# Odds problem - roll a pair of dice **URL:** \ **Category:** Factual Questions\ **Created:** [October 6, 2016, 1:41pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763 "2016-10-06T13:41:48Z")\ **Posts on this page:** 20\ **Page:** 14
**Author:** ![Cliffy\_Biro](https://avatars.discourse-cdn.com/v4/letter/c/57b2e6/32.png) [@Cliffy\_Biro](https://boards.straightdope.com/u/Cliffy_Biro)\ **Post date:** [October 8, 2016, 8:32pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/261 "2016-10-08T20:32:46Z")
> [@Thudlow\_Boink](#): > > Are you familiar with the concept of conditional probability? It’s a very common type of question to ask “What is the probability that B is true, _given that A is true_” (symbolized **P(B|A)**). This is how I would interpret a statement of the form “A is true. What is the probability that B is true?” And it seems to me that your “rule” would wreck discussions of conditional probability: > > Question: A black card is drawn from a deck. What is the probability that it is a spade? > Person used to thinking in terms of conditional probability: If it’s a black card, there are 26 possibilities for what it could be. In that case, the probability that it’s one of the 13 spades is 13/26, or 1/2. > You: If a black card has already been drawn, it is what it is, and probability does not apply. I’m saying that it is meaningless to assign “probabilities” to past events, but I should have more precisely said “to past events whose outcome is known”. So in your card example, if the card has been drawn but I don’t know what it is, that’s OK. I can assign a probability to it because I haven’t seen the outcome of the past event yet. ---
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [October 8, 2016, 8:47pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/262 "2016-10-08T20:47:33Z")
I have checked several “the OP is ambiguous” posts and haven’t found any rational explantion why. “At least one is a 6” rules out a whole lot of the boy/girl nonsense. It seems pretty clear to me. And **Cliffy Biro** : people talk about the odds of past events all the frickin’ time. This is so well entrenched in statistics that it basically requires a separate Universe to presume elsewise. E.g., so a Scientist runs some tests, gets some results and wants to run some standard analysis, but wait! **Cliffy Biro** says you can’t do that! So much for most of Science. As to “proof” by coding. As I mentioned, I’ve seen many algorithms with formal “proofs” of correctness. The most common cause: What the prover thinks the code does and what it actually does are two different things. It introduces another possible way of making a mistake. I’m not saying never do this. But for quite trivial things like this, why? It is far, far less convincing than writing down the 11 possibilities, pointing out they’re all equal and that only one have them has “the other 6.” Ta zonking da. ---
**Author:** ![Cliffy\_Biro](https://avatars.discourse-cdn.com/v4/letter/c/57b2e6/32.png) [@Cliffy\_Biro](https://boards.straightdope.com/u/Cliffy_Biro)\ **Post date:** [October 8, 2016, 9:01pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/263 "2016-10-08T21:01:03Z")
> [@ftg](#): > > people talk about the odds of past events all the frickin’ time. This is so well entrenched in statistics that it basically requires a separate Universe to presume elsewise. E.g., so a Scientist runs some tests, gets some results and wants to run some standard analysis, but wait! **Cliffy Biro** says you can’t do that! So much for most of Science. You can talk about what the probability of a past event _was_, before its outcome was known, sure. You can’t talk about what the \*current \*probability is of an event whose outcome is known. It’s meaningless. Probability measures likelihood, certainty. There is no uncertainty there. ---
**Author:** ![Cliffy\_Biro](https://avatars.discourse-cdn.com/v4/letter/c/57b2e6/32.png) [@Cliffy\_Biro](https://boards.straightdope.com/u/Cliffy_Biro)\ **Post date:** [October 8, 2016, 9:18pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/264 "2016-10-08T21:18:09Z")
Anyway, the ambiguity comes from the fact that we don’t know if you are always told when there is at least one six, or only sometimes. If the latter, there will be cases when at least one six is rolled, but you don’t get to know about it. That affects the calculation. You can no longer assert “given that we know that there is at least one 6, all of 1-6, 2-6, 3-6, …, 6-5, 6-6 are equally likely”, so the basis of the 1/11 calculation disappears. E.g. maybe if it’s 1-6 you never get to hear about the six. ---
