Odds problem - roll a pair of dice

[ftg]

@The_Great_Unwashed

That might explain why you’ve overlooked this and this.

I did see them. I’m just flat out tired of clearing up people’s mistaken notions and faulty comparisons.

Like I said, this thread is rotting now. The ratio of faulty stuff to factual stuff is getting too high to deal with. (As well as unhelpful partial references to earlier posts.)

I wonder why these types of threads just go on and on with people constantly throwing in more error-filled stuff. What is it about Math threads? (Please don’t answer. This is rhetorical.)

Eventually I bail when I realize some people are just completely resistant to simple Math and Logic. I’m around this point now.

[LSLGuy]

I honestly can’t say whether I’m regarded as part of the problem or part of the solution here. A bit of each is probably a fair assessment.

I have to agree with ftg that by the standards of GQ this is painful. And won’t be improved by continuing. I’m out.

[The_Great_Unwashed]

@ftg

I did see them. I’m just flat out tired of clearing up people’s mistaken notions and faulty comparisons.

I was simply hoping that you would clear up your own mistaken notions and faulty comparisons.

For instance you seem to have said that the sequential rolling of two dice (one after the other) is an essentially different experiment to the “simultaneous” rolling of two dice. It isn’t, I’m wondering why you think it is.

You also seemed to have said (it was hard to judge) that the expected number of rolls to have seen at least 2 sixes is less if you roll dice in pairs rather than one at a time. That isn’t right, I’m wondering why you think it is.

[octopus]

@sich_hinaufwinden

That would be tautological, however it’s not what I wrote, not what I meant, and not what I implied. But thanks for trying to put words in my mouth.

Of course fair and 6-sided dice are important to the question. And if they’re not specified you would have to assume those parameters. But here’s the thing - if we treat this as not a trick question, then not only are those reasonable assumptions, but more importantly we all agree to them.

That’s why I was careful to word my post as such:

The issues with the OP are that:

1. It does not state that the roller will reveal only 6s.
2. It does not state that the roller will always reveal a 6 when he can.

While it is reasonable to assume those things, those are not the only reasonable assumptions given the wording in the OP.

And you specifically are the one who asked:

Now that you have your answer you refuse to acknowledge that the wording proposed by CurtC is superior to the OP or give a reason why it isn’t.

And if you are incapable of recognizing the difference between his wording and the OP and think they are equivalent, you should still consider his wording superior since it satisfies both the “1/11 only” and the “not enough info” camps.

Why will you not do that?

As I’ve learned interpretations are subject to a greater deal of subjectivity then I care to attempt to fix. Because at some point all language is either circularly defined or established by the assumption that shared experiences are largely universal. In other words it’s too much effort for no reward to deal with linguistic issues when even if people were 100% clear those reading still couldn’t do the correct calculation.

[pulykamell]

Can we at least agree that if two dice are thrown together until at least one 6 shows up, then the other one being a 6 has a 1/11 probability? And if one die is thrown (or only one die is seen) until a 6 shows, that the chance of the other die being a 6 is 1/6? And that everything else is just a matter of how we interpret the OP?

[MonkeyMensch]

I was going to do a head count of which answers were favored but the number of hypothetical situations and side topics discussed makes that unappealing. I planned, some five or six pages ago, to post a nice explanation of why the probability being discussed in the OP is 1/6. But after a very little re-reading I saw **LSLGuy **had made that exact explanation in post #5! Reading matters!

If at this point some other poster has made this explanation I apologize in advance for the repetition. The OP only states:

"“Two fair dice are rolled together, and you are told that ‘at least one of the dice is a 6.’ A 6 is removed, and you are t(h)en shown the other die. What is the probability of the other die showing a six?”

When the experimenter (I’m calling her that) says truthfully, “At least one of the dice is a 6” she is giving you information about the possible outcomes. When she removes a 6 she effectively takes away that very information. At that point the only information you have is that there is a fair die left behind, randomly rolled, and the probability of a 6 (or of any number, really) is 1/6.

And my prediction of running to at least ten pages has come true. Steve Holt!!

[x-ray_vision]

@MonkeyMensch

When the experimenter (I’m calling her that) says truthfully, “At least one of the dice is a 6” she is giving you information about the possible outcomes.

Here they are. I assume you agree:

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

When she removes a 6 she effectively takes away that very information. At that point the only information you have is that there is a fair die left behind, randomly rolled, and the probability of a 6 (or of any number, really) is 1/6.

