# 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:** 13
**Author:** ![Turble](https://avatars.discourse-cdn.com/v4/letter/t/8dc957/32.png) [@Turble](https://boards.straightdope.com/u/Turble)\ **Post date:** [October 8, 2016, 3:07pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/241 "2016-10-08T15:07:58Z")
I was concerned that **septimus** was in the 1/11 camp because he is so good at stuff like this. I have rethought things. As a result of post #218 by **cerberus** I withdraw all my previous statements and formally switch to the 1/11 camp. :smack: ---
**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, 4:41pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/242 "2016-10-08T16:41:01Z")
> [@jtur88](#): > > 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. Again, nobody but you and \*\*Xema \*\*is suggesting the game host ever lies. Nobody. That’s a gigantic red herring. The point isn’t even that the game host might say different albeit truthful things on different rolls just for the lulz. The point is that the text of the problem does _ **not** _ say: “The game host will always announce either A) ‘There is at least one 6’ when there is such, or announce B) ‘This is a do-over; ignore this roll’ otherwise”. The game host’s rule for his announcement \*could \*be much more complex. And still be applied honestly and consistently to every roll. For some rule sets the answer is 1/6. For others it’s 1/11. For others it’s yet different odds. We’re all left guessing what the rule set really is. _Because the OP’s problem statement is silent on this point._ Once we’ve guessed that, the determination of the resulting odds is trivial. The whole point is that we’re stuck guessing at step 1. So whatever calculation we make at step 2 is a product of guesswork. And hence utterly flawed as a definitive conclusion. ---
**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, 4:59pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/243 "2016-10-08T16:59:34Z")
> [@LSLGuy](#): > > You misunderstood **Cliffy Biro** ’s point. It was not that the observer chooses to lie abut what he sees. I think you may have misunderstood Xema’s point (although I’m not certain). It’s not necessarily that the observer might choose to lie. It’s that there is of course more information that can be garnered that can alter probability and it’s not relevant, although Cliffy Biro seems to think it is. > [@Cliffy\_Biro](#): > > there are infinitely many possible answers, depending on which assumptions you use. > [@Cliffy\_Biro](#): > > The answer depends on how the observer decided to say that one of them is a 6. > [@Cliffy\_Biro](#): > > 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. Do you see what Cliffy is stating there, especially given the last quote from him above? He’s not only stating that the boy/girl problem is either 1/2 or 1/3 depending on how it is phrased, he’s saying that since there is other possible information we could gather, the answer could be anything. Also, he is even saying something close to stating there is a possibility the observer may lie by stating there may be a rule in place that he has to: _“say that at least one of them is black only if both are black”_ Saying that there could be variables like this that will affect the answer makes almost any probability question impossible to answer. You pick someone in a crowd and ask him to either say “country” or “rock” out loud. The person complies. What are the chances he/she said “country”? According to Cliffy, “there are infinitely many possible answers, depending on which assumptions you use” and “the answer depends on how the observer decided to” pick a name. Maybe this person picks first choices 4 out of 5 times consistently. Maybe there’s been a study done proving that when given a choice between saying two words, people pick two syllable words over one syllable words 2 to 1. Or maybe there’s a rule in place that the questioner looks for people that are Country music fans. The additional information will change the probability. But we don’t have additional information. So the probability is based on the information we do have alone. The probability the person in the crowd said “country” is 1/2. ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [October 8, 2016, 5:00pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/244 "2016-10-08T17:00:44Z")