**Author:** ![sich\_hinaufwinden](https://avatars.discourse-cdn.com/v4/letter/s/c5a1d2/32.png) [@sich\_hinaufwinden](https://boards.straightdope.com/u/sich_hinaufwinden)\ **Post date:** [October 8, 2016, 9:19pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/265 "2016-10-08T21:19:09Z")
> [@ftg](#): > > I have checked several “the OP is ambiguous” posts and haven’t found any rational explantion why. “At least one is a 6” rules out a whole lot of the boy/girl nonsense… Imagine a game in which no “target” number is ever specified before playing and one of the dice is to be removed and shown from every roll and it doesn’t matter which of the dice it is or what number it has on it. The dice are simply rolled and one of them shown to the other person. The OP could be describing a roll from that game in which (at least) one of the dice just so happens to be a 6 and decides to show that die. They could have chosen to reveal the other die, whether the second was another 6 or something else, but they happened to reveal the 6 in this particular case. How would this game and the results of that roll be be inconsistent with the way the problem is worded in the OP? The OP doesn’t state that 6 is a “target” number before playing. The OP doesn’t state that rolls without at least one 6 are ignored and the dice rolled again until at least one 6 comes up. The OP doesn’t state that the game only proceeds to asking about probability when at least one 6 is rolled and it is shown to the other person. These are all assumptions that are being made by some posters, but not the only assumptions that could be made about how the game is being played. Ambiguous might not be the right word. I think that “insufficiently defined problem” is a better description. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [October 8, 2016, 9:33pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/266 "2016-10-08T21:33:27Z")
> [@](#): > > Quoth **Xema** : > > It’s not very useful to complicate problems of this type by raising the possibility of an observer operating under strange rules. We cannot avoid the fact that the observer must be operating under some sort of rules. It’s just a question of what those rules are. And sure, you could say “well, just assume that he’s operating under the simplest rule”, but that doesn’t help either, because now we have to decide what rule is simplest. The rule could be “Announce a six if there’s at least one present, or say nothing and re-roll if a six isn’t present”, or it could be “Announce the value of one of the two dice chosen at random”, or it could be “Announce the value of the red die”, or “Announce the value of the die showing the lowest number”, or “Announce the value of the die showing the highest number”. All of those are nice simple rules, and all of them are consistent with what we’re told. Which one should we assume? ---
**Author:** ![x-ray\_vision](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/x-ray_vision/32/351_2.png) [@x-ray\_vision](https://boards.straightdope.com/u/x-ray_vision)\ **Post date:** [October 8, 2016, 9:49pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/267 "2016-10-08T21:49:35Z")
> [@ftg](#): > > I have checked several “the OP is ambiguous” posts and haven’t found any rational explantion why. “At least one is a 6” rules out a whole lot of the boy/girl nonsense. Post 68 gives a good rationale for the ambiguity that _could_ lead different posters to interpret the problem differently. First I’ll post the problem as stated in the OP and then post 68: _“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?”_ > [@Chronos](#): > > I’ll join in with the “problem is ambiguous” crowd, here. 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. > > 2: I roll a pair of dice. I then examine both dice, and if it is the case that at least one of them shows a 6, I announce “At least one die is a 6”, and otherwise I say nothing. On this particular iteration of the experiment, I saw at least one 6, and so I announce as such. What is the probability that both dice are 6? > > Answer: The probability that both dice are 6 is 1 in 11. This is similar to the reason Gardner gave for there being ambiguity in the boy/girl problem. I think Chronos has a better argument for the OP’s original phrasing being ambiguous than Gardner did the boy/girl problem: > [@](#): > > Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys? > > This question is identical to question one, except that instead of specifying that the older child is a boy, it is specified that at least one of them is a boy. In response to reader criticism of the question posed in 1959, Gardner agreed that a precise formulation of the question is critical to getting different answers for question 1 and 2. Specifically, Gardner argued that a “failure to specify the randomizing procedure” could lead readers to interpret the question in two distinct ways: > > From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3. > > From all families with two children, one child is selected at random, and the sex of that child is specified to be a boy. This would yield an answer of 1/2 > **[Boy or girl paradox](https://en.wikipedia.org/wiki/Boy_or_Girl_paradox)** > > The Boy or Girl paradox surrounds a set of questions in probability theory, which are also known as The Two Child Problem, Mr. Smith's Children and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner featured it in his October 1959 "Mathematical Games column" in Scientific American. He titled it The Two Children Problem, and phrased the paradox as follows: > Gardner initially gave the answers .mw-parser-output .sfrac{white-space:nowrap}.... I get Gardner’s point. The phrasing _could_ lead readers to interpret the question in a way different than intended, leading one to get the wrong answer for the right reasons. But one would have to assume a person is selecting a child at random, when a person making selections isn’t stated. As Xema mentions in post 237, it’s more clear not even bringing an observer at all. I think one would have to make unnecessary assumptions in either the phrasing Gardner thought could lead to ambiguity in the boy/girl problem or that Chronos thought could lead to ambiguity in the OP’s original phrasing of this problem, but more so in Gardner’s because the OP’s phrasing does include _being told_ about something by a person, and that person could have only looked at one die. The boy/girl problem just gives information. Assuming it’s by a person that’s giving you partial information before choosing one at random is a stretch. But I’m guessing Gardner was more interested in people understanding probability than arguing whether or not the question as given was ambiguous. ---
