqn]

If you renormalize the probabilities:

[eqn]P(k=1) = Pr(1)*36/11 = 10/11[/eqn]
[eqn]P(k=2) = Pr(2)*36/11 = 1/11[/eqn]

So the probability the second die is 6 if at least one die is 6 is [math]\frac{1}{11}[/math].

50%, either it shows 6 or it doesn't

But doesn't the statement "at least one of the dice is a six" become meaningless once a die showing six is removed? It seems to me that the appearance of that die satisfies the "at least one six" requirement, rendering it irrelevant. The constraint does not appear to pertain to the behavior of the remaining die. The probability BEFORE the six-die is reavealed, however, that both dice show six, would then be 1/11.

Great post

Do we need this apparatus? Or can't we just say:

1. We are looking at the outcomes which contain at least one six.

2. This constraint renders the dice dependent on each other.

3. Still, all outcomes within the sample space have equal probability.

Then, we can subdivide the sample space, into those outcomes which contain only one six, and those which contain two.

There are of course 2*5=10 outcomes in the first camp:

6/1, 6/2, 6/3, 6/4, 6/5, 1/6, 2/6, 3/6, 4/6, 5/6

and in the second camp, there is only one:

6/6

10+1 = ...

1/6 the story does not tell dick on the outcome of the unknown mystery dice.

-1/12

1/6. This is just a roundabout way of asking "what are the odds a random dice roll will roll a 6?"

Have you learned nothing from the Monty Hall problem, people?

The probability can depend on the "story", because it can actually contain information on which the outcome is conditioned.

No, it really isn't.

Ya'll niggas just did OPs homework.

This isnt a version of the monty hall problem

That's correct. But the Monty Hall problem should have taught these people not to discard the possibility that there's some hidden conditional probabilities going on in a "story" like that until they've worked it out themselves.

It's 1/6. Taking away a dice showing 6 gives you more information than being told at least one is 6, because it shows you which one is 6. Being told at least one is 6 is therefore irrelevant.

The story can do that. In this case it does not

1/6 and 1/11 could both be correct, depending on your interpretation of the ambiguous phrasing in this question.

If you interpret the question to mean:

"What proportion of all outcomes which contain at least one six will in fact contain two sixes?"

Then your answer is 1/11.

But the last sentence suggests that the question could also mean:

"Among the possible outcomes of the second die's roll, the fIrst die having already provided a six, what is the proportion of outcomes wherein the second die is also a six?"

Then 1/6 would be correct.

The confusion lies in whether or not we believe that the constraint on the second die is "broken" when the first die meets the six requirement. If not, then there is a dependence between the rolls, and we can't view the second roll in probabilistic isolation.

TL;DR: >>>/phil/

It's to mess around with the math.

The second interpretation assumes too much. It says you are shown the remaining die. It doesn't say you are shown the die that was taken away/which die it was. It could be either one.

kek

Oh ok

You're probably right. Still, I feel that the question is formulated like a riddle, in such a way that makes it appear subtler than it is.

It says that a die showing 6 is removed so clearly its shown to you

Well no, it doesn't say it's shown to you, just that it happens. The question clearly delineates between what is told or shown to you and what actually happens. But either way, just showing it to you doesn't tell you which one it is. There are 11 different, equally likely ways the dice can be rolled and a 6 is shown to you. 1 of those ways results in a 6 remaining.

you're right about this

>Have you learned nothing from the Monty Hall problem, people?

women have no place in math?

Yeah that actually makes sense, i'll agree with this

samefag

...

Fraid not

Assuming the speaker will say "at least one of the dice is a 6" every time if the roll is 66 and half the time if the roll is a 6 and some other number, then the answer is 1/6. That seems like the most likely behaviour to me, but if you assume the speaker will say "at least one of the dice is a 6" every time if the roll has at least one 6 then the answer is 1/11.

Cont'd

That's actually also assuming that the dice that is removed will always be a 6. If you assume who/whatever chose the dice to remove did it at random then the answers are 1/3.5 and 1/6 respectively.

>Assuming the speaker will say "at least one of the dice is a 6" every time if the roll is 66 and half the time if the roll is a 6 and some other number, then the answer is 1/6.
>every time if the roll is 66 and half the time if the roll is a 6 and some other number
Why would they do this? Two dice are rolled, if one of them shows a 6 the speaker will say so. As stated earlier in the thread there are 11 different possible outcomes. Why would the speaker not say there is a 6 if there is one and why is that the most likely behavior?

