Odds problem - roll a pair of dice

[nicky]

My two cents

The problem is the ambiguity of “you are told theres a six”.

2 dice are rolled in secret but one falls on the floor and you see its a six:
1/6 chance the other dice is a six

2 dice are rolled in secret and the other player is obligated to reveal - and only reveal - the existence of one six - 1/11

2 dice are rolled in secret, hidden under cups, passerby comes along, says “watchyadooin ?” and randomly picks up a cup , revealing a 6 - 2/11

[nicky]

Above case #3 is wrong. Its 1/6. Doh!

[Cliffy_Biro]

@misterdls

This is the only proper sample space.

Only 1 of the 11 outcomes allows you to “pick a 6” AND see a 6 on the other die.

Therefore the answer to the original question is 1/11.

For a concrete example of another possible sample space that is consistent with the puzzle statement, see this earlier post, which lists twelve equally likely outcomes of which two are [6 6].

[Cliffy_Biro]

@BigT

The number 6 is given in the problem without any means for it to be different, and thus is fixed. Any simulation must keep the 6, or else it is modeling a different problem. And once you do require there to be a 6, the 1/11 interpretation is the only one that works.

I agree with you that we need only consider cases in which there is a least one six, since the puzzle statement tells us that there was in fact at least one six in this case. (Therefore I still don’t understand sich_hinaufwinden’s preoccupation with cases that don’t include a six, which are irrelevant, just as cases involving three dice or 7-sided dice would be.)

However, the step from “we need only consider possible rolls that include a six, of which there are eleven” to “all eleven such rolls are equally likely” is not justified by the puzzle statement. It requires you to assume things which are not stated.

[ftg]

@nicky

My two cents

The problem is the ambiguity of “you are told theres a six”.

2 dice are rolled in secret but one falls on the floor and you see its a six:
1/6 chance the other dice is a six

2 dice are rolled in secret and the other player is obligated to reveal - and only reveal - the existence of one six - 1/11

2 dice are rolled in secret, hidden under cups, passerby comes along, says “watchyadooin ?” and randomly picks up a cup , revealing a 6 - 2/11

This isn’t the problem in the OP at all. For one thing "and you are told that ‘at least one of the dice is a 6.’ " Not “a 6” but “at least one of the dice is a 6.” It’s in not being careful in stating the problem that causes disputes over the boy/girl and Monty Hall problems.

This OP was sufficiently careful.

You first case is irrelevant to the OP. It’s not the same. The second doesn’t make any sense and the third, well, the third … the less said the better.

The Great Unwashed wrote: “How so? Before you furnish your answer, you may wish to reflect on the fact that it’s physically impossible to throw two dice at exactly the same time.”

It doesn’t matter if they are exactly simultaneous or not. The key is that you only look at the dice after they are both been rolled. This is very different from rolling only one die and then looking at that one die.

These are quite basic- very. very basic concepts in probability.

Bad strawman examples with flawed notions concerning probability aren’t helping.

[Chronos]

Quoth BigT:

I do not understand why such an explanation is still being promoted, when one of the basic rules of word problems is that all required information is given.

Which makes this an illegal word problem, because it doesn’t do so. If I asked “Bob is driving at 60 MPH. How long does it take him to reach his destination?”, nobody would be attempting to solve it. I couldn’t just say “Assume that you have all required information, because that’s what word problems do”, because no matter how many times I say it has all the needed information, it still doesn’t.

I’ll give this yet another try. We are all assuming that the man is honest, because that’s a standard assumption in this sort of problem. That is to say, for any given statement, the man will never say it if it’s not true.

But we cannot go from there to assuming that, for any given statement, the man will always say it if it is true. We know that he doesn’t do that, because there are tons of true statements he could make, but isn’t. If he wanted to tell us the whole truth, he’d just say what both dice were, and there wouldn’t be a problem any more.

So, in particular, let’s take the statement he did make: “At least one die is a six”. He wouldn’t say that if it wasn’t true, but maybe he sometimes wouldn’t say it even if it was true, either. We don’t know, and it’s not reasonable for us to assume.

[LSLGuy]

Reinforcing Chronos at the risk of piling on BigT a bit.

@Chronos

…We don’t know, and it’s not reasonable for us to assume.

It’s especially not reasonable for us to assume when we’re dealing with a deliberately tricky puzzle problem rather than a real world situation.