I admit I haven’t read all the posts, but I’m perplexed on why the correct answer (1/11) isn’t obvious. I have two die. Here are the possible combinations: (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) I now want to play the “game” described in the OP. Example: I roll 2 and 4. Nope, roll again. I roll 3 and 5. Nope, roll again. I roll 4 and 4. Nope, roll again. I roll 1 and 6. Stop! I keep rolling until one or both die is a 6. And then I announce, “One or both die is a 6. Let’s continue…” So in other words, **all combinations that do not contain a 6 can be completely ignored** ; you need to pretend they _do not exist_: [del](1, 1)[/del] [del](1, 2)[/del] [del](1, 3)[/del] [del](1, 4)[/del] [del](1, 5)[/del] (1, 6) [del](2, 1)[/del] [del](2, 2)[/del] [del](2, 3)[/del] [del](2, 4)[/del] [del](2, 5)[/del] (2, 6) [del](3, 1)[/del] [del](3, 2)[/del] [del](3, 3)[/del] [del](3, 4)[/del] [del](3, 5)[/del] (3, 6) [del](4, 1)[/del] [del](4, 2)[/del] [del](4, 3)[/del] [del](4, 4)[/del] [del](4, 5)[/del] (4, 6) [del](5, 1)[/del] [del](5, 2)[/del] [del](5, 3)[/del] [del](5, 4)[/del] [del](5, 5)[/del] (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Another way of thinking of the problem: two “very special” die are rolled, and the _ **only** _ - ONLY!! - combinations they can produce are: (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) The odds of both die being 6 are 1/11. ---
**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, 5:08pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/245 "2016-10-08T17:08:58Z")
> [@x-ray\_vision](#): > > Also, he is even saying something close to stating there is a possibility the observer may lie by stating there may be a rule in place that he has to: > > _“say that at least one of them is black only if both are black”_ How is that “close to a lie”? If there are two black cats, and the observer says “there is at least one black cat”, that is absolutely true and not misleading at all. In the other cases he will say, under this rule, “there is at least one white cat”, again completely truthfully. And yet the statement “there is at least one black cat” would, under this rule, give a probability of 1, that is 100%, that the other cat is black, not 1/3 or 1/2. Assuming you know the rule. In the OP’s question, of course, you don’t know the rule, so all such calculation is futile. > [@](#): > > You pick someone in a crowd and ask him to either say “country” or “rock” out loud. […] But we don’t have additional information. So the probability is based on the information we do have alone. The probability the person in the crowd said “country” is 1/2. The probability cannot be determined. 1/2 is merely your best estimate of the probability. It is not the actual probability. ---
**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, 5:18pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/246 "2016-10-08T17:18:19Z")
> [@Crafter\_Man](#): > > …I now want to play the “game” described in the OP. Example: > > I roll 2 and 4. Nope, roll again… Why roll again? There’s nothing in the OP rules that says you would have to. You’ve made assumptions not specified in the OP. You could in this case say “At least one of the dice is a 4” and remove that die then ask “What is the chance the other die is a 4?” Many people are assuming that the game began by specifying that 6 was a magic number and rolls not including a 6 would be ignored. Other assumptions lead to different answers, such as the one I described above - that no number is magic and that on every roll a die is removed and the question asked what the probability is that the other die is the same. That is consistent with the OP which happens to describe a particular roll in which there was a 6 and it was the one removed. ---
**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, 5:26pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/247 "2016-10-08T17:26:12Z")
> [@Cliffy\_Biro](#): > > How is that “close to a lie”? I’ll retract my statement if you think how it’s labeled is relevant. But the label doesn’t affect the points of my post in total. > [@](#): > > The probability cannot be determined. 1/2 is merely your best estimate of the probability. It is not the actual probability. If you say so. You are taking away from the spirit of discussing probability, IMHO, but proceed as you will. ---
**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, 5:27pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/248 "2016-10-08T17:27:10Z")
> [@sich\_hinaufwinden](#): > > Many people are assuming that the game began by specifying that 6 was a magic number and rolls not including a 6 would be ignored. Yes. I can think of two possible reasons for people making that assumption: - the wording of the question, "“ **at least one** is a 6”, perhaps suggests that 6s are particularly noteworthy? - in many games involving dice, 6 \*is \*noteworthy, what with it being the biggest number. But there are other dice games in which 1, for example, is important. Why can’t we assume that the observer would be keen to note any 1 with the statement “there’s at least one 1”, but otherwise just describe one die at random? That leads to the probability of a throw of 6-6 being 1/5, by my calculations. ---
