# 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:** 3
**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 6, 2016, 6:04pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/41 "2016-10-06T18:04:33Z")
> [@Shagnasty](#): > > In case someone likes a real-world demonstration more than theory, I just modeled this problem in Excel to simulate 200,000 rolls of dice pairs. No matter how many times I run it, the answer always comes out to be right at 1/11th that both dice will be a 6 if either of them is. Yes of course, but your simulation implicitly assumes the first of the two interpretations that I suggested. It’s not really possible to simulate the second interpretation, but it’s still a (somewhat) reasonable interpretation. –Mark ---
**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, 6:14pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/42 "2016-10-06T18:14:14Z")
> [@Skammer](#): > > A math teacher friend of mine shared this problem on Facebook and we are disagreeing about the solution. 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?” > > My friend thinks the answer is 1/6, because the dice rolls are independent events and regardless of what one shows, the other has a 1/6 chance of being a six. I’m quite positive this is incorrect. > > My reasoning is: if you roll a pair of dice, there are 36 possible combinations: (1,1), (1,2), (2,1), etc. Out of these 36 combinations, 11 contain at least one six: each 6 paired with 1-5 on the other die, and double sixes. > > Since we are told that the pair contains at least one six, we have 11 possible combinations with equal chance of occurring. If a six is then removed, only 1 of the remaining 11 dice is another six. So I maintain the answer is 1/11. > > However, the meme that she posted is multiple choice and 1/11 is not one of the available answers, which are: 1/18, 1/36, 2/11 and 1/6. > > At first I jumped on the 2/11 thinking “there are two chances to pick a six from the pair of doubles!” But that logic falls apart I think. I’m back to 1/11. How am I wrong? It’s 1/36. There is only one event that results in a 6 being on the other die of a pair when one 6 is removed and that is double 6s. Double 6s is 1/36 chance. ---
**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, 6:16pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/43 "2016-10-06T18:16:07Z")
> [@TriPolar](#): > > Isn’t the question asking the probability of two sixes being rolled? Yes. This shouldn’t be tricky at all. Yet look at the fun you can have with Monty Hall and the goat. ---
**Author:** ![DrDeth](https://avatars.discourse-cdn.com/v4/letter/d/b487fb/32.png) [@DrDeth](https://boards.straightdope.com/u/DrDeth)\ **Post date:** [October 6, 2016, 6:26pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/44 "2016-10-06T18:26:30Z")
> [@TriPolar](#): > > Isn’t the question asking the probability of two sixes being rolled? Maybe. The question is so badly worded, who knows? They failed to properly _specify the randomizing procedure_. ---
**Author:** ![DrDeth](https://avatars.discourse-cdn.com/v4/letter/d/b487fb/32.png) [@DrDeth](https://boards.straightdope.com/u/DrDeth)\ **Post date:** [October 6, 2016, 6:28pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/45 "2016-10-06T18:28:25Z")
> [@octopus](#): > > It’s 1/36. There is only one event that results in a 6 being on the other die of a pair when one 6 is removed and that is double 6s. Double 6s is 1/36 chance. Not if one die is given to be a six. Ok, the question is badly worded, so let’s look at it another way. One of the dice has to be a six, right? So let us toss two dice, one is normal the other has 6’s on all faces. What is then the chance of the other die rolling a 6? ---
**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, 6:34pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/46 "2016-10-06T18:34:59Z")
> [@DrDeth](#): > > Not if one die is given to be a six. > > Ok, the question is badly worded, so let’s look at it another way. One of the dice has to be a six, right? > > So let us toss two dice, one is normal the other has 6’s on all faces. What is then the chance of the other die rolling a 6? If it’s only asked when one is a 6? Then the answer is 1/6. Independent each have 1/6 chance. But lets do the red die green die list of possibilities. R G 1 6 2 6 3 6 4 6 5 6 6 6 6 1 6 2 6 3 6 4 6 5 6 6 There are 12 different combinations in which a 6 can be given. Six that are green and six that are red 6s. That’s why you can count the 6 6 twice because you are given a different die. You are given either the red or the green so it’s two events. Ok you have twelve possible and equally likely events. Only two still have a 6 left over. 2/12 gives 1/6 which is what should be expected treating dice as individual events and only doing this in the trials you have at least one 6 rolled. ---
