# 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:** 9
**Author:** ![pulykamell](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/pulykamell/32/3166_2.png) [@pulykamell](https://boards.straightdope.com/u/pulykamell)\ **Post date:** [October 7, 2016, 4:12pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/161 "2016-10-07T16:12:19Z")
> [@jtur88](#): > > It doesn’t. I said it defies WHAT WE THINK we know about probability. Sorry, I meant to say think. It just seems like something anybody who plays with dice a lot would have figured out intuitively. ---
**Author:** ![Musicat](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/musicat/32/20189_2.png) [@Musicat](https://boards.straightdope.com/u/Musicat)\ **Post date:** [October 7, 2016, 4:13pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/162 "2016-10-07T16:13:05Z")
> [@Skammer](#): > > So, you have one of 36 possible combinations. Then the omniscient viewer says, “Look, you rolled at least one six.” At that point, what is the chance your other die is a six? > > It’s 1/11. **At that point** is key here. _At that point,_ you are beginning to calculate odds. Not before. Most of the 36 possible double-die roll combinations have been eliminated. _They do not count._ We have no numerator or denominator yet. We are reduced to “what are the chances of rolling a single die and getting N?” Now the numerator is 1 and the denominator is 6. If that wasn’t your intention, you need to seriously revise the premise. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [October 7, 2016, 4:13pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/163 "2016-10-07T16:13:24Z")
> [@Turble](#): > > The question is not ambiguous or poorly worded; it does contain information that is not needed to find the answer. That is a very common thing in math word problems. When solving real world problems it is always necessary to sort out which information is relevant and what can be ignored, and to also figured what is the actual question to be answered. > > Two fair dice are rolled: Irrelevant. It makes no difference how many dice are rolled. > > The dice are rolled together: Irrelevant. It doesn’t matter if the dice are rolled together or if one is rolled on your desk and the other is rolled while dancing naked around a fire of willow sticks in a cemetery on All Hallows Eve. > > One of the dice is a 6: Irrelevant. Rolls of dice are independent events. > > A 6 is removed: Irrelevant. The number on one die in no way alters the outcome of any other die at any place in time or space. > > The probability of the [other] die showing a six: Is the actual question. > > The probability of a fair six-sided die showing a 6 (or any other number from 1 to 5) is always 1/6; it is an independent event and there is no force that can change that to 1/11 or any other number. Any attempt to interpret the question to allow such an answer must be an incorrect interpretation because it leads to the nonsensical event where the odds of an independent event are altered. > > **Musicat** nailed it in posts #89 and #115. I disagree. I believe that the question is fundamentally ambiguous; the fact that it can easily be interpreted as having so much “irrelevant” information is a consequence of this. You’re going through the whole sequence saying “You’re trying to do sleight of hand, but you can’t fool me, you can’t get me to look away from this die that you’re calling the ‘other die’, and the odds of that die rolling a 6 are 1 in 6 and that can’t change…” But the real sleight of hand, as it were, is that I’m possibly not deciding which die I’m calling “the other die” until after I’ve rolled both dice, and THAT changes the odds. ---
**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 7, 2016, 4:19pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/164 "2016-10-07T16:19:31Z")
> [@Musicat](#): > > **At that point** is key here. _At that point,_ you are beginning to calculate odds. Not before. Most of the 36 possible double-die roll combinations have been eliminated. _They do not count._ We have no numerator or denominator yet. We are reduced to “what are the chances of rolling a single die and getting N?” Now the numerator is 1 and the denominator is 6. > > If that wasn’t your intention, you need to seriously revise the premise. I rolled dice for about an hour last night and did the problem as worded. It does seem to be 1/11. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [October 7, 2016, 4:25pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/165 "2016-10-07T16:25:51Z")
