Absolutely a different problem. One uses two dice, the other, three. And that’s just for starters. Saying remove the highest ones isn’t the same as removing a single six. Removing a single six – in the original problem – has no effect upon the value of the remaining one. Removing the highest values of three DOES have an effect upon the value of the remaining dice, and reduces the possibility of the remaining one being a high number.
Exactly how much, I don’t know, but that’s a problem for another day. And thread.
I’m not saying it’s the same problem: for one thing, Thudlow Boink’s question is much less ambiguous. But it (can) illustrate the same principle
Removing a six–if you would remove either die if they happen to be a six, would definitely have an effect on the remaining number. Do you believe that the original problem involves selectively removing a six and not any other number?
Really? The value of one die out of two randomly rolled dice affects the value of the other? That’ll be news to Vegas. If that’s what you believe, we are at an impasse.
Once again, I am at a location where I can’t run a program, and I’d like to simplify the one I wrote yesterday. When I do, I’ll post a version reduced to the simplest logic I can.
No, I don’t believe that the value of the green die will affect the red die, or that the value of the red die will affect the green die. I do, however, believe that the question could be interpreted as a man rolling two dice on the table, looking at the results, removing one and leaving the other. And once you have that, you can say very little about the odds of the die left on the table until you know on exactly what basis the man is making his choice.
I see two distinct topics of conversation here: one is about the wording of the problem. The apparent meaning that the puzzle creator is (obviously to me) trying to convey, to which the correct answer is 1/11, and the stickler interpretation that could perhaps be justified, to which the correct answer is 1/6.
But a separate conversation is going on by people who are taking the same interpretation as me, but who are still insisting that the answer is 1/6.
I think it would be a good next step to re-word the puzzle so that it’s not ambiguous, and then get everyone to agree on the answer. How about this for non-ambiguity?
A pair of fair dice are rolled repeatedly until at least one of them comes up as a six. When that happens, the host of the game removes the “6” die, or if there are two sixes, one of the “6” dice. What is the probability that the remaining die shows a six?
I think this is unambiguous. Does everyone agree that the correct answer is 1/6?
While I don’t find the problem that ambiguous, I thing the “removing” part of this is confusing people because that leads to people trying to solve for “the other die”. There is no this die and the other die.
Maybe we can agree on a more concise wording of the problem?
“Two fair dice are rolled. If at least one is a six, what are the odds they are both sixes?”
Do the same in column B and copy down as far as you like.
Pivot up the result with the row labels and column labels as the dice and count either one.
I just did 2,500. Die 1 is a 6 395 times. Die 2 is a 6 408 times. Of them 61 times both are a 6. So 734 where only one dice is a 6 and only 61 where both are 6.
If I refresh the pivot I get - Die 1 is a 6 385 times. Die 2 is a 6 416 times. Of them 76 times both are a 6. So 728 where only one dice is a 6 and only 76 where both are 6.
I can keep refreshing it but will never get anything like 1/6 of the combinations that include a 6 where both dice are 6.
You may find that the solution looks pretty obvious when seen as a simple pivot.
It isn’t laid out like my SDMB Craps game. It is presented as an observation of a result.
Someone throws two dice but can’t see how they end up and an observer notes there is at least one six in there. He grabs a rolled six and hands it to the thrower and asks, “What is the chance that the other die is a 6?”
“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?”
That’s why the problem says **a **six is removed and not the six is removed because sometimes there are 2 sixes.
The people caught up in the “either of the 2 sixes could be removed,” thus doubling the chances are confused about the nature of the problem. Which 6 is handed to the thrower is immaterial the question is, “is there another 6?”
And still, after all this time there are people posting here claiming the die values are independent events.
They were, at first. But the moment the poser says ‘at least one of the dice is a 6.’ that’s over. Done. They are independent no more. They wouldn’t be independent if you put four thousand volts through them.
