Odds problem - roll a pair of dice

[Amateur_Barbarian]

All endless semanticizing aside, I don’t see how there’s any confusion here. The underlying question is “what are the odds of rolling two sixes in a single roll” - and the answer is 1 in 6.

The original question is nowhere near as convoluted as succeeding arguments try to make it.

[ol]
[li]Two fair dice are rolled.[/li][li]You are told at least one of them is a six.[/li][li]A SIX IS REMOVED - we can assume this validates step 2.[/li][li]What are the odds the remaining die is a six? The same as if the question had been simply phrase as above: 1 in 6.[/li][/ol]
Every other argument throws other assumptions into the mix. When you get to “what if there aren’t even any dice” this no longer belongs in GQ.

[Chingon]

Amazing

[newme]

The original source for the question may be here:

Two fair dice are rolled, and it is revealed that one of the numbers rolled was a 4. What is the probability that the other number rolled was a 6? Note: You are not told which of the numbers rolled is a 4.

The multiple choice options are 1/18, 1/6, 2/11, and 1/36, which are the same as in the OP. That can’t be a coincidence. The expected “correct” answer is 2/11.

However, in the discussion of the question on that site, many people argued that one interpretation would yield the answer of 1/6. Indeed, one of the people who writes the questions for the site admits:

I think that the extent of the dissent voiced in the comments has demonstrated that the wording of the question left too much room for interpretation. I’m not sure if it would be worth changing the wording at this point; probably posting a new question with more precise wording would be preferable, something I may do later.

[LSLGuy]

The 2/11 choice is probably the least plausible / most convoluted choice of them all. I’m unimpressed that the puzzle source thinks that’s the best choice of the bunch.

[Thudlow_Boink]

@Amateur_Barbarian

All endless semanticizing aside, I don’t see how there’s any confusion here. The underlying question is “what are the odds of rolling two sixes in a single roll” - and the answer is 1 in 6.

I’m confused by what you say here. If the question were simply “What is the probability of rolling two sixes in a single roll (of two normal six-sided dice)?” the answer would be 1/36.

[QUOTE]
[li]Two fair dice are rolled.[/li][/QUOTE]
There are 36 possible outcomes, equally likely.

[QUOTE]
[li]You are told at least one of them is a six.[/li][/QUOTE]
This narrows the number of possible outcomes down to 11.

[QUOTE]
[li]A SIX IS REMOVED - we can assume this validates step 2.[/li]
[li]What are the odds the remaining die is a six? The same as if the question had been simply phrase as above: 1 in 6.[/li][/QUOTE]
How are you interpreting these steps to get 1/6? My interpretation, and that of everyone who’s been getting an answer of 1/11, is that at this point the question is equivalent to asking “What is the probability that both dice were sixes?” because this is the only situation in which removing a six would leave a remaining six.

[LSLGuy]

And thus we rewind to post #6 or so and prepare go around the 7 page loop again.

Sigh.

The whole puzzle is *not *equivalent to “what are the odds of throwing double six” where the answer would be 1/36.

The whole puzzle is (arguably) equivalent to rolling one die until you’ve rolled a six, however many times that takes. Then rolling the other die and asking “what are the odds the second one comes up six?” To which the answer is 1/6.

Rolling them simultaneously doesn’t change anything. What changes things is what the observer says and under exactly what circumstances he/she says it.

[Chronos]

It is also (arguably) equivalent to rolling both dice at once until at least one of them is a six, and asking what the odds are that they’re both sixes. To which the answer is 1/11.

And it’s also (arguably) equivalent to a whole bunch of other possible procedures, though those are all more of a stretch.

[jtur88]

[quote=“Amateur_Barbarian, post:341, topic:767763”]

[li]A SIX IS REMOVED - we can assume this validates step 2.[/li][/QUOTE]

This is not important at all, belying your choice of caps. Nothing need be done with it. Al that is necessary is for all to agree on which is “the other die”, and consider the chances of that one being a 6.

[LSLGuy]

@CHronos: As I said a bit ago, I wonder what the original puzzle maker was thinking when he/she asserted 2/11 was the right answer. To be sure there are *conceivable *rule sets that produce that result. What IMO there aren’t, is *simple & plausible *rule sets that do so.

