# 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:** 8
**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, 3:54am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/141 "2016-10-07T03:54:42Z")
> [@septimus](#): > > Problem statement #1: > “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 then shown the other die. What is the probability of the other die showing a six?” > Problem statement #2: > “Two fair dice are rolled together. A 6 is removed, and you are then shown the other die. What is the probability of the other die showing a six?” > How do these problems differ from each other? Do they have the same answer? > > Reread the post you’re responding to. “A six is removed” can be interpreted in multiple ways. > > As for “you cannot show a 6 from a pair of rolled, fair dice when a 6 is not present” — that was my point. Read the two problems at the beginning of this post and tell us if they have the same answer. Are you saying that when the 6 is removed it’s not shown to confirm that statement A is correct? I.e. showing and removing the 6 proves that at least one 6 is rolled. That’s how I interpret this. Who doesn’t have dice and a friend and can’t replicate the scenario? And those people who keep saying 1/11 can’t count. The person rolling, showing, and removing a 6 has 2 6’s to show in the double 6 result. I rolled dice til I got 20 sets of double 6s and I took the result of how many rolls it took that included at least one 6. My ratio is 20/132. Which is 1/6.6. I’m going to keep rolling til I get 20 more. Brb. ---
**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, 4:04am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/142 "2016-10-07T04:04:26Z")
> [@octopus](#): > > Are you saying that when the 6 is removed it’s not shown to confirm that statement A is correct? … > > I rolled dice til I got 20 sets of double 6s and I took the result of how many rolls it took that included at least one 6. > > My ratio is 20/132. Which is 1/6.6. I’m going to keep rolling til I get 20 more. Brb. I’m not sure what you think you’re arguing with me about, but your experimental result _does_ surprise me! Are you saying that you rolled a pair of dice many times and had 20 instances of double-six and 112 instances (132-20) of a single six? Your result is not impossible, but is surprising. On average single-sixes will outnumber double-sixes ten to one! 😉 ---
**Author:** ![jtur88](https://avatars.discourse-cdn.com/v4/letter/j/e9c0ed/32.png) [@jtur88](https://boards.straightdope.com/u/jtur88)\ **Post date:** [October 7, 2016, 4:21am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/143 "2016-10-07T04:21:29Z")
> [@Chronos](#): > > There are possible results for the simultaneous throws that are not possible for your situation, such as a red 2 and a green 6. > . But the chances in simultaneous throw of green-2 and red-6 are exactly the same as red-2 and green-6, so that would not change anything, and can be ignored as a duplicating variation. Your throw of a red-2 gets absorbed as the mirrored green-2, canceling the roll of the other die which cannot be a 6 if the roll does not occur. Only if the odds are different if the 6 is green, would your argument hold sway. Once a die of any color is a 6, the experiment proceeds with an examination of the other die, which will always be independent of the first die. Say the two dice are thrown in separate rooms, and throws continue until one of the operators announces “I have a 6”. The operator in the other room examines his single die roll. What are the chances that it will be a 6? But one of the operators falls asleep, and is awakened by the first one saying “I have a 6”, and suddenly rolls his die for the first time. Do the odds change? ---
**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, 4:42am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/144 "2016-10-07T04:42:32Z")
> [@septimus](#): > > I’m not sure what you think you’re arguing with me about, but your experimental result _does_ surprise me! > > Are you saying that you rolled a pair of dice many times and had 20 instances of double-six and 112 instances (132-20) of a single six? Your result is not impossible, but is surprising. On average single-sixes will outnumber double-sixes ten to one! 😉 Ok! I’m back. It takes awhile to roll that many dice. This time my results look like this. Double 6 At least one 6 Odds Inverse 1 9 0.111111111 9 2 13 0.153846154 6.5 3 41 0.073170732 13.66666667 4 48 0.083333333 12 5 66 0.075757576 13.2 6 80 0.075 13.33333333 7 85 0.082352941 12.14285714 8 86 0.093023256 10.75 9 95 0.094736842 10.55555556 10 97 0.103092784 9.7 11 106 0.103773585 9.636363636 12 135 0.088888889 11.25 13 149 0.087248322 11.46153846 14 150 0.093333333 10.71428571 15 154 0.097402597 10.26666667 16 181 0.08839779 11.3125 17 182 0.093406593 10.70588235 18 201 0.089552239 11.16666667 19 207 0.09178744 10.89473684 20 227 0.088105727 11.35 This looks a lot closer to 1/11. Hmm, maybe you don’t count the green and red die separately and do treat that as one event. Maybe I’m the one who can’t count! Sorry for the ugly formatting but padding spaces seem to be removed. ---
