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#51
10-06-2016, 01:49 PM
 markn+ Guest Join Date: Feb 2015 Location: unknown; Speed: exactly 0 Posts: 1,239
Quote:
 Originally Posted by Turble 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.
1/11 is the answer if you check both dice. If you only check one specific die the answer would be 1/6.

--Mark
#52
10-06-2016, 01:53 PM
 TriPolar Guest Join Date: Oct 2007 Location: rhode island Posts: 38,138
I ran a simulation. Two random numbers between 1 and 6 are defined. If neither is a 6 nothing is recorded. If either is a 6 I recorded the other one. After 1 million rolls I get this:

r(1)=55549
r(2)=55650
r(3)=55540
r(4)=55371
r(5)=55591
r(6)=27832

total=305533

total/27832=10.97775941362460477

so it has to be 1/11.
#53
10-06-2016, 01:53 PM
 DrDeth Charter Member Join Date: Mar 2001 Location: San Jose Posts: 34,863
Quote:
 Originally Posted by octopus If it's only asked when one is a 6? Then the answer is 1/6. Independent each have 1/6 chance.

Yes, and that is one way of reading it.

This debate is pointless.

The question is so poorly worded there are two equally possible answers.

Saying loudly "MY ANSWER IS THE ONE TRUE RIGHT ANSWER" doesnt help when the question allows for two equally right answers.
#54
10-06-2016, 01:55 PM
 DrDeth Charter Member Join Date: Mar 2001 Location: San Jose Posts: 34,863
Quote:
 Originally Posted by TriPolar I ran a simulation. Two random numbers between 1 and 6 are defined. If neither is a 6 nothing is recorded. If either is a 6 I recorded the other one. After 1 million rolls I get this: r(1)=55549 r(2)=55650 r(3)=55540 r(4)=55371 r(5)=55591 r(6)=27832 total=305533 total/27832=10.97775941362460477 so it has to be 1/11.
If one die is always a six, then what is the %? 1 in 6.

You can run simulations all day, but without the right question, you cant get the right answer.
#55
10-06-2016, 01:58 PM
 TroutMan Guest Join Date: Sep 2008 Location: Portland, OR Posts: 3,914
Quote:
 Originally Posted by Turble 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.
This is exactly what I did - if neither die is a six, the roll is discarded. The answer is 1/11.
#56
10-06-2016, 02:07 PM
 Skammer Charter Member Join Date: Aug 2002 Location: Music City USA Posts: 14,057
Okay, thanks for all the discussion. The way both I interpreted the question, as the OP:

1. Two (normal, fair, six-sided) dice are rolled. The result can be one of any 36 possible combinations. You don't know the results.
2. You are told that at least one of the dice is a six. Could be either die or both.
3. You are shown one of the dice, which is the (or a) six. What are the chances the other one is a six?

Thanks to Troutman and Shagnasty and whomever else ran the simulations and confirmed my 1/11 response based on those assumptions.
#57
10-06-2016, 02:09 PM
 Cartoonacy Guest Join Date: Dec 2003 Location: Long Island, NY Posts: 1,520
Quote:
 Originally Posted by octopus 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.
In other words, you're saying that the possibilities are:

R G REMOVED
1 6 Green
2 6 Green
3 6 Green
4 6 Green
5 6 Green
6 6 Green
6 6 Red
6 5 Red
6 4 Red
6 3 Red
6 2 Red
6 1 Red

Hmmm -- not the way I'd have worked it out, but I'll go along with it.

Last edited by Cartoonacy; 10-06-2016 at 02:10 PM.
#58
10-06-2016, 02:10 PM
 octopus Guest Join Date: Apr 2015 Posts: 6,053
Quote:
 Originally Posted by TroutMan This is exactly what I did - if neither die is a six, the roll is discarded. The answer is 1/11.
And I bet the simulation has this flaw. Counting the double six event wrong.

The case in which two sixes are rolled needs to be counted twice because the way that event is handled has two permutations.