Given a puzzle with multiple choice answers it’s typically the case that at least one answer is totally bogus, but there are also at least *two *plausible answers given.

Picking which plausible answer is “best” required detailed penetrating analysis of the problem. Rather than simply jumping to the most obvious possibility while assuming no other plausible interpretations could possibly exist. Doing the latter is how people bite on the [obvious but wrong] choice.

[septimus]

@BigT

… the 1/11 interpretation is the only one that works.

I do not understand why [alternate] explanation is still being promoted, when one of the basic rules of word problems is that all required information is given.

I agree with BigT. Although I do not disagree that the problem statement is somewhat ambiguous, if you assume it is intended to be clear then there is only one interpretation.

IOW, yes, if the puzzle were designed by a pedant then “Not enough information” is a possible correct answer. Given that that is not an option, but 1/6 and 1/11 are possible answers, then clearly 1/11 is the intended answer.

[Peremensoe]

@septimus

I agree with BigT. Although I do not disagree that the problem statement is somewhat ambiguous, if you assume it is intended to be clear then there is only one interpretation.

IOW, yes, if the puzzle were designed by a pedant then “Not enough information” is a possible correct answer. Given that that is not an option, but 1/6 and 1/11 are possible answers, then clearly 1/11 is the intended answer.

But,

@Skammer

1/11 is not one of the available answers, which are: 1/18, 1/36, 2/11 and 1/6.

[Thudlow_Boink]

Yes, 1/11 was not one of the available answers. There are three possible reasons for this, all of which I find plausible enough not to rule out:

1. 1/11 was not the intended answer.
2. 1/11 was the intended answer, and they made a mistake in not including it.
3. 1/11 was the intended answer, and they deliberately left it off the list to try to get people to click on the link to see why the answer they came up with wasn’t one of the choices.

[The_Great_Unwashed]

@octopus

If I say I have 3 black socks, 1 gray sock, and 1 paisley sock in my sock drawer and I draw one black one completely randomly what does anything else matter if the question then asks what’s the probability that the next sock randomly drawn is gray? At some point normal assumptions should be sufficient.

No it would be more like:
Two socks are randomly selected from octopus’s sock draw, and you are told that ‘at least one of the socks has a hole in it.’ A holey sock is thrown behind the wardrobe never to be seen again. What is the probability that the other sock has a hole in it.

(you can replace “has a hole in it” with “black” if you like numbers)

[The_Great_Unwashed]

@ftg

The Great Unwashed wrote: “How so? Before you furnish your answer, you may wish to reflect on the fact that it’s physically impossible to throw two dice at exactly the same time.”

It doesn’t matter if they are exactly simultaneous or not. The key is that you only look at the dice after they are both been rolled. This is very different from rolling only one die and then looking at that one die.

These are quite basic- very. very basic concepts in probability.

Yeah, you’d think I’d know them. Maybe you are misremembering them.

I think you mean to say that the puzzle : after one die being thrown if a six is showing then “at least one die is a six” is announced THEN the other die is thrown etc. is different from the one in the OP.

Too damned right it is different, but there’s no way to read the OP to mean that.

On the other hand, if our experimenter in the puzzle from the OP, threw one die then made himself a cup of tea then posted about it in MPSIMS then answered the door to the man from the Pru then threw the other die then made his announcement that would make no difference whatsoever.

[sich_hinaufwinden]

@Cliffy_Biro

…(Therefore I still don’t understand sich_hinaufwinden’s preoccupation with cases that don’t include a six, which are irrelevant, just as cases involving three dice or 7-sided dice would be.)…

Agreed that you don’t understand. And I don’t know how to make it any simpler so that you might finally grasp it, but I am not the first or only person to point out that:

@Chronos

…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…

(I’ve omitted the 2nd example describing what the 1/11 camp thinks is the only valid interpretation.)

And this post which references that one:

@TroutMan

You’re falling into the same trap I did. Given your assumptions, your solution is exactly right and 1/11 is the correct answer. But the assumption is one I didn’t even realize I was making, that six was picked as the “magic” number before any roles, and therefore any roles that didn’t contain a six would be discarded. The other possibility was best explained to me by Chronos in post 68. You aren’t actually told that six was selected before any roles; it could just be a statement of what one of the dice showed. In that case, the fact that one of the dice is a particular number doesn’t add any new information to the problem, and the other die has a 1/6 chance of being that number.

Those are both saying the exact same thing as what I have been saying.

