Are you saying that when the 6 is removed it’s not shown to confirm that statement A is correct? I.e. showing and removing the 6 proves that at least one 6 is rolled. That’s how I interpret this. Who doesn’t have dice and a friend and can’t replicate the scenario?
And those people who keep saying 1/11 can’t count. The person rolling, showing, and removing a 6 has 2 6’s to show in the double 6 result.
I rolled dice til I got 20 sets of double 6s and I took the result of how many rolls it took that included at least one 6.
My ratio is 20/132. Which is 1/6.6. I’m going to keep rolling til I get 20 more. Brb.
I’m not sure what you think you’re arguing with me about, but your experimental result does surprise me!
Are you saying that you rolled a pair of dice many times and had 20 instances of double-six and 112 instances (132-20) of a single six? Your result is not impossible, but is surprising. On average single-sixes will outnumber double-sixes ten to one!
But the chances in simultaneous throw of green-2 and red-6 are exactly the same as red-2 and green-6, so that would not change anything, and can be ignored as a duplicating variation. Your throw of a red-2 gets absorbed as the mirrored green-2, canceling the roll of the other die which cannot be a 6 if the roll does not occur. Only if the odds are different if the 6 is green, would your argument hold sway.
Once a die of any color is a 6, the experiment proceeds with an examination of the other die, which will always be independent of the first die. Say the two dice are thrown in separate rooms, and throws continue until one of the operators announces “I have a 6”. The operator in the other room examines his single die roll. What are the chances that it will be a 6? But one of the operators falls asleep, and is awakened by the first one saying “I have a 6”, and suddenly rolls his die for the first time. Do the odds change?
I use Monte Carlo simulations a lot — in problems where the number of possibilities is too numerous for exhaustive counting.
Here there are only 11 (or, depending on how you count, 36, or 4, or just 3) possibilities. Why not just list them:
6 & Not-6 … 5 cases
Not-6 & 6 … 5 cases
6 & 6 … 1 case
(Similarly, when I tell you a pair of red Kings will make a better hand than a pair of black Aces 18.553597959240882460% of the time in all-in Texas Hold-em, that’s the result of enumeration, not simulation.)
Sure, those odds are the same, but what’s the duplicating variation to red-6, green-6? You can’t factor out something (in this case, a factor of 2) unless it appears in all terms.
“A 6 is removed.” This might mean “Monty picks one of the two dice at random and shows it to you. It happens to be a six.” Now the answer is 1/6.
It’s similar to the ambiguity in the Monty Hall problem. “Monty opens a curtain. This exposes a goat.” Did Monty deliberately avoid the curtain with the brand-new Maserati?
This is such an obvious mistake! So there are 6 possibilities? That in no way shape or form means they must be equally probable!
This is the second most common mistake made in the Monty Hall problem. (The first is misstating the problem.) You see two remaining doors and think they are equally likely to have the big prize.
Look at any of the tables of eleven possible situations in this thread. Note that “the other is a 6” happens only one out of eleven times. For each of the others, they happen 2 out of eleven times.
Given that the setup leaves 11 possibilities, how on Earth do you assign 1/6 to the (6,6) case but only 1/12 to each of (3,6) and (6,3). How can that possibly be??? The chances of rolling an (a,b) and a (c,d) are the same.
You don’t need to write any stupid code either. The listing of the possibilities, all equally likely, proves the answer without any further debate.
How does that defy what we know about probability? That’s pretty basic. The question in this thread is a little more nuanced and has to do with exactly how it is phrased. I, too, agree that 1/11 is the answer for the way it is phrased in the OP.
Yes, I know what he wrote. Skammer was responding to two posters regarding how he phrased the problem. It is regarding one roll where at least one 6 is showing. I was responding to you regarding how to calculate probability. You wrote:
The only way to get a 1/11 result is by NOT counting some of the results.
Are you not talking about not counting the times a roll shows a 6? If you are, you’re correct. You don’t include the possible rolls of two dice where at least one 6 isn’t showing.
I’m not having anything both ways.
Oh, cmon. If I buy a lottery ticket, I either win or lose. That gives me a 50/50 chance. Gotta run. Off to the convenience store!
It’s not ambiguous as written. You are told that one of the dice is definitely a 6, before that 6 is removed. So you know that the person telling you knows for a fact there is a 6, so 25 of the 36 possibilities we had at the outset are immediately discounted.
That leaves 11 possibilities, only one of which will result in the second die also being a 6. People are looking for complications where there are none.
11? How does a single die with 6 faces allow for 11 possibilities? Remember, we have eliminated everything except a single die. Nothing that comes before this is of any importance.
Yes, it is ambiguous because you don’t know how 6 was selected. Try the problem worded this way: two dice are rolled, a person peeks at one and sees it is 6. He then says, “at least one die is a 6.” What are the odds the other is a 6? The answer is 1/6.
This is a valid interpretation of the problem in the OP, without knowing if 6 was chosen in advance (answer of 1/11) or after looking at a die (1/6).
The question is not ambiguous or poorly worded; it does contain information that is not needed to find the answer. That is a very common thing in math word problems. When solving real world problems it is always necessary to sort out which information is relevant and what can be ignored, and to also figured what is the actual question to be answered.
Two fair dice are rolled: Irrelevant. It makes no difference how many dice are rolled.
The dice are rolled together: Irrelevant. It doesn’t matter if the dice are rolled together or if one is rolled on your desk and the other is rolled while dancing naked around a fire of willow sticks in a cemetery on All Hallows Eve.
One of the dice is a 6: Irrelevant. Rolls of dice are independent events.
A 6 is removed: Irrelevant. The number on one die in no way alters the outcome of any other die at any place in time or space.
The probability of the [other] die showing a six: Is the actual question.
The probability of a fair six-sided die showing a 6 (or any other number from 1 to 5) is always 1/6; it is an independent event and there is no force that can change that to 1/11 or any other number. Any attempt to interpret the question to allow such an answer must be an incorrect interpretation because it leads to the nonsensical event where the odds of an independent event are altered.