If instead the poser said “The total of the dice is 9.” Would you still think of them as independent events? Of course not! Once you say something about a property that one or both might have, then the independence is broken.
Again this is a common problem with people and the Monty Hall problem. They don’t realize that being given more information (shown a door with a goat), that the initial conditions have changed.
Consider the following scenario: Player A secretly writes down all 36 possible rolls of two dice. Player A then erases all the pairs that don’t have a 6. Then Player A erases one 6 from each pair.
Player A then tells Player B what has happened. Player B is asked: How many numbers are on my board and how many of them are a 6? Player B responds 11, one of which is a 6. I.e., 1/11th are 6s.
Now, if the initial 36 pairs were deemed equally likely, what is that probability of that lonely 6 out of 11? 1/11. To assert a higher value requires that for some special reason the other numbers have lower odds. Why?
I didn’t ask you either of those scenarios. I asked what RedSwinglineOne just asked in post 187. If you don’t agree the answer is 1/11 when phrased that way then with the following question, what do you believe the answer is?:
From all families with two children, at least one of whom is a boy, a family is chosen at random, what is the probability that both children are boys?*
Using your methodology, you should believe the answer is 1/2. It isn’t, and no mathematicians believe that it is. If you don’t believe the answer is 1/2, what do you believe the answer is and how did you arrive at it?
Still ambiguous until you say why the observer notes that there is a six.
Maybe the observer is really fascinated by the number 6, but not other numbers. So he only makes the observation when at least one 6 appears. Answer: 1/11
Maybe the observer chose which number to note at random. Answer: 1/6
Maybe the observer is in the habit of making a note on every throw, but always chooses to reveal the lowest number. Answer: 1/1
Maybe the observer only makes a note when it is NOT a double six. Answer: 0/1
None of the those “maybes” are precluded by the OP, hence the ambiguity.
We have a bunch of marbles in a opaque bag. Some are black, some are white, and there is an equal number of each. Let’s say 50 of each.
If I reach into the bag and carefully remove 20 of the black ones only, then hand the bag to someone else and ask him to reach in the same bag, blindly, randomly, and pull out the first marble he feels, are the odds 50-50 that it will be white? No, because we have altered the composition of the bag by selectively removing some colors. The chances of picking a black one at random has been reduced.
The “toss 3 dice, then remove the top 2 values” is very similar. We don’t alter the actual value of the remaining die, but we have reduced the chances that it is a high value, because we have selectively, deliberately removed a large proportion of high values. We have not touched low values, so the remaining dice are more likely to be low values than high.
Contrast this to the OP, where one of 2 dice is removed after a toss – for any reason; it could be because of its value (1,2,3,4,5 or 6) or because it was covered with glue or dipped in paint. The value of the remaining die is not affected by the first one being present or absent under any circumstances. It stands alone, and there is one chance in six that it is of any particular value.
This is a conditional problem, and is not related to the OP whatsoever.
Note the “both” in your proposition. The OP doesn’t ask a probability of “both,” but the probability of “one” and that is why the two questions are dissimilar.
Don’t even think about the Monty Hall problem, as it is a conditional one and not relevant.
Let’s start with your statement that the die values are independent events. At first. I think we can agree on that.
Now we decide to look at both dice and see if one is a six. If none are, we are done and cannot continue with the problem, since that is what the OP says.
Otherwise we are left with 2 dice, one of which (at least) is a six. Are you claiming that "they are “independent no more”? How does the value of one affect the value of the other?
They were independent before, then you saw one was a six and they are no longer independent. How are they connected?
Here’s another way of looking at it. Let’s toss two dice at the same time (it works sequentially, too). What are the chances that ONE will be a six? 1/6, right? What are the chances that the OTHER will be a six? 1/6, right? Will tossing a 4 for the first change the odds for the second? Will tossing a 2 for the second change the odds of the first?
That’s really what the problem is, simply stated. All the other stuff in the OP is just intended to confuse you and lead you down the garden path to the wrong conclusion.