(Breaking a post up to address each point as well as I can.) > [@Musicat](#): > > **At that point** is key here. _At that point,_ you are beginning to calculate odds. Not before. I don’t agree with this. You can begin calculating odds at the beginning, with rolling both dice, and then modify them as new information comes in, “at each point.” > [@](#): > > Most of the 36 possible double-die roll combinations have been eliminated. _They do not count._ True; 25 of them. > [@](#): > > We have no numerator or denominator yet. We can have a denominator; it’s 11, the number of cases which haven’t been eliminated. Depending on exactly what we’re looking at we could have several different numerators. > [@](#): > > **We are reduced to “what are the chances of rolling a single die and getting N?”** > Now the numerator is 1 and the denominator is 6. No, I don’t think we are. If it helps, let’s leave aside the question about “the other die” to start with. I’ve rolled a green die and a red die. **Skammer** tells us that at least one of those dice is a 6. What are the chances that the green die shows a 4? You could say that the chances are 1 in 6, but I don’t agree. We have some information which might relate to the green die; that either it or the red die (or both) is a 6. I say that we’ve got 11 equal possibilities remaining, and in only 1 of them does the green die have a 4. Would you agree with that? ---
**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 7, 2016, 4:43pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/166 "2016-10-07T16:43:44Z")
> [@Musicat](#): > > I don’t find the OP ambiguous at all, to wit: “…you are ten [sic] shown the other die. What is the probability of the other die showing a six?” > > We are talking about ONE (the other) die. What is the probability of it being a 6? > > I don’t know how much clearer we can be. What if it were phrased without asking about “the other die”? I’m curious what you believe the answer is if it were phrased like this?: _Then the omniscient viewer says, “Look, you rolled at least one six.” At that point, what is the chance both dice are 6s?_ ---
**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 7, 2016, 4:46pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/167 "2016-10-07T16:46:00Z")
> [@Musicat](#): > > I don’t find the OP ambiguous at all, to wit: “…you are ten [sic] shown the other die. What is the probability of the other die showing a six?” > > We are talking about ONE (the other) die. What is the probability of it being a 6? > > I don’t know how much clearer we can be. You roll three dice, and remove the highest two. What is the probability that the other die is a 6? ---
**Author:** ![markn\_1](https://avatars.discourse-cdn.com/v4/letter/m/f9ae1b/32.png) [@markn\_1](https://boards.straightdope.com/u/markn_1)\ **Post date:** [October 7, 2016, 5:14pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/168 "2016-10-07T17:14:27Z")
Sigh. As I predicted back in post #24, this thread is generating “endless fruitless debate” because people are disagreeing on the interpretation of the question. For those of you who think the answer is unambiguosly 1/11, consider this situation: I roll a pair of dice. I note the highest number showing (call it X), and say to another person, “At least one of the dice is showing an X. What is the probability that the other is showing an X?” Clearly the answer is 1/6. I repeat this multiple times. Each time the answer is 1/6, although X may be different each time. Now a third person walks in the room just as I’ve rolled a pair with a 6 showing, and hears me say “At least one of the dice is showing a 6.” This is the same as the description in the OP, but the answer here is 1/6 not 1/11. As stated several times in this thread, the answer **depends on the randomization procedure** , which was not stated in the OP. If you assume you know what the randomization procedure was (eg. “ignore throws that don’t show a 6”) you’ll get a specific answer, but it’s not the only answer because THE RANDOMIZATION PROCEDURE WAS NOT DESCRIBED and the answer changes if the randomization procedure changes. –Mark ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [October 7, 2016, 5:22pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/169 "2016-10-07T17:22:19Z")
I’m a perfectionist and don’t like half-correct posts to stand. > [@septimus](#): > > (Similarly, when I tell you a pair of red Kings will make a better hand than a pair of black Aces 18.553597959240882460% of the time in all-in Texas Hold-em, that’s the result of _enumeration_, not _simulation_.) The actual probability is of course a rational number. I get that it is 52949 / 285384 which has a repeating decimal expansion of period 506: .185 53597959240882460123903232136349620160906007344490 23070669694166456423625711322288565581812575337089 67566506881955540604939309842177557256188153505452 30286210859753875480054943514702996664143750175202 53412945364841757071174277464749250133153925938384 77279735374092450873209430101196983713172427325988 84310262663639166876909707622011044767751520757996 24365766826451377792728394023491155776077145179827 88103047122473579457853278389818630336669189583158 13079920387968491576262159055868584083200179407394948560 53597959240882460123903232136349620160906007344490 23070669694166456423625711322288565581812575337089 67566506881955540604939309842177557256188153505452 30286210859753875480054943514702996664143750175202 53412945364841757071174277464749250133153925938384 77279735374092450873209430101196983713172427325988 84310262663639166876909707622011044767751520757996 24365766826451377792728394023491155776077145179827 88103047122473579457853278389818630336669189583158 13079920387968491576262159055868584083200179407394948560 … ---