**Author:** ![x-ray\_vision](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/x-ray_vision/32/351_2.png) [@x-ray\_vision](https://boards.straightdope.com/u/x-ray_vision)\ **Post date:** [October 8, 2016, 9:57pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/268 "2016-10-08T21:57:04Z")
> [@Chronos](#): > > We cannot avoid the fact that the observer must be operating under some sort of rules. I think Xema is saying it’s clearer if there’s no observer at all, not that there is an observer, but he’s not operating under strange rules, as evidenced by the following staement Xema made: \*But clearest of all would be to state it as “Two fair dice are rolled together; at least one of them is a 6.” IOW, there’s no need to introduce an ‘observer’ at all, nor to consider which of many possible recondite strategies he may be using. Just think about a roll of two fair dice, whose result includes at least one 6.\* ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [October 8, 2016, 10:27pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/269 "2016-10-08T22:27:25Z")
You’ve assumed again. It might be implicit, but it’s an assumption. What happens when the result has no 6? You’re guessing what happens. You’re guessing whether that counts somehow. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [October 8, 2016, 10:30pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/270 "2016-10-08T22:30:35Z")
Even if there’s no observer, there’s some mechanism by which we learn that there’s at least one 6. What is that mechanism? ---
**Author:** ![octopus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/octopus/32/3716_2.png) [@octopus](https://boards.straightdope.com/u/octopus)\ **Post date:** [October 8, 2016, 10:46pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/271 "2016-10-08T22:46:48Z")
> [@Chronos](#): > > We cannot avoid the fact that the observer must be operating under some sort of rules. It’s just a question of what those rules are. And sure, you could say “well, just assume that he’s operating under the simplest rule”, but that doesn’t help either, because now we have to decide what rule is simplest. The rule could be “Announce a six if there’s at least one present, or say nothing and re-roll if a six isn’t present”, or it could be “Announce the value of one of the two dice chosen at random”, or it could be “Announce the value of the red die”, or “Announce the value of the die showing the lowest number”, or “Announce the value of the die showing the highest number”. All of those are nice simple rules, and all of them are consistent with what we’re told. Which one should we assume? The rule was, you are shown that the value of one die which is then discarded rolled a 6. It’s pretty unambiguous. Givens are someone rolled two fair dice and selected one that showed a 6. Another given is everyone is acting honestly. Otherwise the answer is incalculable. If a simple problem like this is unsolvable even though everyone in the thread has two dice and some paper why do we even try debating economics, politics, or anything more complex than 2 dice? ---
**Author:** ![octopus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/octopus/32/3716_2.png) [@octopus](https://boards.straightdope.com/u/octopus)\ **Post date:** [October 8, 2016, 10:49pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/272 "2016-10-08T22:49:40Z")
> [@LSLGuy](#): > > You’ve assumed again. It might be implicit, but it’s an assumption. What happens when the result has no 6? You’re guessing what happens. You’re guessing whether that counts somehow. You’re assuming we inhabit a shared reality. At some point assumptions must be made. ---
**Author:** ![Cliffy\_Biro](https://avatars.discourse-cdn.com/v4/letter/c/57b2e6/32.png) [@Cliffy\_Biro](https://boards.straightdope.com/u/Cliffy_Biro)\ **Post date:** [October 8, 2016, 11:01pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/273 "2016-10-08T23:01:12Z")
> [@octopus](#): > > If a simple problem like this is unsolvable even though everyone in the thread has two dice and some paper why do we even try debating economics, politics, or anything more complex than 2 dice? Because you need more than just two dice and some paper. You also need correct assumptions about how the information was acquired. But \*\*Chronos \*\*earlier in the thread and quoted above, and **sich\_hinaufwinden** just above, have given clear illustrations that more than one set of assumptions is consistent with the problem stated in the OP, and these sets of assumptions lead to different answers. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [October 8, 2016, 11:44pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/274 "2016-10-08T23:44:07Z")
> [@Chronos](#): > > Even if there’s no observer, there’s some mechanism by which we learn that there’s at least one 6. What is that mechanism? The simplest “mechanism” is the way the problem is stated - to wit: Two dice are rolled; the result includes at least one six. ---
