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-   -   Odds problem - roll a pair of dice (https://boards.straightdope.com/sdmb/showthread.php?t=806535)

Skammer 10-06-2016 09:41 AM

Odds problem - roll a pair of dice
 
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.

My reasoning is: if you roll a pair of dice, there are 36 possible combinations: (1,1), (1,2), (2,1), etc. Out of these 36 combinations, 11 contain at least one six: each 6 paired with 1-5 on the other die, and double sixes.

Since we are told that the pair contains at least one six, we have 11 possible combinations with equal chance of occurring. If a six is then removed, only 1 of the remaining 11 dice is another six. So I maintain the answer is 1/11.

However, the meme that she posted is multiple choice and 1/11 is not one of the available answers, which are: 1/18, 1/36, 2/11 and 1/6.

At first I jumped on the 2/11 thinking "there are two chances to pick a six from the pair of doubles!" But that logic falls apart I think. I'm back to 1/11. How am I wrong?

Thudlow Boink 10-06-2016 09:51 AM

I agree with your analysis. I've seen that... meme? I don't think that's the right word... and I suspect that the correct answer was left out of the choices deliberately to make it more clickbaitey.

This is a variation of the old "...at least one is a boy" probability question that has been discussed here in numerous threads (e.g. here, here, here, here).

Ignotus 10-06-2016 10:56 AM

"A 6 is eliminated" isn't enough information to give an answer. On what basis is the 6 eliminated? Does it mean that sixes have a half-life, and you measure the odds of one still remaining after the first one has gone kablooey? Or does it mean that you randomly choose one of the combinations including at least one 6, and eliminate a 6 from it?

TriPolar 10-06-2016 10:57 AM

Isn't the question asking the probability of two sixes being rolled?

LSLGuy 10-06-2016 11:04 AM

Quote:

Originally Posted by Skammer (Post 19679483)
...
Since we are told that the pair contains at least one six, we have 11 possible combinations with equal chance of occurring. ...

IMO the above part is right. Then you go wrong here:
Quote:

Originally Posted by Skammer (Post 19679483)
... If a six is then removed, only 1 of the remaining 11 dice is another six. So I maintain the answer is 1/11.
...

How many different ways can we remove a six? Let's assume we have a green and a red die so we can tell them apart. As you said, the possibilities expressed in green, red order are:

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

If the green die is a 6 we're looking at one of the bottom 6 possibilities. 1/6th of which have another 6, the red die.

If the red die is a 6 we're looking at one of the top 6 possibilities. 1/6th of which have another 6, the green die.

Result: it doesn't matter which die is a 6. Red, green, or both. After you remove one of the sixes the chance of the other being a 6 is 1/6.

LSLGuy 10-06-2016 11:10 AM

Late add:

You can restate the problem like this and it becomes really, really obvious:

Quote:

Originally Posted by Restated Problem
Roll two dice together. On those occasions where one die came up a six, what are the odds the other one is a six?

A: Since the two dice don't communicate, the odds of the second one being a 6 are always 1/6. It doesn't matter what the other die is doing.

This is one of those puzzles where the trick is to ignore most of the setup since that's just there to distract you. "Now everybody watch my left hand very carefully as I fiddle with this magic wand..."

watchwolf49 10-06-2016 11:15 AM

We have a condition here, IF one die shows a 6, THEN what are the odds the other die shows a 6. The answer is 1/6. Each valid trial requires that one of the die shows a 6, or else it's not valid, rolling hard 10 or snake-eyes is dismissed.

Now, if the question is what are the odds of rolling a pair of 6's, then that's 1/36 ... we have 36 valid trials of which only one result makes a positive.

ETA: nin'ja'ed by the great LSLGuy

CharmaChameleon 10-06-2016 11:20 AM

The principle always holds in problems such as this:

Entities and their outcomes forever remain independent with respect to each other regardless of spatial, temporal, or any other separation.