**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, 5:35pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/249 "2016-10-08T17:35:08Z")
> [@Cliffy\_Biro](#): > > Yes. I can think of two possible reasons for people making that assumption:… So you admit it is an assumption? And that some assumptions are necessary to calculate an answer? And that there are other assumptions consistent with the OP that lead to different answers? And that in the end, the problem as described is ill-defined, since assumptions have to be made to answer it? And that if you have to choose an answer from those given in the OP, that 1/11 is not one of the answers, but 1/6 is? So maybe assumptions that lead to 1/11 are not consistent with the game actually being played in the OP? ---
**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, 5:43pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/250 "2016-10-08T17:43:13Z")
> [@sich\_hinaufwinden](#): > > So you admit it is an assumption? > > And that some assumptions are necessary to calculate an answer? > > And that there are other assumptions consistent with the OP that lead to different answers? > > And that in the end, the problem as described is ill-defined, since assumptions have to be made to answer it? > > And that if you have to choose an answer from those given in the OP, that 1/11 is not one of the answers, but 1/6 is? > > So maybe assumptions that lead to 1/11 are not consistent with the game actually being played in the OP? I was agreeing with you, but you seem to be arguing? Anyway, you are right of course that 1/11 was not one of the answers given in the OP. But all of the answers in the OP that are greater than or equal to 1/11 are possible, under appropriate assumptions. So it could be 2/11. Don’t ask me to come up with the rule that the observer would have to have been using, though! ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [October 8, 2016, 5:55pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/251 "2016-10-08T17:55:15Z")
> [@sich\_hinaufwinden](#): > > Why roll again? There’s nothing in the OP rules that says you would have to. You’ve made assumptions not specified in the OP. The OP states, > [@](#): > > Two fair dice are rolled together, and you are told that ‘at least one of the dice is a 6.’ There are many, many ways this could occur: I roll 2 and 4. Nope, roll again. I roll 3 and 5. Nope, roll again. I roll 4 and 4. Nope, roll again. I roll 1 and 6. Good! Let’s continue… I roll 5 and 4. Nope, roll again. I roll 2 and 2. Nope, roll again. I roll 6 and 1. Good! Let’s continue… I roll 4 and 6. Good! Let’s continue… I roll 5 and 1. Nope, roll again. I roll 2 and 1. Nope, roll again. I roll 2 and 4. Nope, roll again. I roll 1 and 1. Nope, roll again. I roll 6 and 6. Good! Let’s continue… As I stated in my previous post, _all combinations that do not contain a 6 can be completely ignored; you need to pretend they do not exist._ Here is another way of thinking of the problem: You have a 2-digit, digital display. When you press a button, one of the following will be displayed: (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) There is an equal chance for each of the above to be displayed: The odds of (1, 6) being displayed is 1/11. The odds of (2, 6) being displayed is 1/11. The odds of (3, 6) being displayed is 1/11. The odds of (4, 6) being displayed is 1/11. The odds of (5, 6) being displayed is 1/11. The odds of (6, 1) being displayed is 1/11. The odds of (6, 2) being displayed is 1/11. The odds of (6, 3) being displayed is 1/11. The odds of (6, 4) being displayed is 1/11. The odds of (6, 5) being displayed is 1/11. **The odds of (6, 6) being displayed is 1/11.** And that’s the question in the problem… what are the odds that (6, 6) is displayed. ---
**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, 5:58pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/252 "2016-10-08T17:58:25Z")
> [@](#): > > So it could be 2/11. Don’t ask me to come up with the rule that the observer would have to have been using, though! Actually, I think the rule is quite simple: - if there is a six and a non-six, say “six” with probability 0.45, or the other number with probability 0.55 . By my calculations, under that rule, if the observer says “there is at least one six”, the probability that they are both six is 100/550, that is 2/11 . ---