Permutation one is when die 1 is removed and shown. Permutation two is when die 2 is removed and shown. Those are not the same event.

If one die is red and one die is green there are 6 events where you show a green 6. There are 6 events you show a red 6. These are the 12 events possible. Only two events have a 6 remaining and those are show red 6 while holding a green 6 or show green 6 while holding a red 6. These are distinct events and need to be counted as such.

That is 2/12. Or 1/6. 1/11 can only occur when you treat different events as one event. Showing the red is not the same as showing the green.

Last edited by octopus; 10-06-2016 at 02:12 PM.
#59
10-06-2016, 02:10 PM
 Skammer Charter Member Join Date: Aug 2002 Location: Music City USA Posts: 14,057
Quote:
 Originally Posted by DrDeth If one die is always a six, then what is the %? 1 in 6.
One die that always rolls a six is not a normal, fair die.
#60
10-06-2016, 02:12 PM
 Skammer Charter Member Join Date: Aug 2002 Location: Music City USA Posts: 14,057
octopus, you are wrong. The event where the red die is 6 and the green die is six is the SAME EVENT as the green die being six and the red die being six. You can only count it once. If you play out every one of the 36 possible combinations of two dice you only get (6,6) once.
#61
10-06-2016, 02:12 PM
 TriPolar Guest Join Date: Oct 2007 Location: rhode island Posts: 38,138
Quote:
 Originally Posted by DrDeth If one die is always a six, then what is the %? 1 in 6. You can run simulations all day, but without the right question, you cant get the right answer.
From the OP: "Two fair dice are rolled together"

So one die isn't always a six. When one die is a six it's 1/11.
#62
10-06-2016, 02:20 PM
 TroutMan Guest Join Date: Sep 2008 Location: Portland, OR Posts: 3,914
Quote:
 Originally Posted by Skammer octopus, you are wrong. The event where the red die is 6 and the green die is six is the SAME EVENT as the green die being six and the red die being six. You can only count it once. If you play out every one of the 36 possible combinations of two dice you only get (6,6) once.
Right - octopus and others are thinking of the selection of the die to remove as the event, therefore removing one die versus the other splits it into two cases. But the roll of the dice is the event we are considering, and there's only one 6-6.
#63
10-06-2016, 02:24 PM
 pulykamell Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 43,102
The thread on stackexchange seems to agree with the 1/11 answer given the way this question is phrased.
#64
10-06-2016, 02:25 PM
 CurtC Guest Join Date: Dec 1999 Location: Texas Posts: 6,483
Quote:
 Originally Posted by octopus 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.
No, there are eleven. You counted one of them twice: the (6,6) roll.
#65
10-06-2016, 02:26 PM
 TroutMan Guest Join Date: Sep 2008 Location: Portland, OR Posts: 3,914
I'm not really getting those who argue the question is poorly worded. Tell me where this breaks down or there is ambiguity:

1. Two dice are rolled => there are 36 possible outcomes
2. You are told that at least one of them is a six => this removes 25 of the possible outcomes, leaving 11
3. A die with a six is removed => of the 11 possible outcomes, only one has two sixes, therefore there is 1/11 chance that the die that remains is a six.

I am willing to accept that there is another interpretation, but I'm not seeing it. Arguing that the roll of one die is unrelated to the roll of the other ignores the information we are given that ties them together.

Enlighten me.
#66
10-06-2016, 02:30 PM
 octopus Guest Join Date: Apr 2015 Posts: 6,053
Quote:
 Originally Posted by Skammer One die that always rolls a six is not a normal, fair die.
Yeah. But you said normal, fair die as a given. So that implies that only after a 6 is rolled on at least one die that the question will be asked.

There are with ordered rolls or red and green rolls 6^2 permutations. Correct?

1-1, 1-2. 1-3...6-6. Therefore 36 total possibilities.

