Reply
 
Thread Tools Display Modes
  #1  
Old 10-06-2016, 08:41 AM
Skammer Skammer is offline
Charter Member
 
Join Date: Aug 2002
Location: Music City USA
Posts: 14,031
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?
Advertisements  
  #2  
Old 10-06-2016, 08:51 AM
Thudlow Boink Thudlow Boink is offline
Charter Member
 
Join Date: May 2000
Location: Lincoln, IL
Posts: 23,862
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).
  #3  
Old 10-06-2016, 09:56 AM
Ignotus Ignotus is offline
Guest
 
Join Date: Dec 2012
Posts: 978
"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?
  #4  
Old 10-06-2016, 09:57 AM
TriPolar TriPolar is offline
Guest
 
Join Date: Oct 2007
Location: rhode island
Posts: 37,031
Isn't the question asking the probability of two sixes being rolled?
  #5  
Old 10-06-2016, 10:04 AM
LSLGuy LSLGuy is online now
Charter Member
 
Join Date: Sep 2003
Location: Southeast Florida USA
Posts: 19,691
Quote:
Originally Posted by Skammer View Post
...
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 View Post
... 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.

Last edited by LSLGuy; 10-06-2016 at 10:07 AM.
  #6  
Old 10-06-2016, 10:10 AM
LSLGuy LSLGuy is online now
Charter Member
 
Join Date: Sep 2003
Location: Southeast Florida USA
Posts: 19,691
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..."

Last edited by LSLGuy; 10-06-2016 at 10:14 AM.
  #7  
Old 10-06-2016, 10:15 AM
watchwolf49 watchwolf49 is offline
Guest
 
Join Date: Dec 2009
Location: State of Jefferson
Posts: 7,391
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

Last edited by watchwolf49; 10-06-2016 at 10:17 AM.
  #8  
Old 10-06-2016, 10:20 AM
CharmaChameleon CharmaChameleon is offline
Guest
 
Join Date: Oct 2016
Posts: 15
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.
  #9  
Old 10-06-2016, 10:25 AM
Turble Turble is offline
Guest
 
Join Date: Dec 2007
Location: Eastern PA
Posts: 1,986
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.
  #10  
Old 10-06-2016, 10:31 AM
chrisk chrisk is offline
Charter Member
 
Join Date: Nov 2003
Location: Southern ontario
Posts: 6,566
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.
__________________
Stringing Words Forum
Aspiring writers and authors supporting each other.
Goals and resolutions our particular specialty - also sharing commiseration and triumphs.
Join today!

Last edited by chrisk; 10-06-2016 at 10:35 AM.
  #11  
Old 10-06-2016, 10:33 AM
Cartoonacy Cartoonacy is offline
Guest
 
Join Date: Dec 2003
Location: Long Island, NY
Posts: 1,520
Quote:
Originally Posted by Skammer View Post
"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.
  #12  
Old 10-06-2016, 10:37 AM
Cartoonacy Cartoonacy is offline
Guest
 
Join Date: Dec 2003
Location: Long Island, NY
Posts: 1,520
Quote:
Originally Posted by chrisk View Post
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.
  #13  
Old 10-06-2016, 10:38 AM
Ignotus Ignotus is offline
Guest
 
Join Date: Dec 2012
Posts: 978
Quote:
Originally Posted by LSLGuy View Post
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!"
  #14  
Old 10-06-2016, 10:48 AM
TriPolar TriPolar is offline
Guest
 
Join Date: Oct 2007
Location: rhode island
Posts: 37,031
Quote:
Originally Posted by TriPolar View Post
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.
  #15  
Old 10-06-2016, 10:48 AM
watchwolf49 watchwolf49 is offline
Guest
 
Join Date: Dec 2009
Location: State of Jefferson
Posts: 7,391
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.
  #16  
Old 10-06-2016, 10:53 AM
carnivorousplant carnivorousplant is offline
KB not found. Press any key
Charter Member
 
Join Date: Apr 2000
Location: Central Arkansas
Posts: 53,926
Quote:
Originally Posted by Cartoonacy View Post
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.
  #17  
Old 10-06-2016, 10:54 AM
chrisk chrisk is offline
Charter Member
 
Join Date: Nov 2003
Location: Southern ontario
Posts: 6,566
Quote:
Originally Posted by watchwolf49 View Post
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)
  #18  
Old 10-06-2016, 10:59 AM
chrisk chrisk is offline
Charter Member
 
Join Date: Nov 2003
Location: Southern ontario
Posts: 6,566
Quote:
Originally Posted by Cartoonacy View Post
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.
  #19  
Old 10-06-2016, 10:59 AM
Thudlow Boink Thudlow Boink is offline
Charter Member
 
Join Date: May 2000
Location: Lincoln, IL
Posts: 23,862
Quote:
Originally Posted by Skammer View Post
"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.
  #20  
Old 10-06-2016, 11:07 AM
x-ray vision x-ray vision is offline
Charter Member
 
Join Date: Oct 2002
Location: N.J.
Posts: 5,046
Quote:
Originally Posted by Thudlow Boink View Post
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.