**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, 6:07pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/253 "2016-10-08T18:07:43Z")
> [@Cliffy\_Biro](#): > > I was agreeing with you, but you seem to be arguing? > > Anyway, you are right of course that 1/11 was not one of the answers given in the OP. But all of the answers in the OP that are greater than or equal to 1/11 are possible, under appropriate assumptions. So it could be 2/11. Don’t ask me to come up with the rule that the observer would have to have been using, though! Yes - I see what you had written in previous posts where you acknowledge at least two different sets of assumptions that lead to answers including, but not limited to, 1/11 and 1/6. But in response to my post, you appear to be playing devil’s advocate and making the case for why people could justify certain assumptions to themselves. And that doesn’t help my point of trying (along with you and others) to get some posters here to realize that assumptions do in fact have to be made to calculate any answer - there’s simply not enough information given in the OP to justify a single answer. Some posters seem to think that there is only one way to analyze the problem and so I was replying more for their sake than directly to you. In any event, I’m just discussing - not arguing, if that would imply my response as being combative on any level. ---
**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, 6:10pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/254 "2016-10-08T18:10:21Z")
Got it, sorry for the misunderstanding. ---
**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, 6:16pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/255 "2016-10-08T18:16:05Z")
> [@Crafter\_Man](#): > > …**The odds of (6, 6) being displayed is 1/11.**… I understand how to calculate the probability under assumptions that lead to 1/11. Do you acknowledge that 1/11 is not an answer in the OP? Or do you further assume there is an error in the choices given for answers in the OP? As long as were assuming things, I guess why not continue assuming beyond an ill-defined problem that there is an error in the possible answers. ---
**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, 6:17pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/256 "2016-10-08T18:17:28Z")
> [@Cliffy\_Biro](#): > > Got it, sorry for the misunderstanding. It’s all good - hope you’re having a great weekend! ---
**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, 6:20pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/257 "2016-10-08T18:20:02Z")
> [@LSLGuy](#): > > 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. But Cliffy posted this: > [@](#): > > 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? I don’t think this is the clearest statement of the problem. I think the OP made the problem reasonably clear: > [@](#): > > Two fair dice are rolled together, and you are told that ‘at least one of the dice is a 6.’ 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:** ![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, 6:45pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/258 "2016-10-08T18:45:10Z")
> [@Xema](#): > > But clearest of all would be to state it as “Two fair dice are rolled together; at least one of them is a 6.” That is rather like talking about past events (because you have excluded the possibility of no six being rolled, which cannot apply to fair dice that have yet to be rolled.) Probability does not apply to events that have already happened. The probability of last week’s lottery numbers is not one in fourteen billion or whatever it is, because those numbers did actually come up. If you say “no, I am talking about future rolls, but only those ones with at least one six”, that is reintroducing an assumption that is not in the OP, i.e. that it will always be noted when at least one die is a six. ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [October 8, 2016, 7:11pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/259 "2016-10-08T19:11:09Z")
> [@Cliffy\_Biro](#): > > That is rather like talking about past events (because you have excluded the possibility of no six being rolled, which cannot apply to fair dice that have yet to be rolled.) Probability does not apply to events that have already happened. 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. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [October 8, 2016, 7:54pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/260 "2016-10-08T19:54:49Z")
> [@LSLGuy](#): > > 1,6 > 2,6 > 3,6 > 4,6 > 5,6 > 6,6 > 6,5 > 6,4 > 6,3 > 6,2 > 6,1 That is a complete listing of the possibilities as to what the pair of dice showed after the roll. Now what is the list of possibilities as to what die remains? Well, for each of the above possibilities, we remove a 6: 1 2 3 4 5 6 5 4 3 2 1 And now we ask, what is the probability that the remaining die is a 6? 1/11. [Previous page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=12) [Next page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=14)