Turble 10-06-2016 11:25 AM

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.

chrisk 10-06-2016 11:31 AM

I don't think I agree... though as with many probabilities problems, it may be that I'm missing something or am making an assumption that others aren't. Here's my reasoning:

I roll the 2 dice. To begin with, there are 36 different possibilities, (counting the red die and the green die separately.)

I observe, "Hey, there's a six here." That eliminated 25 possibilities, all the ones composed only of 1 through 5 on both dice, and leaves 11 possibilities left, out of which only one is double-sixes.

Now, for each of these eleven possibilities, I can pick a six and take it off the table. Since I can always do that, doing it doesn't change the original odds of rolling double sixes... the way the observation about having at least one six did. (That changed the odds, because there were many possible rolls in which I couldn't make that observation.)

Thus, even after I've taken that die and removed it from the table, the odds that I originally rolled double sixes is still 1 in 11.

The remaining die still on the table will be a 6 if-and-only-if I rolled double sixes. Thus, the odds that the die on the table is a 6 is also 1 in 11.

No, the dice don't communicate, and yes, the outcome of the dice is independent. But that observation of both dice together does change the answer, as does the process of "looking for" one six and selectively removing it from the table... assuming that I would announce my observation and remove one six every time I roll a result that has one or more sixes. That's the assumption.

Edited to add: If, on the other hand, we assume that I removed one die from the table at random, and it happened to be a die showing a six... then I think the answer of 1 in 6 is correct.

Cartoonacy 10-06-2016 11:33 AM

Quote:

Originally Posted by Skammer (Post 19679483)
"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 then shown the other die. What is the probability of the other die showing a six?"

The wording is distracting. If it had said,
Quote:

"Two fair dice are rolled together, and you are told that 'at least one of the dice is a 6.' What is the probability of the other die showing a six?"
then I would say the answer is 1/11, since there are 11 ways that two dice could show at least one six, but only one of them in which both dice are sixes.

On the other hand, if it said,
Quote:

"Two fair dice are rolled together. A 6 is removed, and you are then shown the other die. What is the probability of the other die showing a six?"
then the answer would be 1/6. After you remove one die, the odds that the other one is a six is the same as rolling that die by itself.

The odds where at least one is a 6 are different from where this one is a 6. The way the problem is worded, I'd say that the act of removing one of the dice makes the odds of the other one being a six 1/6.

Cartoonacy 10-06-2016 11:37 AM

Quote:

Originally Posted by chrisk (Post 19679706)
Edited to add: If, on the other hand, we assume that I removed one die from the table at random, and it happened to be a die showing a six... then I think the answer of 1 in 6 is correct.

Agreed.

Ignotus 10-06-2016 11:38 AM

Quote:

Originally Posted by LSLGuy (Post 19679657)
Late add:

You can restate the problem like this and it becomes really, really obvious:

A: Since the two dice don't communicate, the odds of the second one being a 6 are always 1/6. It doesn't matter what the other die is doing.

This is one of those puzzles where the trick is to ignore most of the setup since that's just there to distract you. "Now everybody watch my left hand very carefully as I fiddle with this magic wand..."

(My emphasis)

Ah, but they do in a way communicate, since you're told the pair contains "at least one 6". So, if neither die showed a 6, they wouldn't be on the table at all! So, if you are one of the dice, you can correctly infer: either myself, or that other die over there, is showing a 6!"

TriPolar 10-06-2016 11:48 AM

Quote:

Originally Posted by TriPolar (Post 19679628)
Isn't the question asking the probability of two sixes being rolled?

Ok, it's not this because you know that one six was rolled. It does sound like the boy girl problem.

watchwolf49 10-06-2016 11:48 AM

I think we're confusing the odds of rolling at least one 6 with the OP. The OP gives us this one 6, we're asked the odds of the other being a 6. We only have six possible trial results; (6,1),(6,2),(6,3),(6,4),(6,5),(6,6); all other combinations are ignored.

