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#51




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Mark 
#52




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




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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




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You can run simulations all day, but without the right question, you cant get the right answer. 


#55




This is exactly what I did  if neither die is a six, the roll is discarded. The answer is 1/11.

#56




Okay, thanks for all the discussion. The way both I interpreted the question, as the OP:
1. Two (normal, fair, sixsided) 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




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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; 10062016 at 02:10 PM. 
#58




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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; 10062016 at 02:12 PM. 
#59




One die that always rolls a six is not a normal, fair die.



#60




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




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So one die isn't always a six. When one die is a six it's 1/11. 
#62




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#63




The thread on stackexchange seems to agree with the 1/11 answer given the way this question is phrased.

#64




No, there are eleven. You counted one of them twice: the (6,6) roll.



#65




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




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? 11, 12. 13...66. Therefore 36 total possibilities. Quote:
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, treefiddy, a million, or a billion dice. 
#67




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There are twelve different events. 6 events where Red comes up as 6 and Green comes up as 16 and you show Red. 6 events where Green comes up as 6 and Red comes up as 16 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




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




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#70




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#71




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#72




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 misstated the problem? Step 1 is critical. Last edited by Musicat; 10062016 at 02:51 PM. 
#73




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 predetermined 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 reroll) and the chances are 1/6 that it is a six. Last edited by jtur88; 10062016 at 02:55 PM. 
#74




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




oops, double post.
Last edited by Skammer; 10062016 at 03:22 PM. 
#76




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Last edited by Musicat; 10062016 at 03:41 PM. 
#77




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#78




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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




Did anybody read the Stackexchange thread I linked to above? The first two answers cover it to my satisfaction, with all the nuances.



#80




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




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#82




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




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#84




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




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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




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; 10062016 at 04:22 PM. 
#87




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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




Please, show us your "unflawed" program, or at least the algorithm it's based on.

#89




Let's return to the OP. I still haven't run my program yet, but I think I know the answer.
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"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; 10062016 at 04:34 PM. Reason: ETA: There is no conditional here, unlike the boy/girl or Monty Hall problem. 


#90




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




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#92




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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:
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




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#94




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




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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:

#96




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What's the explanation for your (C+D)/(2D) formula? 
#97




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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




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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




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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




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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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