Last edited by x-ray vision; 10-06-2016 at 11:08 AM.
  #21  
Old 10-06-2016, 11:10 AM
Thudlow Boink Thudlow Boink is offline
Charter Member
 
Join Date: May 2000
Location: Lincoln, IL
Posts: 23,862
Quote:
Originally Posted by chrisk View Post
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.
  #22  
Old 10-06-2016, 11:10 AM
chrisk chrisk is offline
Charter Member
 
Join Date: Nov 2003
Location: Southern ontario
Posts: 6,566
Quote:
Originally Posted by x-ray vision View Post
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.
  #23  
Old 10-06-2016, 11:16 AM
x-ray vision x-ray vision is offline
Charter Member
 
Join Date: Oct 2002
Location: N.J.
Posts: 5,046
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.

Last edited by x-ray vision; 10-06-2016 at 11:19 AM.
  #24  
Old 10-06-2016, 11:17 AM
markn+ markn+ is offline
Guest
 
Join Date: Feb 2015
Posts: 1,047
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
  #25  
Old 10-06-2016, 11:18 AM
md2000 md2000 is offline
Guest
 
Join Date: Feb 2009
Posts: 13,071
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.

Last edited by md2000; 10-06-2016 at 11:19 AM.
  #26  
Old 10-06-2016, 11:18 AM
Thudlow Boink Thudlow Boink is offline
Charter Member
 
Join Date: May 2000
Location: Lincoln, IL
Posts: 23,862
Quote:
Originally Posted by x-ray vision View Post
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.

Last edited by Thudlow Boink; 10-06-2016 at 11:19 AM.
  #27  
Old 10-06-2016, 11:27 AM
x-ray vision x-ray vision is offline
Charter Member
 
Join Date: Oct 2002
Location: N.J.
Posts: 5,046
I'm blaming my confusion on LSLGuy.

Quote:
Originally Posted by LSLGuy View Post
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
  #28  
Old 10-06-2016, 11:32 AM
LSLGuy LSLGuy is online now
Charter Member
 
Join Date: Sep 2003
Location: Southeast Florida USA
Posts: 19,691
Nope. Anything involving two 6,6 pairs is wrong from the git-go.
  #29  
Old 10-06-2016, 11:39 AM
watchwolf49 watchwolf49 is offline
Guest
 
Join Date: Dec 2009
Location: State of Jefferson
Posts: 7,391
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.

Last edited by watchwolf49; 10-06-2016 at 11:42 AM.
  #30  
Old 10-06-2016, 11:41 AM
Ignotus Ignotus is offline
Guest
 
Join Date: Dec 2012
Posts: 978
Quote:
Originally Posted by markn+ View Post
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?
  #31  
Old 10-06-2016, 12:07 PM
watchwolf49 watchwolf49 is offline
Guest
 
Join Date: Dec 2009
Location: State of Jefferson
Posts: 7,391
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.
  #32  
Old 10-06-2016, 12:07 PM
OldGuy OldGuy is offline
Charter Member
 
Join Date: Dec 2002
Location: Very east of Foggybog, WI
Posts: 4,391
Quote:
Originally Posted by Thudlow Boink View Post
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.
  #33  
Old 10-06-2016, 12:14 PM
TroutMan TroutMan is offline
Guest
 
Join Date: Sep 2008
Location: Portland, OR
Posts: 3,553
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."
  #34  
Old 10-06-2016, 12:15 PM
x-ray vision x-ray vision is offline
Charter Member
 
Join Date: Oct 2002
Location: N.J.
Posts: 5,046
Quote:
Originally Posted by LSLGuy View Post
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.
  #35  
Old 10-06-2016, 12:20 PM
CharmaChameleon CharmaChameleon is offline
Guest
 
Join Date: Oct 2016
Posts: 15
Quote:
Originally Posted by Ignotus View Post
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.)
  #36  
Old 10-06-2016, 12:21 PM
chrisk chrisk is offline
Charter Member
 
Join Date: Nov 2003
Location: Southern ontario
Posts: 6,566
Quote:
Originally Posted by LSLGuy View Post
Nope. Anything involving two 6,6 pairs is wrong from the git-go.
Yes, except that: (To complicate things still further )

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.
  #37  
Old 10-06-2016, 12:24 PM
DrDeth DrDeth is offline
Charter Member
 
Join Date: Mar 2001
Location: San Jose
Posts: 33,560
Quote:
Originally Posted by markn+ View Post
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.