When she removes a 6, we are left with the following possibilities as being the die left:

1
2
3
4
5
6
5
4
3
2
1

1 out of 11 times, the die left over is a 6. If you think I did something wrong above, tell me at what point you think it was.

[sich_hinaufwinden]

@octopus

As I’ve learned interpretations are subject to a greater deal of subjectivity then I care to attempt to fix. Because at some point all language is either circularly defined or established by the assumption that shared experiences are largely universal. In other words it’s too much effort for no reward to deal with linguistic issues when even if people were 100% clear those reading still couldn’t do the correct calculation.

Since I understand the calculations for each of the scenarios discussed you must not be referring to me with that remark.

And you admit now that you will not address the answer to a question that you brought up.

I’ve only asked you that directly one time, so lest I be accused of badgering, I won’t waste either of our time by asking you again.

[Telemark]

@MonkeyMensch

When the experimenter (I’m calling her that) says truthfully, “At least one of the dice is a 6” she is giving you information about the possible outcomes. When she removes a 6 she effectively takes away that very information. At that point the only information you have is that there is a fair die left behind, randomly rolled, and the probability of a 6 (or of any number, really) is 1/6.

No, this is where you go wrong. The set of possible die rolls is not a random set with an equal distribution of 1/6 for each number. What is left is a set of rolls with a probability of 2/11 for numbers 1 through 5, and 1/11 for number 6. This is critical to understand the problem. That information given (one of the die is a 6) doesn’t disappear then the die is removed. That information is built into the probability of the remaining die.

Follow the steps listed by x-ray vision and see if it makes it more clear.

[MonkeyMensch]

@x-ray_vision

1 out of 11 times, the die left over is a 6. If you think I did something wrong above, tell me at what point you think it was.

I’d be happy to. Those are ordered pairs you list and depict permutations, not combinations. This is clear because you list 1,6 and 6,1 both. I agree that after the first roll this represents the set of outcomes in the probability space, all equally likely.

@x-ray_vision

Here they are. I assume you agree:

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

Here’s where the difference arises:

@x-ray_vision

When she removes a 6, we are left with the following possibilities as being the die left:

1
2
3
4
5
6
5
4
3
2
1

You list 11 possibilities left and that’s incorrect. I’ll use REM to denote the die that was removed. Say the second die was removed. Here’s the possible permutations:

(1, REM)
(2, REM)
(3, REM)
(4, REM)
(5, REM)
(6, REM)

Or if the first die was removed:

(REM, 1)
(REM, 2)
(REM, 3)
(REM, 4)
(REM, 5)
(REM, 6)

Those are now two disjoint probability spaces since either the first die or the second die was removed. Only one of those spaces is the case since you can’t remove the first die and/or the second die simultaneously. It’s one or the other. And it’s clear that all values remaining have equal probability. Your error arises because you are conjoining two disjoint probability spaces. To wit, you equate the pairs (6, REM) and (REM, 6) when they’re actually two distinct ordered pairs.

[MonkeyMensch]

I refer **Telemark **to the solution previously given.

Joking aside my response is just, in essence, using notation to elaborate what others have probably said.

[x-ray_vision]

@MonkeyMensch

Say the second die was removed. Here’s the possible permutations:

(1, REM)
(2, REM)
(3, REM)
(4, REM)
(5, REM)
(6, REM)

Or if the first die was removed:

(REM, 1)
(REM, 2)
(REM, 3)
(REM, 4)
(REM, 5)
(REM, 6)

Why are we going to say something that we don’t know happened? We don’t know which die was removed, so we don’t remove possibilities before calculating.

I’m going to post two interpretations of the problem that Chronos posted earlier. Are you interpreting the OP as the first, second, or are you saying the answer is 1/6 either way? Or are you interpreting it another way?

1: I roll a pair of dice. I then choose one die at random, look at the number on the die, and announce “At least one die is a ___”, where the blank is the number I looked at. On this particular iteration of the experiment, the die I looked at was a 6, and I announce as such. What is the probability that the other die is a 6?

2: I roll a pair of dice. I then examine both dice, and if it is the case that at least one of them shows a 6, I announce “At least one die is a 6”, and otherwise I say nothing. On this particular iteration of the experiment, I saw at least one 6, and so I announce as such. What is the probability that both dice are 6?

[Chronos]

Sure, you can call the two 6 cases different cases, but the probability for both of them combined is equal to the probability for any one of the others.