I thought you understood the 1/11 interpretation…

@Cliffy_Biro

…However, the step from “we need only consider possible rolls that include a six, of which there are eleven” to “all eleven such rolls are equally likely” is not justified by the puzzle statement. It requires you to assume things which are not stated.

…but now I’m not so sure about that.

[nicky]

@ftg

This isn’t the problem in the OP at all. For one thing "and you are told that ‘at least one of the dice is a 6.’ " Not “a 6” but “at least one of the dice is a 6.” It’s in not being careful in stating the problem that causes disputes over the boy/girl and Monty Hall problems.

This OP was sufficiently careful.

You first case is irrelevant to the OP. It’s not the same. The second doesn’t make any sense and the third, well, the third … the less said the better.

The Great Unwashed wrote: “How so? Before you furnish your answer, you may wish to reflect on the fact that it’s physically impossible to throw two dice at exactly the same time.”

It doesn’t matter if they are exactly simultaneous or not. The key is that you only look at the dice after they are both been rolled. This is very different from rolling only one die and then looking at that one die.

These are quite basic- very. very basic concepts in probability.

Bad strawman examples with flawed notions concerning probability aren’t helping.

Thank you so much for patronising me.

“youre told” is not sufficiently careful.

[sich_hinaufwinden]

@Cugel

Nah. Pretty clear the spirit of the problem is to get people to guess the “obvious” 1/6, instead of the actual 1/11.

While it may be the case that what you say is the spirit of the problem, they failed to write it in a way that 1/11 is the only reasonable interpretation.

[The_Great_Unwashed]

@ftg

The OP is more like roll both dice until at least one of them is a 6, what are the It takes fewer rolls on average to get 1+ 6 out of two dice than a 6 out of 1 dice.

I didn’t see this first time around.

Are you saying that it takes fewer rolls to have seen more than one six throwing two dice repeatedly than it takes to have seen more than one six throwing a single die repeatedly? (I suppose you’re counting the first as “two rolls”, if not, well duh!)

That just cannot be right, any sequence of single throws can be thought of as a sequence of pairs of throws:

1 , 3 , 5 , 6 , 5 , 2 , 6 , 2
just the same as
{1 , 3} , {5 , 6} , {5 , 2} , {6 , 2}

[Telemark]

@Thudlow_Boink

Yes, 1/11 was not one of the available answers. There are three possible reasons for this, all of which I find plausible enough not to rule out:

1. 1/11 was not the intended answer.
2. 1/11 was the intended answer, and they made a mistake in not including it.
3. 1/11 was the intended answer, and they deliberately left it off the list to try to get people to click on the link to see why the answer they came up with wasn’t one of the choices.

Or:

4. They took a problem that did have 2/11 as the correct answer and for some reason changed the problem so 1/11 was the correct answer but weren’t smart enough to notice. BTW, this is what actually happened.

[Thudlow_Boink]

@Telemark

Or:

4. They took a problem that did have 2/11 as the correct answer and for some reason changed the problem so 1/11 was the correct answer but weren’t smart enough to notice. BTW, this is what actually happened.

Ah. What was the original problem? (My apologies if it has already been mentioned in this thread.)

[sich_hinaufwinden]

@Thudlow_Boink

Ah. What was the original problem? (My apologies if it has already been mentioned in this thread.)

Post 355:

@newme

The answer to the original puzzle was 2/11, because it is slightly different from the OP’s puzzle. Here again is the original puzzle, where you are told one of the numbers is a four not a six:

For some reason, the people at Brilliant.org changed the four to a six when they posed the question on facebook (the OP’s version), but they forgot to change the answer. I presume in the changed version, the answer they want to hear is 1/11.

Note that this link takes you to a Monty Hall type problem. I’m not sure where you’re supposed to go from there. I tried answering the problem to see if it would be the next problem or something but then it said I would have to sign up so I bailed.

[Some_Call_Me.Tim]

This probably won’t help anyone but bridge players already familiar with it, but this is similar to the principle of restricted choice. There, in one of the possible scenarios the other person must choose a particular option and in the other scenario the other person chooses randomly between equivalent options. When observing this, even through the two scenarios started as equiprobable, the fact that the other person only had a 50% chance of selecting that particular option between equivalent ones makes the odds you’re within that scenario less likely.

Here, the fact that if there were initially two sixes but the other person had to choose between equivalent options makes that scenario half as likely as the scenario where they had no choice as there was but one six.