**Author:** ![octopus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/octopus/32/3716_2.png) [@octopus](https://boards.straightdope.com/u/octopus)\ **Post date:** [October 8, 2016, 11:49pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/275 "2016-10-08T23:49:51Z")
> [@Cliffy\_Biro](#): > > Because you need more than just two dice and some paper. You also need correct assumptions about how the information was acquired. But \*\*Chronos \*\*earlier in the thread and quoted above, and **sich\_hinaufwinden** just above, have given clear illustrations that more than one set of assumptions is consistent with the problem stated in the OP, and these sets of assumptions lead to different answers. It’s pretty clear. Of course since language itself is at it’s core circular definitions and presumed shared assumptions there’s no way to “prove” it. Without a doctoral dissertation how is 2 fair dice are rolled a six is shown and discarded not sufficiently clear? ---
**Author:** ![The\_Great\_Unwashed](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/the_great_unwashed/32/4791_2.png) [@The\_Great\_Unwashed](https://boards.straightdope.com/u/The_Great_Unwashed)\ **Post date:** [October 9, 2016, 12:22am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/276 "2016-10-09T00:22:20Z")
> [@Cliffy\_Biro](#): > > You can talk about what the probability of a past event _was_, before its outcome was known, sure. You can’t talk about what the \*current \*probability is of an event whose outcome is known. It’s meaningless. Probability measures likelihood, certainty. There is no uncertainty there. Really? I have just rolled a die. What _is_ the probability that I rolled a 4? – you seem to be saying that that question is meaningless!? It isn’t. ---
**Author:** ![x-ray\_vision](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/x-ray_vision/32/351_2.png) [@x-ray\_vision](https://boards.straightdope.com/u/x-ray_vision)\ **Post date:** [October 9, 2016, 12:46am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/277 "2016-10-09T00:46:05Z")
> [@LSLGuy](#): > > You’ve assumed again. It might be implicit, but it’s an assumption. What happens when the result has no 6? You’re guessing what happens. You’re guessing whether that counts somehow. Is that question for me? If so, which iteration of the problem are you asking me about? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [October 9, 2016, 12:55am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/278 "2016-10-09T00:55:09Z")
> [@](#): > > Quoth **Xema** : > > The simplest “mechanism” is the way the problem is stated - to wit: Two dice are rolled; the result includes at least one six. And how did we learn that? Did we see both dice? In that case, we don’t need to guess about whether they’re both sixes: We already know. Did we look at one random die? In that case, there’s one specific die we know about, which has no effect at all on the one we don’t know about, and so the other one has a 1 in 6 chance of being another six. Do we not actually know that one of them is a six, even though it happens to be true? Then, given that knowledge, the best we can say is that the chance of double-six is 1 in 36. ---
**Author:** ![newme](https://avatars.discourse-cdn.com/v4/letter/n/e0b2c6/32.png) [@newme](https://boards.straightdope.com/u/newme)\ **Post date:** [October 9, 2016, 1:08am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/279 "2016-10-09T01:08:24Z")
> [@octopus](#): > > Without a doctoral dissertation how is 2 fair dice are rolled a six is shown and discarded not sufficiently clear? It is not sufficiently clear because\* the manner in which a six is shown \*determines the probability of both dice being sixes. ---
**Author:** ![x-ray\_vision](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/x-ray_vision/32/351_2.png) [@x-ray\_vision](https://boards.straightdope.com/u/x-ray_vision)\ **Post date:** [October 9, 2016, 1:21am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/280 "2016-10-09T01:21:01Z")
> [@Chronos](#): > > And how did we learn that? Did we see both dice? In that case, we don’t need to guess about whether they’re both sixes: We already know We didn’t see anything. We are just told that two dice are rolled and the result includes at least one six and we’re not guessing anything. We’re calculating probability. > [@](#): > > Did we look at one random die? In that case, there’s one specific die we know about, which has no effect at all on the one we don’t know about, and so the other one has a 1 in 6 chance of being another six. We didn’t look at anything. > [@](#): > > Do we not actually know that one of them is a six, even though it happens to be true? Then, given that knowledge, the best we can say is that the chance of double-six is 1 in 36. WTF? Xema said “Two dice are rolled; the result includes at least one six.” Yes, we _actually know_ one of them is 6. This thread has passed the point of silly. Jack has 4 balls. Jill has 2. He gives Jill 2 balls. How many balls does Jill have? _Who is the observer? Is he far-sighted? Do we not actually know that Jack started off with 4 balls? Is one man’s ball another man’s orange?_ [Previous page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=13) [Next page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=15)