[misterdls]

Rolling a Green die and a Red die gives the following sample space:

G1,R1 G2,R1 G3,R1 G4,R1 G5,R1 G6,R1
G1,R2 G2,R2 G3,R2 G4,R2 G5,R2 G6,R2
G1,R3 G2,R3 G3,R3 G4,R3 G5,R3 G6,R3
G1,R4 G2,R4 G3,R4 G4,R4 G5,R4 G6,R4
G1,R5 G2,R5 G3,R5 G4,R5 G5,R5 G6,R5
G1,R6 G2,R6 G3,R6 G4,R6 G5,R6 G6,R6
36 possibilities, equally likely, obviously.

The roller truthfully says that a 6 appears on one or both of the dice. This reduces
the sample space to only the 11 possible combinations bolded above and listed below:

G6,R1
G6,R2
G6,R3
G6,R4
G6,R5
G6,R6
G1,R6
G2,R6
G3,R6
G4,R6
G5,R6

Again, obvious.

Note that:

1. There are 11, not 12, equally possible combinations which contain one or two 6’s.

2. The 6-6 combination can only appear once, as G6,R6, as opposed to, for example,
   1-2, where 1-2 can appear as either G1,R2 or as G2,R1. Saying there are 12
   combinations is saying there are two 6-6 combinations. There are not. This is
   fundamental. There are no “other circumstances” that can give anything different
   than 11 combinations.

3. There is nothing “peculiar” about any of this; it all flows from the original
   problem. There’s no reason to consider other factors, such as the roller’s motive,
   or any might-have been, or things we are not told. We have a simple,
   straightforward problem to solve. And we have before us all the facts we need.

We can put each die face of the 11 listed combinations into one of two “modes”. Either
the die face shows 6 or it doesn’t. If it shows 6 we’ll say it’s a “mode6” face.
If it isn’t a 6 then we will call it a “modex” face. The actual number shown on a
modex die in the box isn’t important in our deliberations here.

Now consider the initial dice throw. After the throw and BEFORE seeing if either die
is a 6 (and therefore mode6), the content of each of the 36 combinations will ALWAYS
consist of one of the following mode combinations:

Green Red
mode6 mode6
mode6 modex
modex mode6
modex modex.

4 combinations, each equally likely. If one of the 4 combinations are removed
from this list, the remaining 3 combinations remain equally likely.

When the roller excludes the possibility of the fourth combination (modex-modex) by
stating, truthfully, that at least one of the dice is showing a 6, AS HE ALWAYS DOES, the sample space we are given when
we are now facing the following 3 possibilities comprising our sample space:

Green Red
mode6 mode6
mode6 modex
modex mode6

All equally likely.

In plain words, either 1. the Green shows a 6 and the Red also shows a 6, or
2. the Green shows a 6 and the Red does not show a 6, or 3. the Green does not show
a 6 and the Red shows a 6.

These are all and the ONLY possible configurations of the two dice in the box at this
point.

Note that there are 4 mode6 faces and 2 modex faces. This is the direct result from
the roller’s exclusion of the modex-modex combination, and is normal and expected, and
will be the critical factor in determining the final answer.

Let’s label each combination with a case letter, to more easily distinguish them:

Green Red
mode6 mode6 Case A
mode6 modex Case B
modex mode6 Case C

The roller removes one die, and it’s a 6, a mode6 face. So it could be any of
these four faces:

Green mode6 from Case A
Red mode6 from Case A
Green mode6 from Case B
Red mode6 from Case C

Bingo.

Counting the Cases, we see there are 2 Case A’s, 1 Case B, and 1 Case c. Two of the
four mode6 faces come from Case A where both faces are 6’s.

The answer to the question “What is the probability that the die remaining in the box
is a 6” is 2/4 or 1/2.

QED.

This problem is very similar to the Boy-Girl problem where the probability of a
certain child’s sex is a boy (or girl, as the case may be) equals 1/2.

I really do like working on (simple) probability problems. Kind of a hobby. I’ve
done this work simply because I enjoy it. I had no intent to embarrass or anger
anyone.

Best wishes to EVERYONE.

[x-ray_vision]

@misterdls

Green Red
mode6 mode6
mode6 modex
modex mode6
modex modex.

4 combinations, each equally likely.

WTF?

[Grestarian]

@Skammer

A math teacher friend of mine shared this problem on Facebook and we are disagreeing about the solution. Here is the problem verbatim:

“Two fair dice are rolled together, and you are told that ‘at least one of the dice is a 6.’ A 6 is removed, and you are ten shown the other die. What is the probability of the other die showing a six?”

My friend thinks the answer is 1/6, because the dice rolls are independent events and regardless of what one shows, the other has a 1/6 chance of being a six. I’m quite positive this is incorrect.