Quote:
 Originally Posted by Skammer octopus, you are wrong. The event where the red die is 6 and the green die is six is the SAME EVENT as the green die being six and the red die being six. You can only count it once. If you play out every one of the 36 possible combinations of two dice you only get (6,6) once.
Right you only have 1 event in which two sixes are rolled. That's obvious. What's not obvious is when you list each event individually and notice that you can choose which 6 to show in the case of two sixes. There are two such choices possible. Those must be counted. Showing the red 6 is different than showing the green 6. Treating the dice this way makes it easier to come to the right answer which is each die has a 1/6 chance of ever being a 6 regardless of what the other die shows.

Probability is basically number of permutations that match your criteria divided by total number of permutations. The trick is counting them properly.

Here's another question. You have a million dice. You roll them all and select one randomly. What's the probability you pick one that shows a 6? Calculate that. What effect does the 999,999 dice you didn't select have on what you did select? The answer is 1/6 regardless of 2, tree-fiddy, a million, or a billion dice.
#67
10-06-2016, 02:40 PM
 octopus Guest Join Date: Apr 2015 Posts: 6,053
Quote:
 Originally Posted by CurtC No, there are eleven. You counted one of them twice: the (6,6) roll.
No I didn't count them twice.

There are twelve different events.

6 events where Red comes up as 6 and Green comes up as 1-6 and you show Red.
6 events where Green comes up as 6 and Red comes up as 1-6 and you show Green.

Showing Red is obviously a different event than showing Green. Get some dice and try it. Red is obviously a different shade. Picking it out is a different event. Try it you'll see. Or use a quarter and a nickel and do coin flips.

Q N

T T
T H
H T
H H

If the question was I flipped a quarter and a nickel and one coin was head what is the probability of the other coin being head?

Add them up Q has two head events. Nickel has two head events. There are 4 possible ways I can combine the act of showing showing you a head with a flip of two coins.

Show Q N
Quarter H T
Quarter H H
Nickel T H
Nickel H H

4 ways of showing yet only two events out of the 4 have double head. Thus you get the expected value of 1/2 for a fair coin coming up heads. Showing a nickel is not the same thing as showing a quarter. You must treat them as separate events.

Tell me what physical property of the universe changes the probability of a quarter coming up heads because a nickel was flipped? Some form of monetary quantum entanglement that I'm not aware of?

You guys can thank professor octopus when you do the experiment.
#68
10-06-2016, 02:41 PM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 75,194
I'll join in with the "problem is ambiguous" crowd, here. There are multiple possible experiments that are all consistent with the problem statement, and which have different answers. Here are two examples:

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?

Answer: The probability that the other die is a 6 is 1 in 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?

Answer: The probability that both dice are 6 is 1 in 11.
#69
10-06-2016, 02:41 PM
 TroutMan Guest Join Date: Sep 2008 Location: Portland, OR Posts: 3,914
Quote:
 Originally Posted by octopus Right you only have 1 event in which two sixes are rolled. That's obvious. What's not obvious is when you list each event individually and notice that you can choose which 6 to show in the case of two sixes. There are two such choices possible. Those must be counted. Showing the red 6 is different than showing the green 6.
No, this isn't correct. Thinking of a red and green die separately might seem easier, but it's wrong. You don't know what die was removed, and it doesn't matter. Removing one die or the other doesn't magically split this single event into two. If you roll two dice, there are only 36 outcomes, and only one of those has two sixes.