Any ideas?

[Amateur_Barbarian]

@LSLGuy

The whole puzzle is *not *equivalent to “what are the odds of throwing double six” where the answer would be 1/36.

You’re absolutely correct. However, no form of the question is “what are the odds of rolling double six.” There is NO question until we are told (let’s assume with complete validity) that “at least one of them is a six.” The question is THEN - “what are the odds the other die is also a six?” In other words, what are the odds that the roll was ANY double… which is one in six. Identifying the first die is a red herring.

The whole puzzle is (arguably) equivalent to rolling one die until you’ve rolled a six, however many times that takes. Then rolling the other die and asking “what are the odds the second one comes up six?” To which the answer is 1/6.

Rolling them simultaneously doesn’t change anything.

Absolutely wrong. Rolling one die and then another is one odds calculation; rolling both is a different one, even though the odds happen to be the same in both cases (1 in 6.)

What the observer says, if honest, is entirely irrelevant except to identify WHICH of the six possible doubles might have been rolled. That it is ONE of those six does not change the fundamental starting odds of one in six.

“One in eleven” is a phantom created by combining conditions irrelevant to the question.

Sigh all you like.

[Cugel]

A few posts in, I made my own simulation

Dim occ(6)
For a=1 To 100000
d1=Random(5)+1
d2=Random(5)+1
If d1=6:occ(d2)=occ(d2)+1:EndIf
If d2=6:occ(d1)=occ(d1)+1:EndIf
Next a

Comes out 1/6. But reading a bit further and seeing 1/11, I realised d2 isn’t evalutaed when d1=6, and Wally! 1/11.

[Some_Call_Me.Tim]

@Amateur_Barbarian

Sigh all you like.

Sigh.

Let me describe a different problem which I propose is analogous(if simpler):
“Two fair coins are flipped together. You are told that ‘at least one of the coins is heads.’ A heads is removed, and you are then shown the other coin. What is the probability that other coin shows heads?”

Now, the same arguments apply here as to the dice. You might argue that the middle two sentences about the shown coin are irrelevant, and that therefore there’s a fair 1 in 2 chance of the other coin being heads.

The other position is that when two fair coins are flipped the chance of getting one head and one tail is 50-50, and the chance of two heads is 1 in 4. The scenario in which the other coin is a tails is twice as likely as the scenario in which the other coin is a heads. Therefore the odds the other coin is tails is 2 in 3, and the odds it is heads is 1 in 3.

The 1 in 2 logic is analogous to the 1 in 6 logic in the dice case, whereas the 1 in 3 logic is analogous to the 1 in 11 logic for dice. Hopefully this narrowing of the options available makes the differences between the two more obvious, and it would take less time to try this experimentally because the odds are more skewed.

[Thudlow_Boink]

@LSLGuy

The whole puzzle is (arguably) equivalent to rolling one die until you’ve rolled a six, however many times that takes. Then rolling the other die and asking “what are the odds the second one comes up six?” To which the answer is 1/6.

@Amateur_Barbarian

There is NO question until we are told (let’s assume with complete validity) that “at least one of them is a six.” The question is THEN - “what are the odds the other die is also a six?” In other words, what are the odds that the roll was ANY double… which is one in six. Identifying the first die is a red herring.

This is the interpretation I’d lean towards if the original wording had said “one of the dice is a 6” instead of “at least one of the dice is a 6.”

If it had simply said “One of the dice is a 6,” it would be natural and reasonable (IMHO) to interpret this as “I’ll tell you what one of the dice was: it happened to be a six. Now if I remove that one, what’s the probability that the other one is a 6?” And then your reasoning would be correct, and the answer would be 1/6.

But (again IMHO) the words “at least” don’t fit that interpretation. To me, “at least one of the dice is a 6” means “Out of all the possible ways the pair of dice could come up, the result was one of the ones in which at least one of the two dice shows a 6.” There are 11 such possibilities, and in only one of them would the remaining die be a 6 if a 6 were removed.