The operative phrase in the OP is "'at least one of the dice is a 6.'", it doesn't matter which die is a 6.

carnivorousplant 10-06-2016 11:53 AM

Quote:

Originally Posted by Cartoonacy (Post 19679711)
The odds where at least one is a 6 are different from where this one is a 6. The way the problem is worded, I'd say that the act of removing one of the dice makes the odds of the other one being a six 1/6.

Agreed.

chrisk 10-06-2016 11:54 AM

Quote:

Originally Posted by watchwolf49 (Post 19679761)
I think we're confusing the odds of rolling at least one 6 with the OP. The OP gives us this one 6, we're asked the odds of the other being a 6. We only have six possible trial results; (6,1),(6,2),(6,3),(6,4),(6,5),(6,6); all other combinations are ignored.

The operative phrase in the OP is "'at least one of the dice is a 6.'", it doesn't matter which die is a 6.

But if we start not knowing if the red or green die, (for example,) is a 6, then we have 11 possible results at the early stage: (1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

chrisk 10-06-2016 11:59 AM

Quote:

Originally Posted by Cartoonacy (Post 19679711)
The way the problem is worded, I'd say that the act of removing one of the dice makes the odds of the other one being a six 1/6.

I would say that the wording is not only distracting, it's ambiguous. We have no idea whether there was intention or not in removing the 6, so we can't tell if it changes the odds or how.

Thudlow Boink 10-06-2016 11:59 AM

Quote:

Originally Posted by Skammer (Post 19679483)
"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?"

As in the case of the infamous Monty Hall problem, one way of settling this would be experimentally.

Roll two fair dice. At least one of the dice should be a 6, so if this is not the case, re-roll until it is.

Now, remove a 6.

Does the remaining die show a 6? Write down YES or NO.

Repeat this experiment many times, and then calculate the number of YESes you wrote down, dividing by the total number of trials.

x-ray vision 10-06-2016 12:07 PM

Quote:

Originally Posted by Thudlow Boink (Post 19679799)
As in the case of the infamous Monty Hall problem, one way of settling this would be experimentally.

Roll two fair dice. At least one of the dice should be a 6, so if this is not the case, re-roll until it is.

Now, remove a 6.

Does the remaining die show a 6? Write down YES or NO.

Repeat this experiment many times, and then calculate the number of YESes you wrote down, dividing by the total number of trials.

Right, but how is it that a 6 being removed is decided?

We have:

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

Is on pair of dice randomly chosen to have a 6 removed?: 1 out of 11 pairs.

Or is one die out of ever die showing a 6 in the running to be chose?: 1 out of 12 dice.

Thudlow Boink 10-06-2016 12:10 PM

Quote:

Originally Posted by chrisk (Post 19679797)
I would say that the wording is not only distracting, it's ambiguous. We have no idea whether there was intention or not in removing the 6, so we can't tell if it changes the odds or how.

I suppose you could interpret it to mean "A die is removed, and that die happens to be a 6," in which case the answer would be 1/6. But the more natural (IMHO) interpretation is that it's part of the setup that a 6 specifically be removed.

chrisk 10-06-2016 12:10 PM

Quote:

Originally Posted by x-ray vision (Post 19679821)
Right, but as has been mentioned, how is it that a 6 being removed is decided?

We have:

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

Is on pair of dice randomly chosen to have a 6 removed?: 1 out of 11 pairs.

Or is one die out of ever die showing a 6 in the running to be chose?: 1 out of 12 dice.

I don't think this makes sense... those 22 dice, 11 pairs, represent a possibility space. They're not all on the table at the same time, as it were, and you can pick any of the 12 6's to remove. You have to remove a 6 from each pair in the possibility space--assuming you're looking for a 6 to remove.
From 10 of the pairs, you have no choice under that assumption
From the last pair, it doesn't really matter what choice you make.

x-ray vision 10-06-2016 12:16 PM

I see your point about them not actually being on the table at the same time.