Quote:
 Here's another question. You have a million dice. You roll them all and select one randomly. What's the probability you pick one that shows a 6? Calculate that. What effect does the 999,999 dice you didn't select have on what you did select? The answer is 1/6 regardless of 2, tree-fiddy, a million, or a billion dice.
While this is correct, it isn't pertinent to the question asked. In our case, you are told that one of a pair of dice is a 6. This additional information eliminates some of the potential outcomes and it is no longer valid to consider the two dice as independent.
#70
10-06-2016, 02:44 PM
 TroutMan Guest Join Date: Sep 2008 Location: Portland, OR Posts: 3,914
Quote:
 Originally Posted by Chronos I'll join in with the "problem is ambiguous" crowd, here. There are multiple possible experiments that are all consistent with the problem statement, and which have different answers. Here are two examples: 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? Answer: The probability that the other die is a 6 is 1 in 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? Answer: The probability that both dice are 6 is 1 in 11.
OK, this form of ambiguity I understand and agree with. I started with the assumption that 6 was a specific number we were looking for in advance, but the question doesn't actually state that.
#71
10-06-2016, 02:47 PM
 Skammer Charter Member Join Date: Aug 2002 Location: Music City USA Posts: 14,057
Quote:
 Originally Posted by octopus Right you only have 1 event in which two sixes are rolled. That's obvious. What's not obvious is when you list each event individually and notice that you can choose which 6 to show in the case of two sixes. There are two such choices possible. Those must be counted. Showing the red 6 is different than showing the green 6.
It doesn't matter which die you show; there is only one out of 11 cases where they are both six. Selecting the other 6 from the pair change anything.

Quote:
 Here's another question. You have a million dice. You roll them all and select one randomly. What's the probability you pick one that shows a 6? Calculate that. What effect does the 999,999 dice you didn't select have on what you did select? The answer is 1/6 regardless of 2, tree-fiddy, a million, or a billion dice.
Sure, but that is a completely different experiment.
#72
10-06-2016, 02:50 PM
 Musicat Charter Member Join Date: Oct 1999 Location: Sturgeon Bay, WI USA Posts: 20,037
I just wrote a program to test this, but I can't run it until later today. Meanwhile, it seems vital to define the problem. Here's how I see it (for one iteration).

1. Roll two dice. If either is a six, continue, otherwise, DISCARD THIS ROLL and do not include it in any numbers. Repeat this step.

2. To continue...now we have two dice, one or both must be sixes. Add one to our total tally. This will become the denominator for our final odds calculation.

3. The rest is just keeping track of how many (2nd die) faces that turn up each time, from 1..6. The tally of sixes here will be the numerator for the final odds calc.

Does everyone agree so far, or have I mis-stated the problem? Step 1 is critical.

Last edited by Musicat; 10-06-2016 at 02:51 PM.
#73
10-06-2016, 02:52 PM
 jtur88 Guest Join Date: Aug 2011 Location: Cebu, Philippines Posts: 13,872
I presume the dice are rolled and rerolled until the condition is met, that there is a six. So the 36 possible rolls are not applicable, since many do not meet the condition.

Once there is a roll that meets the condition, and a pre-determined die (one with a six) that influences the condition is removed from the table, the remaining die remains perfectly random (not being involved in the decision to remove a die aor to not re-roll) and the chances are 1/6 that it is a six.

Last edited by jtur88; 10-06-2016 at 02:55 PM.
#74
10-06-2016, 03:14 PM
 Skammer Charter Member Join Date: Aug 2002 Location: Music City USA Posts: 14,057
Musicat and jtur: based on your interpretation, your answer will be 1/6. That was not my interpretation of the problem.

You: "Roll two dice until one of them is a six. What is the chance the other is a six?" Answer: 1/6
Me: "Roll two dice. If one of them is a six, what is the chance that the other is a six?" Answer: 1/11
#75
10-06-2016, 03:21 PM
 Skammer Charter Member Join Date: Aug 2002 Location: Music City USA Posts: 14,057
oops, double post.

Last edited by Skammer; 10-06-2016 at 03:22 PM.
#76
10-06-2016, 03:40 PM
 Musicat Charter Member Join Date: Oct 1999 Location: Sturgeon Bay, WI USA Posts: 20,037
Quote:
 Originally Posted by Skammer Musicat and jtur: based on your interpretation, your answer will be 1/6. That was not my interpretation of the problem. You: "Roll two dice until one of them is a six. What is the chance the other is a six?" Answer: 1/6 Me: "Roll two dice. If one of them is a six, what is the chance that the other is a six?" Answer: 1/11
I fail to see the difference between lines You & Me. Either way, it's an IF statement in a loop. UNTIL is the same as IF in the structure you are proposing; both are conditional and have the same true/false decision to make. (I'm thinking computer program logic.)