**Author:** ![TriPolar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tripolar/32/3008_2.png) [@TriPolar](https://boards.straightdope.com/u/TriPolar)\ **Post date:** [October 7, 2016, 5:26pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/170 "2016-10-07T17:26:18Z")
> [@](#): > > 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? Assuming _ten_ was meant to be _then_, the only ambiguity I see is the phrase “you are told that ‘at least one of the dice is a 6’”. That could mean whoever is telling you is a liar. So in that case it’s like saying two fair dice are rolled, one of them is removed, what is the probability of it being a 6? So the answer is that it is either 1/6 or 1/11 depending on whether or not the die removed is actually a 6. Given that the choices are " 1/18, 1/36, 2/11 and 1/6" we have to assumed we were lied to and the answer is 1/6, and who ever posed the problem is just being annoying. That’s my conclusion, and I’m sticking to it, because it really doesn’t matter. ETA: Arggh! Even then there’s no ambiguity, it states a 6 is removed, so there’s no lie. The answer is 1/11 and the answer is none of the above. Unless we assume it’s all a lie it’s not ambiguous. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [October 7, 2016, 5:28pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/171 "2016-10-07T17:28:30Z")
> [@markn\_1](#): > > I roll a pair of dice. I note the highest number showing (call it X), and say to another person, “At least one of the dice is showing an X. What is the probability that the other is showing an X?” Clearly the answer is 1/6. I repeat this multiple times. Each time the answer is 1/6, although X may be different each time. I’m not sure that’s completely clear, especially if I know you’re picking the highest number to tell me about. If X=1, then my answer is 1. For X=2, I’d say 1/3, for 3, 1/5, for 4, 1/7, for 5, 1/9 and 1/11 for 6. Now, since you’re more likely tell me a higher number in proportion, the weighted average for my answer comes back to 1 out of 6. ---
**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 7, 2016, 5:36pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/172 "2016-10-07T17:36:58Z")
> [@markn\_1](#): > > I roll a pair of dice. I note the highest number showing (call it X), and say to another person, “At least one of the dice is showing an X. What is the probability that the other is showing an X?” Please clarify. What do you mean by “the other”? Is “the other is showing an X” equivalent to “both are showing an X,” since we are already given that at least one is? If not, what does it mean? > [@](#): > > Clearly the answer is 1/6. No, this is not “clearly” at all. Can you show your work? ETA after seeing **chrisk** ’s post: _Does_ the other person know that X is the highest number showing? As I interpret it, a question of the form “A is true. What is the probability that B is true?” is a conditional probability question. It means, “_Given that A is true_, what is the probability that B is true?” It’s asking for P(B|A), not P(B). ---
**Author:** ![DrDeth](https://avatars.discourse-cdn.com/v4/letter/d/b487fb/32.png) [@DrDeth](https://boards.straightdope.com/u/DrDeth)\ **Post date:** [October 7, 2016, 5:39pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/173 "2016-10-07T17:39:47Z")
> [@markn\_1](#): > > Sigh. As I predicted back in post #24, this thread is generating “endless fruitless debate” because people are disagreeing on the interpretation of the question. They wont stop, no matter how many times this is pointed out. :rolleyes: ---
**Author:** ![don\_t\_ask](https://avatars.discourse-cdn.com/v4/letter/d/e68b1a/32.png) [@don\_t\_ask](https://boards.straightdope.com/u/don_t_ask)\ **Post date:** [October 7, 2016, 5:40pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/174 "2016-10-07T17:40:35Z")
I can’t see that there is any ambiguity in the stated question at all. Ironically the idea of removing a 6 is an attempt to eliminate any ambiguity caused by saying “if one die is 6.” So it becomes like a casino game. You bet $100 and roll two dice. No 6 showing and the croupier pushes both dice back to you and you are coming out again. Any sixes showing the croupier takes one off the table. If the remaining die isn’t a 6 you lose your $100. If it is a 6 you win the payout which should be $1100. For those people who believe the correct odds are one in six I will happily set up this game and pay them $700 every time the remaining die on the table is a 6. ---