Then I don't understand what you meant by this:

We have no idea whether there was intention or not in removing the 6, so we can't tell if it changes the odds or how.

markn+ 10-06-2016 12:17 PM

This problem is going to produce endless fruitless debate, because, like the boy-girl problem, the specified conditions are ambiguous. The details of the randomization procedure that produced the roll with at least one 6 have not been specified, so there are at least two possible answers, depending on how the roll was produced. I would plead with everyone who is about to reply here with the answer that seems "obvious" to them to read Wikipedia's article on the boy-girl problem first, to save a lot of pointless argumentation.

Here's one interpretation of the given problem:

"Two dice are rolled numerous times. In the cases where a six appears on at least one of the dice, the observer is told that one die is a six and asked the probability that the other is a six."
In this case, 25/36 of the rolls will not produce a six and are ignored. Of the remaining cases, 1 in 11 will have two sixes, so the answer is 1/11.

Here's another interpretation of the problem:
"Two dice are rolled once. It just so happens that at least one die shows a six. The observer is asked for the probability that the other is a six."

In this case the probability that the second die is a six is the same as the probability that any arbitrary thrown die is a six, 1/6.

Clearly different people in this thread are already interpreting the problem in different ways, and therefore producing different results.

--Mark

md2000 10-06-2016 12:18 PM

Yes, you have 11 possible combinations - but the case presented to you is that "one of the sixes is removed". As the posters talking about red and green dice point out, there are in fact 2 cases of 6,6 - one where we remove the red and the green die is 6, and one where we remove the green and the red die is 6. so the odds at 2 in 12, not 1 in 11. It's not the odds that 6,6 is rolled, it's the odds that the case presented to you matches the criteria - at least one 6.

Thudlow Boink 10-06-2016 12:18 PM

Quote:

Originally Posted by x-ray vision (Post 19679848)
Right. But how the choice is made to remove a 6 affects the answer to the question being asked. Is it decided by taking one randomly from one of the 11 pairs, or by taking one randomly from the 12 dice showing a 6?

At the point the 6 is removed, there are only two dice, and they've already been rolled. At that point, if there's any choice left about which 6 to remove, it doesn't matter which it is—either way, the remaining die will show a 6.

x-ray vision 10-06-2016 12:27 PM

I'm blaming my confusion on LSLGuy. :D

Quote:

Originally Posted by LSLGuy (Post 19679644)
IMO the above part is right. Then you go wrong here:How many different ways can we remove a six? Let's assume we have a green and a red die so we can tell them apart. As you said, the possibilities expressed in green, red order are:

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

If the green die is a 6 we're looking at one of the bottom 6 possibilities. 1/6th of which have another 6, the red die.

If the red die is a 6 we're looking at one of the top 6 possibilities. 1/6th of which have another 6, the green die.

Result: it doesn't matter which die is a 6. Red, green, or both. After you remove one of the sixes the chance of the other being a 6 is 1/6.

He has every combination twice because order matters, but he doesn't for the 6,6 combination. The order of 6s matters too. So it should be:

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

Answer: 1 out of 6

LSLGuy 10-06-2016 12:32 PM

Nope. Anything involving two 6,6 pairs is wrong from the git-go.

watchwolf49 10-06-2016 12:39 PM

In the boy/girl paradox, if at least one child is a boy, and we remove that boy, what are the odds of having two boys left over ... the answer is zero. Or are we asking, after we remove the boy, what are the odds of the one left being a boy ... that's 1/2.

We don't care about the odds of having at least one 6, that's 11/36. A least one 6 is a certainty, we pick the die up and examine the remain die in isolation, the odds of a 6 is 1/6.

Ignotus 10-06-2016 12:41 PM

Quote:

Originally Posted by markn+ (Post 19679854)
This problem is going to produce endless fruitless debate, because, like the boy-girl problem, the specified conditions are ambiguous. The details of the randomization procedure that produced the roll with at least one 6 have not been specified, so there are at least two possible answers, depending on how the roll was produced. I would plead with everyone who is about to reply here with the answer that seems "obvious" to them to read Wikipedia's article on the boy-girl problem first, to save a lot of pointless argumentation.

