# 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:** 7
**Author:** ![jtur88](https://avatars.discourse-cdn.com/v4/letter/j/e9c0ed/32.png) [@jtur88](https://boards.straightdope.com/u/jtur88)\ **Post date:** [October 6, 2016, 11:47pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/121 "2016-10-06T23:47:39Z")
Instead of rolling the dice simultaneously, do it sequentially. Keep throwing the red one until it is a 6. Then rolll the green one. What are the odds that the green one will be six? Just superimpose the two events in an overelap time frame… In what way would the results be skewed by the dice not being thrown simultaneously? Or by the dice being thrown in separate rooms by collaborating experimenters who then communicate the events to each other with a walkie talkie? ---
**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 6, 2016, 11:48pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/122 "2016-10-06T23:48:27Z")
> [@TroutMan](#): > > You’re falling into the same trap I did. Given your assumptions, your solution is exactly right and 1/11 is the correct answer. But the assumption is one I didn’t even realize I was making, that **six was picked as the “magic” number before any roles** , and therefore any roles that didn’t contain a six would be discarded. The other possibility was best explained to me by Chronos in [post 68](http://boards.straightdope.com/sdmb/showthread.php?p=19680453#post19680453). You aren’t actually told that six was selected before any roles; it could just be a statement of what one of the dice showed. In that case, the fact that one of the dice is a particular number doesn’t add any new information to the problem, and the other die has a 1/6 chance of being that number. It says a 6 is revealed. That means a six was rolled. It’s sort of a given. ---
**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 6, 2016, 11:49pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/123 "2016-10-06T23:49:29Z")
> [@jtur88](#): > > Instead of rolling the dice simultaneously, do it sequentially. Keep throwing the red one until it is a 6. Then rolll the green one. What are the odds that the green one will be six? > > In what way would the results be skewed by the dice not being thrown simultaneously? Or by the dice being thrown in separate rooms by collaborating experimenters? All those are 1/6. ---
**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, 12:11am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/124 "2016-10-07T00:11:03Z")
> [@jtur88](#): > > Instead of rolling the dice simultaneously, do it sequentially. Keep throwing the red one until it is a 6. Then rolll the green one. What are the odds that the green one will be six? Just superimpose the two events in an overelap time frame… > > In what way would the results be skewed by the dice not being thrown simultaneously? Or by the dice being thrown in separate rooms by collaborating experimenters who then communicate the events to each other with a walkie talkie? Again, the difference is similar to it is in the boy/girl problem: > [@](#): > > Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls? > Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys? > **[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}.... ---
**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, 12:13am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/125 "2016-10-07T00:13:41Z")
Different answers for each boy/girl problem. Also different answers depending on how this problem is phrased (see post 68). ---
**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, 1:08am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/126 "2016-10-07T01:08:02Z")
> [@x-ray\_vision](#): > > Different answers for each boy/girl problem. Also different answers depending on how this problem is phrased (see post 68). (A)“Mr. Smith says: ‘I have two children and at least one of them is a boy.’ Given this information, what is the probability that the other child is a boy?” (B)“Mr. Smith says: ‘I have two children and it is not the case that they are both girls.’ Given this information, what is the probability that both children are boys?” So what’s the paradox? It’s a paradox because people don’t read what is written? Child 1 is C1. Child 2 is C2. Here’s a table of what the children can be. C1 C2 B B B G G B G G Now we have to exclude G-G right? So we are left with C1 C2 B B B G G B You have 6 possible known, unknown pairs Child 1 being a boy, child 2 boy Child 1 being a boy, child 2 girl Child 1 being a girl, child 2 boy. Child 2 being a boy, child 1 boy Child 2 being a girl, child 1 boy Child 2 being a boy, child 1 girl. 2 of these 6 are the other child being a boy if the known child is a boy. So the answer is 1/3. Example 2 is the same as example 1 just written differently. Still get 1/3. I honestly don’t see what’s hard about that. ---