**Author:** ![markn\_1](https://avatars.discourse-cdn.com/v4/letter/m/f9ae1b/32.png) [@markn\_1](https://boards.straightdope.com/u/markn_1)\ **Post date:** [October 7, 2016, 5:57pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/175 "2016-10-07T17:57:15Z")
> [@chrisk](#): > > I’m not sure that’s completely clear, especially if I know you’re picking the highest number to tell me about. If X=1, then my answer is 1. For X=2, I’d say 1/3, for 3, 1/5, for 4, 1/7, for 5, 1/9 and 1/11 for 6. You’re right, I shouldn’t have said “highest number”. I should have said “I pick one of the dice at random and announce its value”. The rest of my argument holds. –Mark ---
**Author:** ![Musicat](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/musicat/32/20189_2.png) [@Musicat](https://boards.straightdope.com/u/Musicat)\ **Post date:** [October 7, 2016, 6:12pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/176 "2016-10-07T18:12:21Z")
> [@Thudlow\_Boink](#): > > You roll three dice, and remove the highest two. What is the probability that the other die is a 6? Now you’ve moved the goalposts and changed the formula and odds. If you remove the highest two, the chances of the remaining one being a high number are reduced. This is not the original premise at all. Do you see the difference? ---
**Author:** ![MonkeyMensch](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@MonkeyMensch](https://boards.straightdope.com/u/MonkeyMensch)\ **Post date:** [October 7, 2016, 6:21pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/177 "2016-10-07T18:21:36Z")
What a cool problem! I think the problem is unambiguous (excepting the typo for “then”) and my first answer was 1/11. I was wrong. The list of combinations showing boxcars twice are, in a sense, correct. There are two ways that a six can be removed when double six is rolled and that case (“a six is removed”) needs to be counted twice. The right answer, as others have pointed out, is 1/6. But, boy, it’s a subtle difference. Also, this thread is bound for 10 pages plus, easy… ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [October 7, 2016, 6:22pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/178 "2016-10-07T18:22:20Z")
> [@Musicat](#): > > Now you’ve moved the goalposts and changed the formula and odds. If you remove the highest two, the chances of the remaining one being a high number are reduced. This is not the original premise at all. Do you see the difference? If I remove one **because** it’s a 6, then the chances of the remaining one being a six are reduced. It’s the same principle! ---
**Author:** ![Musicat](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/musicat/32/20189_2.png) [@Musicat](https://boards.straightdope.com/u/Musicat)\ **Post date:** [October 7, 2016, 6:24pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/179 "2016-10-07T18:24:49Z")
> [@x-ray\_vision](#): > > What if it were phrased without asking about “the other die”? I’m curious what you believe the answer is if it were phrased like this?: > > _Then the omniscient viewer says, “Look, you rolled at least one six.” At that point, what is the chance both dice are 6s?_ I fail to see the difference between saying “look, you rolled at least one six” and “we now remove the six.” The remaining “other” die can be any value from 1…6 and is unaffected by the status or even presence of the first die. How about this version? _…the omniscient viewer says, “Look, you rolled at least one six.” At that point, what is the chance the other die is 2?_ Answer: 1/6 I dare you to try it, either with real dice or a computer simulation, assuming you program it correctly, of course. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [October 7, 2016, 6:28pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/180 "2016-10-07T18:28:22Z")
> [@Musicat](#): > > I fail to see the difference between saying “look, you rolled at least one six” and “we now remove the six.” The remaining “other” die can be any value from 1..6 and is unaffected by the status or even presence of the first die. > > How about this version? _…the omniscient viewer says, “Look, you rolled at least one six.” At that point, what is the chance the other die is 2?_ Answer: 1/6 > > I dare you to try it, either with real dice or a computer simulation, assuming you program it correctly, of course. What is your definition of programming it correctly? That is, to what specifications should we set up the simulation? [Previous page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=8) [Next page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=10)