Here's one interpretation of the given problem:

"Two dice are rolled numerous times. In the cases where a six appears on at least one of the dice, the observer is told that one die is a six and asked the probability that the other is a six."
In this case, 25/36 of the rolls will not produce a six and are ignored. Of the remaining cases, 1 in 11 will have two sixes, so the answer is 1/11.

Here's another interpretation of the problem:
"Two dice are rolled once. It just so happens that at least one die shows a six. The observer is asked for the probability that the other is a six."

In this case the probability that the second die is a six is the same as the probability that any arbitrary thrown die is a six, 1/6.

Clearly different people in this thread are already interpreting the problem in different ways, and therefore producing different results.

--Mark

Is the observer aware of us observing him? If so, does that knowledge somewhat skew his judgment?

Is my cat dead or alive?

watchwolf49 10-06-2016 01:07 PM

100 valid trials:
1 = 11
2 = 21
3 = 14
4 = 15
5 = 25
6 = 14

Fuzzy data ... tends to the 1/6 prediction ... need more trials I would say.

OldGuy 10-06-2016 01:07 PM

Quote:

Originally Posted by Thudlow Boink (Post 19679799)
As in the case of the infamous Monty Hall problem, one way of settling this would be experimentally.
.....
Repeat this experiment many times, and then calculate the number of YESes you wrote down, dividing by the total number of trials.

I tend to avoid these discussions now in the Dope, as some posters are never satisfied with probability proofs or the discussion gets sidetracked on the exact meaning of things like "at random". But this is good advice. If you don't like someone else's calculation do a simple simulation.

Doing this not only gives you your answer, it makes you think precisely what you mean by any of the instructions because you have to program it into the simulation in a precise way that the computer understands.

TroutMan 10-06-2016 01:14 PM

I ran a simulation 100K times. I ignored any roll without at least one six. For the remaining rolls, here are the counts:

Code:

Die  occurrences    %
1        5428      17.7%
2        5627      18.4%
3        5533      18.1%
4        5536      18.1%
5        5690      18.6%
6        2801        9.1%

So assuming the interpretation that the question is asking about the odds over multiple rolls, the answer is clearly 1/11.

I understand markn+'s point about the answer depending on how you interpret it, but the second interpretation of "on this one roll where one die just happened to be 6, what are the odds the other one is a 6" doesn't work for me. The whole point of probabilities and odds is assuming what happens over multiple events. Pretending that we're talking only about this single roll isn't a probability. It smacks too much of "either it's a 6 or it's not, so the odds are 50/50."

x-ray vision 10-06-2016 01:15 PM

Quote:

Originally Posted by LSLGuy (Post 19679904)
Nope. Anything involving two 6,6 pairs is wrong from the git-go.

Then the answer is 1/11.

And someone is willing to bet on this with actual dice rolls.


http://math.stackexchange.com/questi...ity-both-are-2

It seems like the OP's problem is theoretically the same.

CharmaChameleon 10-06-2016 01:20 PM

Quote:

Originally Posted by Ignotus (Post 19679936)
Is the observer aware of us observing him? If so, does that knowledge somewhat skew his judgment?

Is my cat dead or alive?

It's almost impossible not to ask, "Is that Schrödinger's cat?" (But, I refrain.)

Just as the "observer" in the problem cannot change the outcome of the initial roll by any action on his/her part, we, as "meta-observers", cannot "meta-change" any "meta-outcome" in the problem space, as all of that is inviolable from change caused by any event residing in internal or external (base or "meta") space. (The dice are rolled, and cannot be un-rolled.)

chrisk 10-06-2016 01:21 PM

Quote:

Originally Posted by LSLGuy (Post 19679904)
Nope. Anything involving two 6,6 pairs is wrong from the git-go.