@Turble

Charma got it. The roll of each die is an independent process. You can completely ignore the fact that one of them shows a 6. The chance of any one 6-sided showing a 6 is always 1/6 no matter how many dice you roll and no matter what the other dice show.

For Ogg’s sakes! It’s not a math, statistics, or probability challenge. It’s a test of reading ability! :eek:
“…fair dice are rolled …what is the probability…die showing a six?”
[We are assuming these are standard cubes, not D&D/wargaming dice]

Haldurson said it best: The red herring doesn’t change the fact that the probability of any single 6-sided die roll being a 6 is 1/6

[ul]
[li]You could just as easily say the probability of any 6-sided die-roll being a 4 is 1/6[/li][li]You could just as easily say the probability of any 10-sided die-roll being a 2 is 1/10[/li][li]You could just as easily say the probability of any ten billion-sided die-roll being a 2 is (okay, once the darned thing stopped rolling)…[/li][/ul]

I roll 400 octahedrons, each with numbers 1 through 8 on each face. I remove 399 of those dice. What’s the probability that the one I left behind shows a 5*?

There is no Lady Luck. There is no force or entity with any vested interest in making dice rolls sequential or logical or pretty. Why make things more complex than they already are?

—G!

• Now the real challenge!
  What’s the probability that the one left behind shows a 9? :smack:

[Thudlow_Boink]

@Grestarian

I roll 400 octahedrons, each with numbers 1 through 8 on each face. I remove 399 of those dice. What’s the probability that the one I left behind shows a 5*?

That depends: How did you decide which 399 to remove? Did you, for example, deliberately remove all the ones that showed 5s?

[Haldurson]

@Thudlow_Boink

That depends: How did you decide which 399 to remove? Did you, for example, deliberately remove all the ones that showed 5s?

If I said instead that I use some pre-specified algorithm to remove dice, and one die remains, you still have to assume that the odds are 1 out of 8 because you don’t know what the criteria is. Information about that remaining die doesn’t exist, therefore, it’s irrelevant to the question.

If I non-randomly selected a face on a die, and I asked you what the odds are that I chose an 8 on that die, without knowing my criteria for selecting that die, you’d be correct if you said that it’s 1 out of 8. On the other hand, if you knew the criteria, then the odds might be different.

Once you add knowledge to the mix, you can change those odds. But without such knowledge, you have to assume that all possible results are equally likely. It really doesn’t matter if the dice are picked deliberately or not, unless you have access to the actual criteria used.

[Cliffy_Biro]

@Haldurson

If I non-randomly selected a face on a die, and I asked you what the odds are that I chose an 8 on that die, without knowing my criteria for selecting that die, you’d be correct if you said that it’s 1 out of 8. […] without such knowledge, you have to assume that all possible results are equally likely.

I’m not sure that there’s a rational basis for assuming that the probability is 1/8, or any particular value.

Say we have a million bags each containing 100 balls that are either black or white, in proportions that may differ in each bag. We don’t know how many black balls are in each bag and have no reason to expect any particular number. For all we know, every single ball in every bag could be black, or they could all be white, or whatever. Anything’s possible.

A goes to take a ball out of the first bag, saying “we have no information, so it’s a 1/2 chance that the ball will be black.” Meanwhile B says “I reckon that it is a 1/3 chance.” Afterwards, they open up the bag and see that there were 70 black balls in there, so for this bag it was actually a 7/10 chance. Neither of them was right this time.

They do the same with all the bags, with A doggedly sticking to 1/2 while B persists with 1/3. After they’ve done the million bags, given what we know, we can only expect that A will have been correct about the same number of times as B. So A’s guess of the probability was no better than B’s.

[Haldurson]

@Cliffy_Biro

Say we have a million bags each containing 100 balls that are either black or white, in proportions that may differ in each bag. We don’t know how many black balls are in each bag and have no reason to expect any particular number. For all we know, every single ball in every bag could be black, or they could all be white, or whatever. Anything’s possible.

I roll a die, and only I know what is on that die. For me, it’s either 0% or 100% that the number showing is a 6. As far as you are concerned, thoiugh, it’s 1 out of 6. Knowledge changes everything. In your example, you are cheating by looking ahead in time. If all you know is that each ball may be either black or white, then yes, as you say, anything is possible. if I’m a betting man, and without foreknowledge, If I bet on I pick from a random ball from a random bag on what color it’s going to be, and someone else without foreknowledge of the results were betting against me, if you wanted to make the bet fair it would be 50-50. If I were cheating, obviously, the odds would be different. What you are suggesting is that someone is cheating.