[sich_hinaufwinden]

@Thudlow_Boink

This is the interpretation I’d lean towards if the original wording had said “one of the dice is a 6” instead of “at least one of the dice is a 6.”

If it had simply said “One of the dice is a 6,” it would be natural and reasonable (IMHO) to interpret this as “I’ll tell you what one of the dice was: it happened to be a six. Now if I remove that one, what’s the probability that the other one is a 6?” And then your reasoning would be correct, and the answer would be 1/6.

But (again IMHO) the words “at least” don’t fit that interpretation. To me, “at least one of the dice is a 6” means “Out of all the possible ways the pair of dice could come up, the result was one of the ones in which at least one of the two dice shows a 6.” There are 11 such possibilities, and in only one of them would the remaining die be a 6 if a 6 were removed.

Based on your interpretation of that phrase, in the game you think they are playing, what would have happened if neither of the dice had been a 6 on that particular roll?

[newme]

@LSLGuy

@CHronos: As I said a bit ago, I wonder what the original puzzle maker was thinking when he/she asserted 2/11 was the right answer. To be sure there are *conceivable *rule sets that produce that result. What IMO there aren’t, is *simple & plausible *rule sets that do so.

Any ideas?

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:

Two fair dice are rolled, and it is revealed that one of the numbers rolled was a 4. What is the probability that the other number rolled was a 6? Note: You are not told which of the numbers rolled is a 4.

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.

[octopus]

@Cliffy_Biro

Yes. But it’s important that the assumptions are justified, hence my repeated use of that word and its opposite, which you may have noticed.

What I have learned in this thread is not to assume anything.

[octopus]

@Amateur_Barbarian

All endless semanticizing aside, I don’t see how there’s any confusion here. The underlying question is “what are the odds of rolling two sixes in a single roll” - and the answer is 1 in 6.

The original question is nowhere near as convoluted as succeeding arguments try to make it.

[ol]
[li]Two fair dice are rolled.[/li][li]You are told at least one of them is a six.[/li][li]A SIX IS REMOVED - we can assume this validates step 2.[/li][li]What are the odds the remaining die is a six? The same as if the question had been simply phrase as above: 1 in 6.[/li][/ol]
Every other argument throws other assumptions into the mix. When you get to “what if there aren’t even any dice” this no longer belongs in GQ.

Rolling two 6’s in a singey roll of two dice is 1/36.

[septimus]

@Amateur_Barbarian

All endless semanticizing aside, I don’t see how there’s any confusion here…

[ol]
[li]Two fair dice are rolled.[/li][li]You are told at least one of them is a six.[/li][li]A SIX IS REMOVED - we can assume this validates step 2.[/li][li]What are the odds the remaining die is a six? The same as if the question had been simply phrase as above: 1 in 6.[/li][/ol]

Wrong.

I have two children and I tell you at least one of them is a girl. What is the chance both are girls?

Do the people getting OP’s problem wrong also get the two-children problem wrong? They are really the exact same problem. The correct answer is 1/(2N-1) not 1/N, where N=6 and N=2 for the respective problems.

[Cliffy_Biro]

@Amateur_Barbarian

Rolling one die and then another is one odds calculation; rolling both is a different one, even though the odds happen to be the same in both cases (1 in 6.)

It’s not a coincidence that the answer is the same, because calculation-wise they are equivalent. You have two independent die rolls. It doesn’t matter if it’s one die rolled twice or two dice rolled once.

[Colophon]

how hard do people want to make this?

Let’s work backwards and see if it makes things easier.

Out of all possible rolls of two dice, there is only one possible outcome where, when a six is removed, the other die is a six.

Agreed?

ONE.

So, now let’s ask, how many possible outcomes are there in which at least one of the two dice shows a six. (Because these are the only outcomes that are valid in the question, correct?)

Answer: ELEVEN.

So out of ELEVEN possible outcomes, only ONE will result in the other die being a six. Therefore the probability is quite clearly 1/11, just as the simulations show.

There is no ambiguity.

Can the 1/6 camp please explain how they get their answer? They seem to suggest that there are only SIX possible outcomes that result in at least one of the two dice showing a six. What do they imagine these six outcomes are?