**Author:** ![newme](https://avatars.discourse-cdn.com/v4/letter/n/e0b2c6/32.png) [@newme](https://boards.straightdope.com/u/newme)\ **Post date:** [October 7, 2016, 4:52am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/145 "2016-10-07T04:52:01Z")
> [@Skammer](#): > > Speaking as the OP, I’ll say no, you are both misunderstanding the (admittedly ambiguously worded) problem I posed. Although I did not come up with the wording myself. > > Two dice are rolled. The implication is that ANY of the 36 possible combinations may have turned up. **Not, you keep rolling until you have a six.** Why even bother rolling two dice if you only count the results when one of them lands on a specific face? That’s just like rolling one die in the first place and is a trivial question. > > 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. > [@x-ray\_vision](#): > > If we knew more information, such as what motivated the omniscient viewer to say that at least one six was rolled, then the probability may change. But we don’t know. The question as stated in post 130 is not ambiguous. > Right.\*\* We don’t count any of the 36 possible rolls that don’t include at least one 6.\*\* Did you read post 130 carefully? See the bolded part? **Skammer: "** \*\*Not, you keep rolling until you have a six." x-ray: “\*\*\*\*We don’t count any of the 36 possible rolls that don’t include at least one 6.” **You can’t have it both ways.The way I interpret** Skammer **in post 130 is** roll two dice, and use whatever comes up. No throwing anything out. \*\* ---
**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, 5:18am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/146 "2016-10-07T05:18:59Z")
I use Monte Carlo simulations a lot — in problems where the number of possibilities is too numerous for exhaustive counting. Here there are only 11 (or, depending on how you count, 36, or 4, or just 3) possibilities. Why not just list them: 6 & Not-6 … 5 cases Not-6 & 6 … 5 cases 6 & 6 … 1 case (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_.) ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [October 7, 2016, 10:18am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/147 "2016-10-07T10:18:48Z")
I can’t believe this thread is still going. People are looking for ambiguity where there is none. Roll two dice. Is at least one of them a six? P (25/36) No --\> end game P (11/36) Yes --\> There are now 11 possibilities: 6 1 6 2 6 3 6 4 6 5 6 6 1 6 2 6 3 6 4 6 5 6 Is the other die a six? P (10/11) No P (1/11) Yes. Answer is 1/11. Where is the ambiguity please? ---
**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 7, 2016, 11:20am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/148 "2016-10-07T11:20:11Z")
> [@](#): > > Quoth **jtur88** : > > But the chances in simultaneous throw of green-2 and red-6 are exactly the same as red-2 and green-6, so that would not change anything, and can be ignored as a duplicating variation. Sure, those odds are the same, but what’s the duplicating variation to red-6, green-6? You can’t factor out something (in this case, a factor of 2) unless it appears in all terms. ---
**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, 11:25am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/149 "2016-10-07T11:25:11Z")
> [@Colophon](#): > > Answer is 1/11. Where is the ambiguity please? “A 6 is removed.” This might mean “Monty picks one of the two dice at random and shows it to you. It happens to be a six.” Now the answer is 1/6. It’s similar to the ambiguity in the Monty Hall problem. “Monty opens a curtain. This exposes a goat.” Did Monty deliberately avoid the curtain with the brand-new Maserati? ---
**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 7, 2016, 11:45am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/150 "2016-10-07T11:45:15Z")
> [@Musicat](#): > > “What is the probability of the…die showing a six?” Six possibilities, one possible outcome. **Answer: 1/6** > > If you don’t agree with me, please show me where I have defined the problem wrong, or how the first parts of this problem affect the final question. This is such an obvious mistake! So there are 6 possibilities? That in _no way shape or form means they must be equally probable_! This is the second most common mistake made in the Monty Hall problem. (The first is misstating the problem.) You see two remaining doors and think they are equally likely to have the big prize. Look at any of the tables of eleven possible situations in this thread. Note that “the other is a 6” happens only one out of eleven times. For each of the others, they happen 2 out of eleven times. Given that the setup leaves 11 possibilities, how on Earth do you assign 1/6 to the (6,6) case but only 1/12 to _each_ of (3,6) and (6,3). How can that possibly be??? The chances of rolling an (a,b) and a (c,d) are the same. You don’t need to write any stupid code either. The listing of the possibilities, all equally likely, proves the answer without any further debate. ---
**Author:** ![jtur88](https://avatars.discourse-cdn.com/v4/letter/j/e9c0ed/32.png) [@jtur88](https://boards.straightdope.com/u/jtur88)\ **Post date:** [October 7, 2016, 12:02pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/151 "2016-10-07T12:02:08Z")
I think 1/11 is right. In Monopoly, a roll of 11 occurs twice as often as a 12. It defies what we think we know about probability, but it happens. ---
**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, 12:14pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/152 "2016-10-07T12:14:45Z")
> [@jtur88](#): > > In Monopoly, a roll of 11 occurs twice as often as a 12. It defies what we think we know about probability, but it happens. How does that defy what we know about probability? That’s pretty basic. The question in this thread is a little more nuanced and has to do with exactly how it is phrased. I, too, agree that 1/11 is the answer for the way it is phrased in the OP. ---