**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, 1:51am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/127 "2016-10-07T01:51:54Z")
> [@](#): > > Quoth **jtur88** : > > Instead of rolling the dice simultaneously, do it sequentially. Keep throwing the red one until it is a 6. Then rolll the green one. What are the odds that the green one will be six? Just superimpose the two events in an overelap time frame… > > In what way would the results be skewed by the dice not being thrown simultaneously? Or by the dice being thrown in separate rooms by collaborating experimenters who then communicate the events to each other with a walkie talkie? There are possible results for the simultaneous throws that are not possible for your situation, such as a red 2 and a green 6. **octopus** , the two statements you give of the boy-girl problem are equivalent, but those aren’t the same two statements that **x-ray vision** just gave. ---
**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, 1:53am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/128 "2016-10-07T01:53:53Z")
> [@Chronos](#): > > **octopus** , the two statements you give of the boy-girl problem are equivalent, but those aren’t the same two statements that **x-ray vision** just gave. That’s why I gave up. ---
**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, 1:54am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/129 "2016-10-07T01:54:50Z")
> [@TriPolar](#): > > From the OP: “Two fair dice are rolled together” > > So one die isn’t always a six. When one die is a six it’s 1/11. _“at least one of the dice is a 6”,_ thus one die is always a six. Look, the question is so poorly worded it is unanswerable. ---
**Author:** ![Skammer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/skammer/32/143_2.png) [@Skammer](https://boards.straightdope.com/u/Skammer)\ **Post date:** [October 7, 2016, 2:35am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/130 "2016-10-07T02:35:36Z")
> [@Musicat](#): > > That’s not the OP’s problem. It didn’t say anything about BOTH dice, but what would happen IF one condition was already met. That condition is no longer a factor, as we are not considering multiple conditions, nor conditional conditions. > [@jtur88](#): > > Instead of rolling the dice simultaneously, do it sequentially. Keep throwing the red one until it is a 6. Then rolll the green one. What are the odds that the green one will be six? Just superimpose the two events in an overelap time frame.. 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. ---
**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, 2:36am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/131 "2016-10-07T02:36:10Z")
> [@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. > > **octopus** , the two statements you give of the boy-girl problem are equivalent, but those aren’t the same two statements that **x-ray vision** just gave. Yeah, that’s true but I was reading the wiki article that he linked and was addressing one of the examples given as a paradox. With the examples x-ray vision gave the answers would be. A) Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls? Possibility table. Older child is OC. Younger child is YC. OC YC G B G G 2 permutations to satisfy first condition of OC being a girl. 1 of the permutation satisfies the condition of 2 girls. So: 1/2 B) Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys? 1/3 Same table as my previous post. Where is the paradox? Different questions have different answers. ---
**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, 2:37am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/132 "2016-10-07T02:37:48Z")
> [@octopus](#): > > Yeah, that’s true but I was reading the wiki article that he linked and was addressing one of the examples given as a paradox. Then maybe you should start a new thread? ---
**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, 2:42am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/133 "2016-10-07T02:42:50Z")
> [@x-ray\_vision](#): > > Then maybe you should start a new thread? What’s hard about calculating number of events that meet criteria A divided by number of events that meet criteria B? Plus I answered your questions. Which since they are different than the OP’s shouldn’t they be in their own thread as well? ---
**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, 2:48am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/134 "2016-10-07T02:48:22Z")
> [@octopus](#): > > What’s hard about calculating number of events that meet criteria A divided by number of events that meet criteria B? No one here is discussing how hard something is. > [@](#): > > Plus I answered your questions. Which since they are different than the OP’s shouldn’t they be in their own thread as well? My questions are not different. They are regarding your confusion with answering the OP’s question depending how the OP’s question is interpreted. The answer is 1/11 in the way he just phrased it in post 130. ---
**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, 2:58am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/135 "2016-10-07T02:58:33Z")
> [@x-ray\_vision](#): > > No one here is discussing how hard something is. > My questions are not different. They are regarding your confusion with answering the OP’s question depending how the OP’s question is interpreted. The answer is 1/11 in the way he just phrased it in post 130. How do you interpret the questions you asked other than the obvious. Older child being a girl and figuring out the probability that the 2nd is as well isn’t ambiguous at all. Being shown that one die is a 6 after a pair of dice have been rolled is ambiguous in what way precisely? If I flip a coin and roll a die and I show you heads on the coin how does that change what the die rolled? It’s still a 1/6 chance on the die if you get shown a head, an ace of spades, a Draw 4 card, or another 6. Do that with a coin. Flip one and roll a die. What’s the probability if heads comes up you roll a 6? How’s it different with 2 different dice? This is a repeatable experiment. ---