Last edited by Musicat; 10-06-2016 at 03:41 PM.
#77
10-06-2016, 03:41 PM
 octopus Guest Join Date: Apr 2015 Posts: 6,053
Quote:
 Originally Posted by Skammer Musicat and jtur: based on your interpretation, your answer will be 1/6. That was not my interpretation of the problem. You: "Roll two dice until one of them is a six. What is the chance the other is a six?" Answer: 1/6 Me: "Roll two dice. If one of them is a six, what is the chance that the other is a six?" Answer: 1/11
Well the fact that you were showed a 6 means that one of them is a 6. You are only being asked in the cases where at least one is a 6.
#78
10-06-2016, 03:43 PM
 octopus Guest Join Date: Apr 2015 Posts: 6,053
Quote:
 Originally Posted by Skammer It doesn't matter which die you show; there is only one out of 11 cases where they are both six. Selecting the other 6 from the pair change anything. Sure, but that is a completely different experiment.
Same principle. The ONLY thing that REALLY matters is the dice are fair. If the dice are fair the other details are irrelevant. Each die has a 1/6 regardless of the state of the other die to show a 6.

You aren't actually even doing the procedure right. If you did, with colored dice, instead of trying to figure it out in your head with imprecise notions, you'd get 1/6 which is the theoretically right answer.
#79
10-06-2016, 03:50 PM
 pulykamell Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 43,102
Did anybody read the Stackexchange thread I linked to above? The first two answers cover it to my satisfaction, with all the nuances.
#80
10-06-2016, 03:51 PM
 Musicat Charter Member Join Date: Oct 1999 Location: Sturgeon Bay, WI USA Posts: 20,037
Quote:
 Originally Posted by Skammer You: "Roll two dice until one of them is a six. What is the chance the other is a six?" Answer: 1/6 Me: "Roll two dice. If one of them is a six, what is the chance that the other is a six?" Answer: 1/11
I think I see the problem here. As I understand the problem, anytime we roll two dice and NEITHER is a six, I would discard this roll completely as if it never happened, where you would include it in the total roll count. This is the denominator I spoke about a few posts ago. With different denominators, we get different ratios.
#81
10-06-2016, 03:57 PM
 octopus Guest Join Date: Apr 2015 Posts: 6,053
Quote:
 Originally Posted by pulykamell The thread on stackexchange seems to agree with the 1/11 answer given the way this question is phrased.
Most of the folks are saying 1/6. Someone even has a table where the post below corrects the interpretation of the table to count the double 6 properly.
#82
10-06-2016, 04:07 PM
 N9IWP Charter Member Join Date: Aug 2001 Location: Southeast MN Posts: 5,801
Quote:
 Originally Posted by Musicat There's your problem. No die has a ten on any face.
Actually is it stated anywhere that the dice are 6 sided? That is the most common configuration, but not teh only one.

If one die is >= 6 sides, and the other is < 6:
If one die shows 6, the odds are 0 that other one does (assuming it is numbered normally, and not 3,4,5,6 or something)

Brian
#83
10-06-2016, 04:12 PM
 octopus Guest Join Date: Apr 2015 Posts: 6,053
Quote:
 Originally Posted by N9IWP Actually is it stated anywhere that the dice are 6 sided? That is the most common configuration, but not teh only one. If one die is >= 6 sides, and the other is < 6: If one die shows 6, the odds are 0 that other one does (assuming it is numbered normally, and not 3,4,5,6 or something) Brian
Good point. D&D dice make probability fun.
#84
10-06-2016, 04:12 PM
 Turble Guest Join Date: Dec 2007 Location: Eastern PA Posts: 2,066
I just rolled two dice. One of them shows a 6.

What are the chances the other die shows a 6?
What are the chances the other die shows a 3?