Yes, except that: (To complicate things still further :D )

If you assume that either die is removed randomly, then each of the 11 cases is twinned, like so:
(1, 6)
(1, 6)
(2, 6)
(2, 6)
(3, 6)
(3, 6)
(4, 6)
(4, 6)
(5, 6)
(5, 6)
(6, 1)
(6, 1)
(6, 2)
(6, 2)
(6, 3)
(6, 3)
(6, 4)
(6, 4)
(6, 5)
(6, 5)
(6, 6)
(6, 6)

Where the bolded number is kept on the table. Out of those 22 cases, we can remove 10 on the grounds that a number other than 6 is removed, leaving:

(1, 6)
(2, 6)
(3, 6)
(4, 6)
(5, 6)
(6, 1)
(6, 2)
(6, 3)
(6, 4)
(6, 5)
(6, 6)
(6, 6)
So, out of 12 remaining cases, in 2 there's a 6 still on the table, so the odds are 2/12 or 1/6.

DrDeth 10-06-2016 01:24 PM

Quote:

Originally Posted by markn+ (Post 19679854)
This problem is going to produce endless fruitless debate, because, like the boy-girl problem, the specified conditions are ambiguous. The details of the randomization procedure that produced the roll with at least one 6 have not been specified, so there are at least two possible answers, depending on how the roll was produced. I would plead with everyone who is about to reply here with the answer that seems "obvious" to them to read Wikipedia's article on the boy-girl problem first, to save a lot of pointless argumentation.

Yes and here is the relevant quote from Martin Gardner "Gardner argued that a "failure to specify the randomizing procedure" could lead readers to interpret the question in two distinct ways:..."


(spoken in the voice of the Captain) What we got here is failure to specify the randomizing procedure.

watchwolf49 10-06-2016 01:26 PM

Quote:

Originally Posted by TroutMan (Post 19680036)
I ran a simulation 100K times. I ignored any roll without at least one six. For the remaining rolls, here are the counts: ... [snip]

I just realized I'm throwing the Black Death Dice that came with my +1 (to damage) two-handed sword ... [smile] ... `detect magic` doesn't work on it. Troutman is throwing fair dice, the Law of Large Numbers can be applied here which gives 1/11 odds.

chrisk 10-06-2016 01:35 PM

Quote:

Originally Posted by DrDeth (Post 19680061)
Yes and here is the relevant quote from Martin Gardner "Gardner argued that a "failure to specify the randomizing procedure" could lead readers to interpret the question in two distinct ways:..."


(spoken in the voice of the Captain) What we got here is failure to specify the randomizing procedure.

*Golf clap for combining Martin Gardner and Cool Hand Luke in one post.* :D

Shagnasty 10-06-2016 01:53 PM

In case someone likes a real-world demonstration more than theory, I just modeled this problem in Excel to simulate 200,000 rolls of dice pairs. No matter how many times I run it, the answer always comes out to be right at 1/11th that both dice will be a 6 if either of them is.

markn+ 10-06-2016 02:04 PM

Quote:

Originally Posted by Shagnasty (Post 19680140)
In case someone likes a real-world demonstration more than theory, I just modeled this problem in Excel to simulate 200,000 rolls of dice pairs. No matter how many times I run it, the answer always comes out to be right at 1/11th that both dice will be a 6 if either of them is.

Yes of course, but your simulation implicitly assumes the first of the two interpretations that I suggested. It's not really possible to simulate the second interpretation, but it's still a (somewhat) reasonable interpretation.

--Mark

octopus 10-06-2016 02:14 PM

Quote:

Originally Posted by Skammer (Post 19679483)
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.

My reasoning is: if you roll a pair of dice, there are 36 possible combinations: (1,1), (1,2), (2,1), etc. Out of these 36 combinations, 11 contain at least one six: each 6 paired with 1-5 on the other die, and double sixes.