**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, 3:05am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/136 "2016-10-07T03:05:35Z")
> [@Skammer](#): > > Here is the problem verbatim: > > “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?” As others agree, the problem statement is ambiguous. (Moreover, since “a six is removed,” the pre-knowledge that “at least one of the dice is a 6” is redundant.) Did the guy remove a die at random and it happened to be a six? (Then the answer is 1/6.) Did he remove any duplicated die? (The the answer is 1.) Did he remove a single six, whenever there was at least one six? (The the answer is 1/11.) If he always removes a single six when he can, but chooses the red six when both red and green are present, did he show you the color of the removed six? (Now the answers are 1/6 or 0, for red, green, resp.) As usual, “running a simulation” makes sense only when the problem is well-defined. But if it’s well-defined, why run a simulation when you can just treat 11 cases? (or even just 3 or 4 cases.) ---
**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, 3:06am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/137 "2016-10-07T03:06:27Z")
> [@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. This wording is still ambiguous. You haven’t told us WHY the omniscient viewer chose to say “six”. If the roll is six - one, why did he choose to say six instead of one? What would the omniscient viewer say if the roll is one - two? The only way to get a 1/11 result is by NOT counting some of the results. ---
**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, 3:11am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/138 "2016-10-07T03:11:48Z")
> [@septimus](#): > > As others agree, the problem statement is ambiguous. (Moreover, since “a six is removed,” the pre-knowledge that “at least one of the dice is a 6” is redundant.) > > Did the guy remove a die at random and it happened to be a six? (Then the answer is 1/6.) Did he remove any duplicated die? (The the answer is 1.) Did he remove a single six, whenever there was at least one six? (The the answer is 1/11.) If he always removes a single six when he can, but chooses the red six when both red and green are present, did he show you the color of the removed six? (Now the answers are 1/6 or 0, for red, green, resp.) > > As usual, “running a simulation” makes sense only when the problem is well-defined. But if it’s well-defined, why run a simulation when you can just treat 11 cases? (or even just 3 or 4 cases.) If you do the process as described. That is have one person roll two dice and show the other person a 6 whenever a 6 is present in the pair of dice you can figure out an unambiguous answer. Obviously, you cannot show a 6 from a pair of rolled, fair dice when a 6 is not present. Where is the ambiguity? Just do this with someone else as explained and it’s not hard. ---
**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, 3:26am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/139 "2016-10-07T03:26:32Z")
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? > [@septimus](#): > > Did the guy remove a die at random and it happened to be a six? (Then the answer is 1/6.) Did he remove any duplicated die? (The the answer is 1.) Did he remove a single six, whenever there was at least one six? (The the answer is 1/11.) If he always removes a single six when he can, but chooses the red six when both red and green are present, did he show you the color of the removed six? (Now the answers are 1/6 or 0, for red, green, resp.) > [@octopus](#): > > If you do the process as described. That is have one person roll two dice and show the other person a 6 whenever a 6 is present in the pair of dice you can figure out an unambiguous answer. > > Obviously, you cannot show a 6 from a pair of rolled, fair dice when a 6 is not present. Where is the ambiguity? Just do this with someone else as explained and it’s not hard. 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. ---
**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, 3:30am UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/140 "2016-10-07T03:30:02Z")
> [@newme](#): > > This wording is still ambiguous. You haven’t told us WHY the omniscient viewer chose to say “six”. If the roll is six - one, why did he choose to say six instead of one? What would the omniscient viewer say if the roll is one - two? 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. > [@](#): > > The only way to get a 1/11 result is by NOT counting some of the results. Right. We don’t count any of the 36 possible rolls that don’t include at least one 6. [Previous page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=6) [Next page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=8)