Thinking the answers will be different is nonsensical.

Those who have written simulations giving the 1/11 result have written flawed programs. Musicat appears to be on the right track.
#85
10-06-2016, 04:15 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,521
Quote:
 Originally Posted by Musicat I think I see the problem here. As I understand the problem, anytime we roll two dice and NEITHER is a six, I would discard this roll completely as if it never happened, where you would include it in the total roll count.
No, I don't think that's the problem; everyone seems to agree on that.

The interpretation where 1/6 is the correct answer: Roll two dice. If the red die is a 6, what is the probability that the green die is a 6? Equivalently, if the green die is a 6, what is the probability that the red die is a 6.

The interpretation where 1/11 is the correct answer: Roll two dice. If at least one of the dice is a 6, what is the probability that both are 6s?

In both interpretations, we are not considering rolls where neither is a 6.
#86
10-06-2016, 04:20 PM
 Skammer Charter Member Join Date: Aug 2002 Location: Music City USA Posts: 14,057
No, not wrong track - just a different understanding of the problem (which matches my understanding, as the OP).

Roll two dice. You could get any combination at all. 36 possibilities. If you roll and I said "out of all the things you could have rolled, you rolled at least one six" then the chances the other die is a six is 1/11. By the way, the chance it is a three is 2/11.

Now, if I said "That roll doesn't count because you didn't roll a six. Roll again" until you rolled a six, the chance that the other die is also a six is 1/6. The chance of a three is also 1/6.

ETA: Thudlow Boink is also correct. If I said "the red die is a six. What are the odds that the green die is a six?" the answer is 1/6. But that's not the question.

Last edited by Skammer; 10-06-2016 at 04:22 PM.
#87
10-06-2016, 04:21 PM
 CurtC Guest Join Date: Dec 1999 Location: Texas Posts: 6,483
Quote:
 Originally Posted by octopus No I didn't count them twice. There are twelve different events. 6 events where Red comes up as 6 and Green comes up as 1-6 and you show Red. 6 events where Green comes up as 6 and Red comes up as 1-6 and you show Green.
You've counted the (Red=6,Green=6) event twice here.

A nice thing with probability is that you can list out all the mutually exclusive events that have an equal chance of happening, and then you can just count them to determine the probability you're looking for. There are 36 ways for that pair of red and green dice to be rolled. Eleven of them will have at least one six showing. Those are:

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

When you did it, you listed the (6,6) event twice.
#88
10-06-2016, 04:23 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,521
Quote:
 Originally Posted by Turble Those who have written simulations giving the 1/11 result have written flawed programs.
Please, show us your "unflawed" program, or at least the algorithm it's based on.
#89
10-06-2016, 04:32 PM
 Musicat Charter Member Join Date: Oct 1999 Location: Sturgeon Bay, WI USA Posts: 20,037
Let's return to the OP. I still haven't run my program yet, but I think I know the answer.
Quote:
 Originally Posted by Skammer "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?"
Now break that down:

"Two fair dice are rolled together..." One roll, and one roll only.

"...and you are told that 'at least one of the dice is a 6.'" We now have a simple odds situation. Of two dice, one is a 6. If it is not, we don't have a question and this exercise is finito.

"A 6 is removed,..." So this die is no longer involved. Fuggitaboudit. Only one die remaining, in so far, an indeterminate state.

"...and you are shown the other die." We have a simple problem, which can be reduced to the question, "what are the odds of a single die, in a single throw, being a 6?" Nothing else that happened before matters, unlike the boy/girl problem, or the Monty Hall door problem.

"What is the probability of the...die showing a six?" Six possibilities, one possible outcome. Answer: 1/6

If you don't agree with me, please show me where I have defined the problem wrong, or how the first parts of this problem affect the final question.