If the dice they removed isn't always the six, the probabilities can be split into two: The situation where the speaker removes the first die and the situation where the speaker removed the second die. Each of the 11 possibilities can have these two outcomes, and below I have tallied up whether or not there is a six left after removing the first or second die.
1,6-y,n
2,6-y,n
3,6-y,n
4,6-y,n
5,6-y,n
6,1-n,y
6,2-n,y
6,3-n,y
6,4-n,y
6,5-n,y
6,6-y,y
There are 12 situations where the remaining die shows a 6, out of a total 22 situations. Meaning the probability of finding a 6 on the remaining die is 12/22 or 54%
12/22

there is no "hidden" information. the probability being 1/11 instead of 1/6 stems from the fact that getting two sixes is twice as unlikely as getting any other number and a six

>Why would they do this? Two dice are rolled, if one of them shows a 6 the speaker will say so. As stated earlier in the thread there are 11 different possible outcomes. Why would the speaker not say there is a 6 if there is one and why is that the most likely behavior?

Suppose the roll is a six and a five. I would expect a typical speaker to be equally likely to say "at least one of the dice is a 6" as they are to say "at least one of the dice is a 5". The question does not tell you that sixes are special to the speaker and they will always report on them.

>If the dice they removed isn't always the six, the probabilities can be split into two: The situation where the speaker removes the first die and the situation where the speaker removed the second die. Each of the 11 possibilities can have these two outcomes, and below I have tallied up whether or not there is a six left after removing the first or second die.

Hmm. I'm quite sure this doesn't work, but my brain isn't coming up with why. I got my answer by just calculating the conditional probability:

Let E represent "you are told that at least one of the dice is a 6 and then a 6 dice is removed"
Let D represent "the remaining die is a 6"

P(D) = 1/36

P(E) = 1*(1/36) + (1/2)*(10/36)
= 1/36 + 10/72
= 12/72

P(D|E) = P(E|D)*P(D)/P(E)
= 1*(1/36)/(12/72)
= 72/432
= 1/6

I'm pretty sure this is correct because I modelled it with a short program: repl.it/DcJA/0

>getting two sixes is twice as unlikely as getting any other number and a six
That depends on what you assume about the behaviour of the speaker/dice remover.

isn't it true regardless of the problem? a 1 and a 6 you can get as (1, 6) or (6, 1) while a 6 and a 6 is only possible as (6, 6)

>I would expect a typical speaker to be equally likely to say "at least one of the dice is a 6" as they are to say "at least one of the dice is a 5". The question does not tell you that sixes are special to the speaker and they will always report on them.
You're assuming too much. The question only states "you are told that at least one of the dice is a 6." The question does not go into the behavior of the speaker, but that doesn't matter. The question says you're told at least one of the dice is a 6, you should ignore all other cases because they are not mentioned in the question.

>Hmm. I'm quite sure this doesn't work, but my brain isn't coming up with why.
Maybe that's because you aren't using your brain and you're using the formula instead.
For example, why don't you explain to me where you are getting P(E) from?
>Let E represent "you are told that at least one of the dice is a 6 and then a 6 dice is removed"
>P(E) = 1*(1/36) + (1/2)*(10/36)
I am not seeing how you get from the first line to the second.
You also never showed why P(E|D)=1? You just seem to be all over the place, and it's hard to pinpoint your mistakes when you forgo showing your reasoning.

You're correct. My mistake.

>You're assuming too much. The question only states "you are told that at least one of the dice is a 6." The question does not go into the behavior of the speaker, but that doesn't matter. The question says you're told at least one of the dice is a 6, you should ignore all other cases because they are not mentioned in the question.
You're confusing a person stating a fact with the fact itself. The behaviour of the speaker absolutely matters. You cannot interpret what hearing a statement means without knowing the circumstances under which it will be said.

>Maybe that's because you aren't using your brain and you're using the formula instead.
It's a pretty good formula though :)

>For example, why don't you explain to me where you are getting P(E) from?
Breaking the space of dice rolls into all 36 equal possibilities:

1/36 chance of two sixes, in which case you are told about a 6 and a 6 is removed with probability 1

10/36 chance of a combination of 6 and some other number, in which case you are told about a 6 and a 6 is removed with probability 1/2

In the remaining 25/36 outcomes in the space of dice rolls there is 0 probability of you being told about the 6 and a 6 being removed

>You also never showed why P(E|D)=1?
If D occurs then both dice are 6, so you will always be told about a 6 and have one removed i.e. E will always also occur

>You just seem to be all over the place, and it's hard to pinpoint your mistakes when you forgo showing your reasoning.

My fault, sorry. Hopefully it's clearer now?

Ah but now you're assuming that you have to be told something about the dice at all. You have to take into account the probability of not being told anything, or being lied to, or the person putting on a tutu and doing a rendition of Swan Lake :^)

>[math] Pr [/math]
>not [math] \mathbb{P} [/math]

how you activate math formatting on your browser please?
mine doesn't do anything good

You don't *have* to be told something about the dice, but in this case you have been. So you need to be able to say what that means in terms of dice rolls.

