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.
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.
The thread on stackexchange seems to agree with the 1/11 answer given the way this question is phrased.
No, there are eleven. You counted one of them twice: the (6,6) roll.
I’m not really getting those who argue the question is poorly worded. Tell me where this breaks down or there is ambiguity:
- Two dice are rolled => there are 36 possible outcomes
- You are told that at least one of them is a six => this removes 25 of the possible outcomes, leaving 11
- 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.
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.
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.
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.
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.
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.
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.
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.
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.
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).
-
Roll two dice. If either is a six, continue, otherwise, DISCARD THIS ROLL and do not include it in any numbers. Repeat this step.
-
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.
-
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.
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.
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
oops, double post.
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.)
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.
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.
Did anybody read the Stackexchange thread I linked to above? The first two answers cover it to my satisfaction, with all the nuances.
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.