Last edited by DrDeth; 10-06-2016 at 12:25 PM.
  #38  
Old 10-06-2016, 12:26 PM
watchwolf49 watchwolf49 is offline
Guest
 
Join Date: Dec 2009
Location: State of Jefferson
Posts: 7,391
Quote:
Originally Posted by TroutMan View Post
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.

Last edited by watchwolf49; 10-06-2016 at 12:30 PM. Reason: Crow for dinner tonight I guess ... [sigh]
  #39  
Old 10-06-2016, 12:35 PM
chrisk chrisk is offline
Charter Member
 
Join Date: Nov 2003
Location: Southern ontario
Posts: 6,566
Quote:
Originally Posted by DrDeth View Post
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.*
  #40  
Old 10-06-2016, 12:53 PM
Shagnasty Shagnasty is offline
Charter Member
 
Join Date: May 2000
Posts: 26,771
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.
  #41  
Old 10-06-2016, 01:04 PM
markn+ markn+ is offline
Guest
 
Join Date: Feb 2015
Posts: 1,047
Quote:
Originally Posted by Shagnasty View Post
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

Last edited by markn+; 10-06-2016 at 01:08 PM.
  #42  
Old 10-06-2016, 01:14 PM
octopus octopus is offline
Guest
 
Join Date: Apr 2015
Posts: 5,439
Quote:
Originally Posted by Skammer View Post
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.
  #43  
Old 10-06-2016, 01:16 PM
octopus octopus is offline
Guest
 
Join Date: Apr 2015
Posts: 5,439
Quote:
Originally Posted by TriPolar View Post
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.
  #44  
Old 10-06-2016, 01:26 PM
DrDeth DrDeth is offline
Charter Member
 
Join Date: Mar 2001
Location: San Jose
Posts: 33,560
Quote:
Originally Posted by TriPolar View Post
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.
  #45  
Old 10-06-2016, 01:28 PM
DrDeth DrDeth is offline
Charter Member
 
Join Date: Mar 2001
Location: San Jose
Posts: 33,560
Quote:
Originally Posted by octopus View Post
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?
  #46  
Old 10-06-2016, 01:34 PM
octopus octopus is offline
Guest
 
Join Date: Apr 2015
Posts: 5,439
Quote:
Originally Posted by DrDeth View Post
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.
  #47  
Old 10-06-2016, 01:39 PM
watchwolf49 watchwolf49 is offline
Guest
 
Join Date: Dec 2009
Location: State of Jefferson
Posts: 7,391
Quote:
Originally Posted by octopus View Post
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 ...
  #48  
Old 10-06-2016, 01:41 PM
Colophon Colophon is online now
Guest
 
Join Date: Sep 2002
Location: Hampshire, England
Posts: 13,331
Quote:
Originally Posted by octopus View Post
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.

Last edited by Colophon; 10-06-2016 at 01:45 PM.
  #49  
Old 10-06-2016, 01:42 PM
Musicat Musicat is offline
Charter Member
 
Join Date: Oct 1999
Location: Sturgeon Bay, WI USA
Posts: 19,901
Quote:
Originally Posted by Skammer View Post
"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.
  #50  
Old 10-06-2016, 01:42 PM
Turble Turble is offline
Guest
 
Join Date: Dec 2007
Location: Eastern PA
Posts: 1,986
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.
Reply

Bookmarks

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is Off
HTML code is Off

Forum Jump


All times are GMT -5. The time now is 06:31 AM.

Powered by vBulletin® Version 3.8.7
Copyright ©2000 - 2017, vBulletin Solutions, Inc.

Send questions for Cecil Adams to: cecil@chicagoreader.com

Send comments about this website to: webmaster@straightdope.com

Terms of Use / Privacy Policy

Advertise on the Straight Dope!
(Your direct line to thousands of the smartest, hippest people on the planet, plus a few total dipsticks.)

Publishers - interested in subscribing to the Straight Dope?
Write to: sdsubscriptions@chicagoreader.com.

Copyright © 2017 Sun-Times Media, LLC.

 
Copyright © 2017