Suppose you know that speaker A loves the number 6 and will always report if at least one 6 is rolled. And suppose you know that speaker B likes the number 6 as well as the number 4, so they will always report on either of those if at least one is rolled (choosing randomly if both are rolled).

If the dice are cast for speaker A and they say that at least one six was rolled, that is different information than you would get if the dice were cast for speaker B and B says that at least one six was rolled. It's even different information if neither of them says anything, because the situations in which that happens are different due to their different behaviour.

Let's recall what you defined E to be:
>Let E represent "you are told that at least one of the dice is a 6 and then a 6 dice is removed"
Let us also define A as "at least one of the dice is a 6" and B to be "a 6 is removed."
Then we have P(E)=P(B|A) I think the the reason we are getting different answers is because you think P(B)=.5 while I think P(B)=1. In the case of OP's question, P(B)=1 from the statement "a dice showing 6 is removed" so your interpretation is not correct.

There are 6×6=36 points in the probability space. If we know that at least one of them has a six component, we move to the fibre over {6} of one of the projections (invariant modulo isomorphism), so now we have a space of six possibilities. The probability that the remaining one is 6 is one point in this space, and assuming a uniform distribution from fair dice, we have a probability of 1/6.

>You don't *have* to be told something about the dice, but in this case you have been. So you need to be able to say what that means in terms of dice rolls.
Just as you don't have to be told at least one is a 6, but in this case you have been. If you are going to treat what you are told as a variable, then you should also treat being told anything as a variable, being lied to, etc. Ultimately all this means is you are assuming what you are told has no connection to what actually happened, since you could be told anything.

So the question then becomes, what is the chance that the dice remaining is 6 given that a 6 was removed?

>In the case of OP's question, P(B)=1 from the statement "a dice showing 6 is removed" so your interpretation is not correct.
Just because an event occurred doesn't mean the probability of it happening is 1. If I flip a coin and the result is heads, that doesn't mean the probability of the coin landing on heads is 1.

>Then we have P(E)=P(B|A)
It would be P(A AND B), which is P(B|A)*P(A). I think I understand what you're getting at though.

>I think the the reason we are getting different answers is because you think P(B)=.5 while I think P(B)=1. In the case of OP's question, P(B)=1 from the statement "a dice showing 6 is removed" so your interpretation is not correct.
I think both interpretations are reasonable. I've given answers for both above, 1/11 if a 6 dice is always removed and 1/6 if the dice that is removed is chosen randomly. I think we disagree about the result in the latter case, not sure about the first case.

My calculation for the first case:

Let E represent "you are told that at least one of the dice is a 6 and then a 6 dice is removed"
Let D represent "the remaining die is a 6"

P(D) = 1/36

# 11/36 likelihood of at least one 6 across all possibly dice combinations, and when that happens you are told about the 6 and a 6 is removed with probability 1
P(E) = 1*(11/36)

P(D|E) = P(E|D)*P(D)/P(E)
= 1*(1/36)/(11/36)
= 1/11

Sure. In principle, someone could approach a question like this with all sorts of prior beliefs about how an unknown speaker is expected to act, which could result in any possible "correct" answer.

In practice, I think most people coming at this question will have one of the two priors that I gave above: the speaker always announces 6s or the speaker is announcing one of the two rolls at random.

OK, I just wanted you to realize that those are not the only two possibilities, and that one is not more rational than the other because it assumes the least. If we assume the least then the question becomes

P(66|R)

Where R is the removal of a 6

P(66|R) = P(R|66) P(66) / P(R)

= P(R|66) (1/36) / ( P(R|66) (1/36) + P(R|16) (1/36) + ... )

= 1 / ( 1 + ( P(R|16) + P(R|26) + ... + P(R|65) ) / P(R|66) )

Now if we assume that P(R|16) = P(R|26) = ... = P(R|65), we have

P(66|R) = 1/(1+10P(R|16)/P(R|66))

Now we can see that if P(R|16) = P(R|66) = 1, we get 1/11

If 2P(R|16) = P(R|66), we get 1/6

So these answers can also be seen as interpretations of the chance of removal of a 6 from two dice showing 6.

1/36
prove me wrong

It's 1/6
Simple probability. The dice roll independent of each other. You 1/11 people are forgetting that.

So what. The thought process is displayed in that post.

What do you mean? If you are trying to post then use the preview and then use the tags.

People saying 1/11 are simply wrong. You were shown the result of one dice. It's no longer a matter of probability what that dice roll was. It's not one roll of two dice. It's now one unknown roll of one dice. The "at least one is six" is extraneous.

The "first" die is not removed. One of them is.

you 1/6 people are not realizing that 5 out of the 6 outcomes are each weighted at 2, because there are 2 ways of getting a die that isn't 6, but there is only 1 way of getting a die that is 6, so 1/(5*2+1) = 1/11.

2 ways each*