Since we are told that the pair contains at least one six, we have 11 possible combinations with equal chance of occurring. If a six is then removed, only 1 of the remaining 11 dice is another six. So I maintain the answer is 1/11.

However, the meme that she posted is multiple choice and 1/11 is not one of the available answers, which are: 1/18, 1/36, 2/11 and 1/6.

At first I jumped on the 2/11 thinking "there are two chances to pick a six from the pair of doubles!" But that logic falls apart I think. I'm back to 1/11. How am I wrong?

It's 1/36. There is only one event that results in a 6 being on the other die of a pair when one 6 is removed and that is double 6s. Double 6s is 1/36 chance.

octopus 10-06-2016 02:16 PM

Quote:

Originally Posted by TriPolar (Post 19679628)
Isn't the question asking the probability of two sixes being rolled?

Yes. This shouldn't be tricky at all. Yet look at the fun you can have with Monty Hall and the goat.

DrDeth 10-06-2016 02:26 PM

Quote:

Originally Posted by TriPolar (Post 19679628)
Isn't the question asking the probability of two sixes being rolled?

Maybe. The question is so badly worded, who knows?


They failed to properly specify the randomizing procedure.

DrDeth 10-06-2016 02:28 PM

Quote:

Originally Posted by octopus (Post 19680193)
It's 1/36. There is only one event that results in a 6 being on the other die of a pair when one 6 is removed and that is double 6s. Double 6s is 1/36 chance.

Not if one die is given to be a six.

Ok, the question is badly worded, so let's look at it another way. One of the dice has to be a six, right?

So let us toss two dice, one is normal the other has 6's on all faces. What is then the chance of the other die rolling a 6?

octopus 10-06-2016 02:34 PM

Quote:

Originally Posted by DrDeth (Post 19680232)
Not if one die is given to be a six.

Ok, the question is badly worded, so let's look at it another way. One of the dice has to be a six, right?

So let us toss two dice, one is normal the other has 6's on all faces. What is then the chance of the other die rolling a 6?

If it's only asked when one is a 6? Then the answer is 1/6. Independent each have 1/6 chance.

But lets do the red die green die list of possibilities.

R G
1 6
2 6
3 6
4 6
5 6
6 6


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

There are 12 different combinations in which a 6 can be given. Six that are green and six that are red 6s. That's why you can count the 6 6 twice because you are given a different die. You are given either the red or the green so it's two events. Ok you have twelve possible and equally likely events. Only two still have a 6 left over. 2/12 gives 1/6 which is what should be expected treating dice as individual events and only doing this in the trials you have at least one 6 rolled.

watchwolf49 10-06-2016 02:39 PM

Quote:

Originally Posted by octopus (Post 19680202)
Yes. This shouldn't be tricky at all. Yet look at the fun you can have with Monty Hall and the goat.

Goats are cool ...

Colophon 10-06-2016 02:41 PM

Quote:

Originally Posted by octopus (Post 19680193)
It's 1/36. There is only one event that results in a 6 being on the other die of a pair when one 6 is removed and that is double 6s. Double 6s is 1/36 chance.

But you are ignoring the fact that 25 out of those 36 chances won't have a six at all, so they can be eliminated from the game.

Edit: that page linked by x-ray vision gives the clearest explanation:

Quote:

Another way of phrasing this is to explicitly use conditional probability. You are asking for P[both dice are 2|at least one die is 2]. Since the first event implies the second, this probability is equal to P[both dice are 2]/P[at least one die is 2]. Since P[both dice are 2]=1/36 and P[at least one die is 2]=11/36, we have P[both dice are 2|at least one die is 2]=(1/36)/(11/36)=1/11.

Musicat 10-06-2016 02:42 PM

Quote:

Originally Posted by Skammer (Post 19679483)
"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?"

There's your problem. No die has a ten on any face.

Turble 10-06-2016 02:42 PM

I suspect that those who are running simulations are testing one simulated die and discarding the roll if that die is not a 6. Please try it by checking both dice for a 6 before discarding the roll.


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