Last edited by Musicat; 10-06-2016 at 04:34 PM. Reason: ETA: There is no conditional here, unlike the boy/girl or Monty Hall problem.
#90
10-06-2016, 04:38 PM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 75,194
The phrase "the other die" reveals a lot: It's only meaningful in one interpretation of the problem. If I roll a red die and a green die, and both happen to come up 6, and I say "at least one die is a 6", which die is "the other one"? The one I didn't look at? But I looked at both. The one that didn't come up 6? But both did. The green one? Why?

In any setup of the problem where "the other die" is meaningful, the answer is 1/6. In the setups where the answer is 1/11, that phrase doesn't make sense.
#91
10-06-2016, 04:38 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,521
Quote:
 Originally Posted by Musicat "...and you are shown the other die." We have a simple problem, which can be reduced to the question, "what are the odds of a single die, in a single throw, being a 6?" Nothing else that happened before matters, unlike the boy/girl problem, or the Monty Hall door problem.
If nothing that happened before matters, then what determines which die is the "other die," the one that you are shown?
#92
10-06-2016, 04:40 PM
 octopus Guest Join Date: Apr 2015 Posts: 6,053
Quote:
 Originally Posted by CurtC You've counted the (Red=6,Green=6) event twice here. A nice thing with probability is that you can list out all the mutually exclusive events that have an equal chance of happening, and then you can just count them to determine the probability you're looking for. There are 36 ways for that pair of red and green dice to be rolled. Eleven of them will have at least one six showing. Those are: (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) (6,1) (6,2) (6,3) (6,4) (6,5) When you did it, you listed the (6,6) event twice.
That's because showing a red 6 and having a green 6 is a different event than being shown a green 6 and having a red. You can make two choices about which 6 to show in the double 6 case.

You are only showing cases where a 6 occurs. Use two different colored dice or denominations of coins and it's pretty evident they are different events.

Quote:
 Originally Posted by Thudlow Boink Please, show us your "unflawed" program, or at least the algorithm it's based on.
In Excel you can do this.

Make a column representing one die =RANDBETWEEN(1,6) do that for another column. Third column (C) is true if either are a 6. Fourth column (D) is true if both are a 6.

(C+D)/(2D) = 1 / 6

Or get a penny and a nickel and flip them 100 times and do the experiment by showing if you count the times you can show two heads by the times you get one head while discarding the events where you get double tails you get a 1/2 ratio which is analogous to 1/6 for the dice.

The reason for two different denominations is just the clarity that two different coins have in the choice of showing one to be heads has on the number of events possible. Whereas with interchangeable pennies or dice that detail is lost easily.
#93
10-06-2016, 04:43 PM
 octopus Guest Join Date: Apr 2015 Posts: 6,053
Quote:
 Originally Posted by Chronos The phrase "the other die" reveals a lot: It's only meaningful in one interpretation of the problem. If I roll a red die and a green die, and both happen to come up 6, and I say "at least one die is a 6", which die is "the other one"? The one I didn't look at? But I looked at both. The one that didn't come up 6? But both did. The green one? Why? In any setup of the problem where "the other die" is meaningful, the answer is 1/6. In the setups where the answer is 1/11, that phrase doesn't make sense.
What? What else are you looking at but the die that wasn't revealed? Aka, the "other die." And look the green and red is just to illustrate that even though the numbers are the same and they are both fair dice looking at one of them and seeing that it's a 6 is different than looking at the other and seeing that it's a 6. It could be looking at numbers in column A vs looking at numbers in column B of a spreadsheet. Those are two different peeks.
#94
10-06-2016, 04:44 PM
 borschevsky Guest Join Date: Sep 2001 Location: Canada Posts: 1,901
Quote:
 Originally Posted by octopus That's because showing a red 6 and having a green 6 is a different event than being shown a green 6 and having a red. You can make two choices about which 6 to show in the double 6 case.
But these events aren't as likely as the others. There are two ways to throw 1&6 (red 1 + green 6, and red 6 + green 1), but only one way to throw 6&6 (red 6 + green 6).
#95
10-06-2016, 04:48 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,521
Quote:
 Originally Posted by Chronos The phrase "the other die" reveals a lot: It's only meaningful in one interpretation of the problem. If I roll a red die and a green die, and both happen to come up 6, and I say "at least one die is a 6", which die is "the other one"? The one I didn't look at? But I looked at both. The one that didn't come up 6? But both did. The green one? Why? In any setup of the problem where "the other die" is meaningful, the answer is 1/6. In the setups where the answer is 1/11, that phrase doesn't make sense.
Are you assuming the Axiom of Choice?