**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, 1:45pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/153 "2016-10-07T13:45:59Z")
> [@newme](#): > > Did you read post 130 carefully? See the bolded part? > > **Skammer: "** \*\*Not, you keep rolling until you have a six." > > x-ray: “\*\*\*\*We don’t count any of the 36 possible rolls that don’t include at least one 6.” > > **You can’t have it both ways.The way I interpret** Skammer **in post 130 is** roll two dice, and use whatever comes up. No throwing anything out. > \*\* Yes, I know what he wrote. Skammer was responding to two posters regarding how he phrased the problem. It is regarding one roll where at least one 6 is showing. I was responding to you regarding how to calculate probability. You wrote: _The only way to get a 1/11 result is by NOT counting some of the results._ Are you not talking about not counting the times a roll shows a 6? If you are, you’re correct. You don’t include the possible rolls of two dice where at least one 6 isn’t showing. I’m not having anything both ways. > [@ftg](#): > > This is such an obvious mistake! So there are 6 possibilities? That in _no way shape or form means they must be equally probable_! Oh, cmon. If I buy a lottery ticket, I either win or lose. That gives me a 50/50 chance. Gotta run. Off to the convenience store! ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [October 7, 2016, 3:32pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/154 "2016-10-07T15:32:29Z")
> [@septimus](#): > > “A 6 is removed.” This might mean “Monty picks one of the two dice at random and shows it to you. It happens to be a six.” Now the answer is 1/6. > > It’s similar to the ambiguity in the Monty Hall problem. “Monty opens a curtain. This exposes a goat.” Did Monty deliberately avoid the curtain with the brand-new Maserati? It’s not ambiguous as written. You are told that one of the dice is definitely a 6, before that 6 is removed. _So you know that the person telling you knows for a fact there is a 6_, so 25 of the 36 possibilities we had at the outset are immediately discounted. That leaves 11 possibilities, only one of which will result in the second die also being a 6. People are looking for complications where there are none. ---
**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, 3:42pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/155 "2016-10-07T15:42:06Z")
> [@Colophon](#): > > It’s not ambiguous as written. You are told that one of the dice is definitely a 6, before that 6 is removed. _So you know that the person telling you knows for a fact there is a 6_, so 25 of the 36 possibilities we had at the outset are immediately discounted. > > That leaves 11 possibilities, only one of which will result in the second die also being a 6. People are looking for complications where there are none. 11? How does a single die with 6 faces allow for 11 possibilities? Remember, we have eliminated everything except a single die. Nothing that comes before this is of any importance. ---
**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, 3:45pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/156 "2016-10-07T15:45:23Z")
> [@Musicat](#): > > Remember, we have eliminated everything except a single die. Nothing that comes before this is of any importance. You have asserted this. That doesn’t mean the rest of us are convinced. ---
**Author:** ![TroutMan](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/troutman/32/6721_2.png) [@TroutMan](https://boards.straightdope.com/u/TroutMan)\ **Post date:** [October 7, 2016, 3:47pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/157 "2016-10-07T15:47:27Z")
> [@Colophon](#): > > It’s not ambiguous as written. You are told that one of the dice is definitely a 6, before that 6 is removed. _So you know that the person telling you knows for a fact there is a 6_, so 25 of the 36 possibilities we had at the outset are immediately discounted. > > That leaves 11 possibilities, only one of which will result in the second die also being a 6. People are looking for complications where there are none. Yes, it is ambiguous because you don’t know how 6 was selected. Try the problem worded this way: two dice are rolled, a person peeks at one and sees it is 6. He then says, “at least one die is a 6.” What are the odds the other is a 6? The answer is 1/6. This is a valid interpretation of the problem in the OP, without knowing if 6 was chosen in advance (answer of 1/11) or after looking at a die (1/6). ---
**Author:** ![jtur88](https://avatars.discourse-cdn.com/v4/letter/j/e9c0ed/32.png) [@jtur88](https://boards.straightdope.com/u/jtur88)\ **Post date:** [October 7, 2016, 3:48pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/158 "2016-10-07T15:48:53Z")
> [@pulykamell](#): > > How does that defy what we know about probability? That’s pretty basic. The question in this thread is a little more nuanced and has to do with exactly how it is phrased. I, too, agree that 1/11 is the answer for the way it is phrased in the OP. It doesn’t. I said it defies WHAT WE THINK we know about probability. ---
**Author:** ![Turble](https://avatars.discourse-cdn.com/v4/letter/t/8dc957/32.png) [@Turble](https://boards.straightdope.com/u/Turble)\ **Post date:** [October 7, 2016, 3:53pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/159 "2016-10-07T15:53:19Z")
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. ---
**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, 3:54pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/160 "2016-10-07T15:54:25Z")
> [@Thudlow\_Boink](#): > > You have asserted this. That doesn’t mean the rest of us are convinced. 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. > [@Turble](#): > > **Musicat** nailed it in posts #89 and #115. Thank you, Sir. 🙂 [Previous page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=7) [Next page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=9)