Your friend is absolutely correct. It is stated that the dice are fair. That means that both of them have a 1/6 chance of rolling a 6. The result of the second die is not dependent upon the first.
Unless I am missing something that changes the original premise, I would say the odds that both are sixes in this case is 1/6. Saying “what are the odds they are both sixes” is the same, in this case only, of saying “what are the odds that the one that is [not known] is a six?” And the odds of any unknown single die roll, unrestrained by any other factor or condition, is 1/6.
Do you own a pair of dice?
Doesn’t everybody? I doubt if I have the time to devote to an experiment with sufficient trials, if that’s what you are suggesting. Can’t we let a computer do the work?
Dice won’t trick you.
What if we don’t have power? Can’t turn on a computer can we? :eek:
https://s25.postimg.org/45ptftjfj/S5030859.jpgfree screen capture
Yet you said this:
What was the point of saying that if you think you get the same answer either way?
So how is this:
*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? *
different from this:
Two fair dice are rolled. If at least one is a six, what are the odds they are both sixes?
We don’t need dice or a computer. Let’s look at the question you said the answer is 1/6 to in the above exchange.
“Two fair dice are rolled. If at least one is a six, what are the odds they are both sixes?”
These are the possible rolls, all with an equal chance of occurring:
1,6
2,6
3,6
4,6
5,6
6,6
6,5
6,4
6,3
6,2
6,1
Do you agree those are all of the possible rolls all with an equal chance of occurring? If not, why not? If you agree, how is it that the probability of 2 6s being face up in an occurrence where at least one 6 is face up not 1/11?
I think you missed something there. There are 36 possible combos with 11 chances of rolling a six.
I didn’t miss anything. All possible rolls aren’t needed when we have the information that at least one six has been rolled.
I agree with markn+'s repeated insistence that it depends on assumptions about how the statement was made. In fact, there are infinitely many possible answers, depending on which assumptions you use.
To my mind, the problem can be most clearly stated as follows:
Two dice are rolled privately. An observer then states the value of one of the dice.
We are considering only the cases in which the observer says “6”.
What is the probability that both dice are 6?
The answer depends on how the observer decided to say that one of them is a 6.
Possible observer rules:
-
Pick a die at random and say its value: Probability that they’re both 6 = 1/6
-
If either die is a 6, say 6: Probability that they’re both 6 = 1/11
These are the two assumed rules that most people are using. But there are many other possible rules. For example:
-
If there is a 3, say 3. Otherwise, if there is a 6, say 6.
p = 1/9 -
If there is a 4 or 5, and a 6, say 6 with probability 2/3. Otherwise, pick a die at random and say its value:
p = 3/20 (by my quick calculations, could be wrong)
etc.
The first two rules are perhaps the most obvious, but they are not the only possible ones.
Because you learned something about the value of the dice that either one or both might have.
Read the bold part carefully: One or both.
Surely you must agree that if the person said something about both dice, e.g., “The dice add up to 9.” they are no longer independent? Right? After all, if one is a 4 the other has to be a 5. That’s not independent.
Similarly, if you are told something about one or both that the maximum is a 3, then you know that one is a 3 and the other is a 1, 2, or 3. That’s not independent.
OTOH, saying “The red die is a 6.” in the red/green dice case indeed doesn’t say anything about the value of the green dice. It is still independent. Note that here you are only saying something about one (not one or both) dice.
(Likewise, if one said “The capital of Texas is Austin.” that’s a statement of neither dice and independence still applies. There’s a couple other scenarios I’ll spare you.)
RedSwinglineOne gives a fine analysis proving without any question whatsoever the same applies to “One is a 6.” If you can’t understand something so very, very basic (complete with a list!), then you will never, ever be able to understand basic probability and … well … you’re just never going to get this.
People who continue to oppose 1/11: What are you thinking when you see the list of 11 equal probabilities??? How can anyone possible argue anything against that???
Like with all the Monty Hall threads, I don’t understand people wasting time writing code for this. The analysis is so trivial why bother? And from my decades of experience, code doesn’t sway a lot of people. (Sometimes it didn’t sway me. I saw a lot of incorrect programs over the years. Many with formal “proofs”.)
Someone else in this thread has already pointed out that writing code can be useful, not for its output, but because the process of writing code forces you to be explicit about how you’re interpreting the problem.
Does it make it easier to intuitively solve, if the number of variables is reduced from 6 to 2?
Thee are two cats in a room, and you are told that at least one of them is black. I remove a black cat. What are the odds that the remaining cat is black?
We know that it is not true that both cats are white. So the three possibilities are BW, WB, or BB. A black one is removed, leaving, with equal chance, a white, a white, or a black. 1/3.
It’s not very useful to complicate problems of this type by raising the possibility of an observer operating under strange rules.
The OP stated:
By far the best way to proceed is to assume that, indeed, at least one is a 6.
Yes, it might be the case that the person who told you this is a quirky individual who would say he saw a 6 only if he actually saw both a 3 and a 4 - the OP didn’t specifically eliminate that possibility. But that’s really not an interesting case worth considering.
That’s equivalent to the boy/girl version, and the answer is not necessarily 1/3. Could be 1/3 or 1/2, or in fact anything between 1/3 and 1, depending on how the statement that at least one of them is black was arrived at.
For example, if the rule was “say that at least one of them is black only if both are black”, then the probability that both are black, given that that statement was actually made, becomes 1.
In other words, the assertion “leaving, with equal chance, a white, a white, or a black” (my emphasis) is not valid. It is not necessarily true that those possibilities have equal chance. It is only true under certain assumptions.
Nothing in my argument contradicts the fact of there being at least one 6, and I did not suggest the possibility of the observer making an untrue or misleading statement.
I did say that there are two reasonable assumptions. They lead to two reasonable answers, 1/11 and 1/6. Neither is definitively right or wrong.
If you’re a good programmer.
If you’re not, you’ll carry the same sloppy half-understood (or completely unrecognized) assumptions forward and so your code will do something. Whether that’s the right *something *is a different matter.
So far the folks rushing to code are the same ones glossing right over the ambiguities.
ETA: I hadn’t seen Cliffy Biro’s direct response to **Xema **when I posted. Not trying to pile on here.
You misunderstood Cliffy Biro’s point. It was not that the observer chooses to lie abut what he sees.
It’s that he *might *choose to do something other than simply always say “There is at least one 6” or “This is a do-over; ignore this roll” depending on whether at least one 6 is actually factually observed.
The whole crux of this puzzle is that the odds depend *totally *on the observer’s decision process for what he announces under what circumstances. (And parenthetically on the fact that they’re using two 6-sided fair dice).
And this is the one part of the puzzle that the puzzle writer breezily glossed over. Thereby triggering unrecognized assumptions in the audience.
Why? To provoke threads like this. The hardest part of changing people’s minds is getting them to recognize unrecognized assumptions. “As they assume so shall they conclude” was one of my favorite professor’s favorite sayings.
Occam’s Razor is a good rule in the real world. When parsing jokes, riddles, and math puzzles, not so much.
If you are told that one of them is a six, and then that die is removed, you are adding ambiguities by saying “what if he lied?”
However, what if the operator tells you that one is a six, and that he removed that die, but in fact, he removed a random die anyway, regardless of whether it was a six or not? Does that change the odds that the die left on the table is a six?
If so, then we have invented a mathematical lie detector machine. Repeat the procedure a hundred times. If the die on the table is a six in about 16 trials, he was lying. If it is a six in nine trials, he was telling the truth.