See my formulation quoted below. The "other die" is the one that's remaining after a 6 has been removed. In the case where two sixes were originally rolled, it's irrelevant which is the one that is removed.
Quote:
 Originally Posted by Thudlow Boink 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.
#96
10-06-2016, 04:52 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,521
Quote:
 Originally Posted by octopus In Excel you can do this. Make a column representing one die =RANDBETWEEN(1,6) do that for another column. Third column (C) is true if either are a 6. Fourth column (D) is true if both are a 6. (C+D)/(2D) = 1 / 6
The way I was interpreting the problem, it's equivalent to "If at least one of the dice is a 6, what is the probability that both are sixes?" The answer to this question would be D/C, which turns out experimentally to be quite close to 1/11.

What's the explanation for your (C+D)/(2D) formula?
#97
10-06-2016, 05:07 PM
 octopus Guest Join Date: Apr 2015 Posts: 6,053
Quote:
 Originally Posted by borschevsky But these events aren't as likely as the others. There are two ways to throw 1&6 (red 1 + green 6, and red 6 + green 1), but only one way to throw 6&6 (red 6 + green 6).
Yeah but in a chart of events where a die is being a revealed and a die is being concealed there are 12 permutations which I listed above will do so again.

Revealed Red result Green Result

Red 6 1
Red 6 2
Red 6 3
Red 6 4
Red 6 5
Red 6 6
Green 1 6
Green 2 6
Green 3 6
Green 4 6
Green 5 6
Green 6 6

Out of the 12 possibilities of a particular die being revealed and a particular die being concealed there are only 2 that have two 6s. 2/12 = 1/6.

If you do this as a real life experiment with two people. You will see that 1/6 of the time when at least one six is rolled that the concealed die is a 6. Do it with real dice and two people.
#98
10-06-2016, 05:08 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,384
Quote:
 Originally Posted by Musicat Let's return to the OP. I still haven't run my program yet, but I think I know the answer.Now break that down: "Two fair dice are rolled together..." One roll, and one roll only. "...and you are told that 'at least one of the dice is a 6.'" We now have a simple odds situation. Of two dice, one is a 6. If it is not, we don't have a question and this exercise is finito. "A 6 is removed,..." So this die is no longer involved. Fuggitaboudit. Only one die remaining, in so far, an indeterminate state.
Then you must have removed a die from one of the following 11 rolls, all having an equal chance of having occurred:

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

There is a 1/11 chance of the other die being a 6.
#99
10-06-2016, 05:13 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,521
Quote:
 Originally Posted by octopus Yeah but in a chart of events where a die is being a revealed and a die is being concealed there are 12 permutations which I listed above will do so again.
When the dice are thrown, there are 11 possible outcomes, all equally likely.

In the case of (6, 6), there are two possible ways that one of the dice could be revealed, but that doesn't make that outcome more likely to have come up in the first place.
#100
10-06-2016, 05:16 PM
 borschevsky Guest Join Date: Sep 2001 Location: Canada Posts: 1,901
Quote:
 Originally Posted by octopus If you do this as a real life experiment with two people. You will see that 1/6 of the time when at least one six is rolled that the concealed die is a 6. Do it with real dice and two people.
If you do it with real dice, how do you decide which die to remove when you roll two sixes? That is, how do you know which of the following events you're in?

Red 6 6
Green 6 6

Do you agree with this statement: If you roll two dice, you'll get a 1 and a 6 twice as often as you'll get two sixes.

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