**Author:** ![watchwolf49](https://avatars.discourse-cdn.com/v4/letter/w/e9c0ed/32.png) [@watchwolf49](https://boards.straightdope.com/u/watchwolf49)\ **Post date:** [October 6, 2016, 6:39pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/47 "2016-10-06T18:39:34Z")
> [@octopus](#): > > Yes. This shouldn’t be tricky at all. Yet look at the fun you can have with Monty Hall and the goat. Goats are cool … ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [October 6, 2016, 6:41pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/48 "2016-10-06T18:41:36Z")
> [@octopus](#): > > It’s 1/36. There is only one event that results in a 6 being on the other die of a pair when one 6 is removed and that is double 6s. Double 6s is 1/36 chance. But you are ignoring the fact that 25 out of those 36 chances won’t have a six at all, so they can be eliminated from the game. Edit: that page linked by **x-ray vision** gives the clearest explanation: > [@](#): > > Another way of phrasing this is to explicitly use conditional probability. You are asking for P[both dice are 2|at least one die is 2]. Since the first event implies the second, this probability is equal to P[both dice are 2]/P[at least one die is 2]. Since P[both dice are 2]=1/36 and P[at least one die is 2]=11/36, we have P[both dice are 2|at least one die is 2]=(1/36)/(11/36)=1/11. ---
**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 6, 2016, 6:42pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/49 "2016-10-06T18:42:31Z")
> [@Skammer](#): > > “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?” There’s your problem. No die has a ten on any face. ---
**Author:** ![Turble](https://avatars.discourse-cdn.com/v4/letter/t/8dc957/32.png) [@Turble](https://boards.straightdope.com/u/Turble)\ **Post date:** [October 6, 2016, 6:42pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/50 "2016-10-06T18:42:52Z")
I suspect that those who are running simulations are testing one simulated die and discarding the roll if that die is not a 6. Please try it by checking **both** dice for a 6 before discarding the roll. ---
**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 6, 2016, 6:49pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/51 "2016-10-06T18:49:11Z")
> [@Turble](#): > > I suspect that those who are running simulations are testing one simulated die and discarding the roll if that die is not a 6. Please try it by checking **both** dice for a 6 before discarding the roll. 1/11 is the answer if you check both dice. If you only check one specific die the answer would be 1/6. –Mark ---
**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 6, 2016, 6:53pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/52 "2016-10-06T18:53:48Z")
I ran a simulation. Two random numbers between 1 and 6 are defined. If neither is a 6 nothing is recorded. If either is a 6 I recorded the other one. After 1 million rolls I get this: r(1)=55549 r(2)=55650 r(3)=55540 r(4)=55371 r(5)=55591 r(6)=27832 total=305533 total/27832=10.97775941362460477 so it has to be 1/11. ---
**Author:** ![DrDeth](https://avatars.discourse-cdn.com/v4/letter/d/b487fb/32.png) [@DrDeth](https://boards.straightdope.com/u/DrDeth)\ **Post date:** [October 6, 2016, 6:53pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/53 "2016-10-06T18:53:51Z")
> [@octopus](#): > > If it’s only asked when one is a 6? Then the answer is 1/6. Independent each have 1/6 chance. Yes, and that is one way of reading it. This debate is pointless. The question is so poorly worded there are two equally possible answers. Saying loudly “MY ANSWER IS THE ONE TRUE RIGHT ANSWER” doesnt help when the question allows for two equally right answers. 😛 ---
**Author:** ![DrDeth](https://avatars.discourse-cdn.com/v4/letter/d/b487fb/32.png) [@DrDeth](https://boards.straightdope.com/u/DrDeth)\ **Post date:** [October 6, 2016, 6:55pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/54 "2016-10-06T18:55:05Z")
> [@TriPolar](#): > > I ran a simulation. Two random numbers between 1 and 6 are defined. If neither is a 6 nothing is recorded. If either is a 6 I recorded the other one. After 1 million rolls I get this: > > r(1)=55549 > r(2)=55650 > r(3)=55540 > r(4)=55371 > r(5)=55591 > r(6)=27832 > > total=305533 > > total/27832=10.97775941362460477 > > so it has to be 1/11. If one die is always a six, then what is the %? 1 in 6. You can run simulations all day, but without the right question, you cant get the right answer. ---
**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 6, 2016, 6:58pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/55 "2016-10-06T18:58:52Z")
> [@Turble](#): > > I suspect that those who are running simulations are testing one simulated die and discarding the roll if that die is not a 6. Please try it by checking **both** dice for a 6 before discarding the roll. This is exactly what I did - if **neither** die is a six, the roll is discarded. The answer is 1/11. ---
**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 6, 2016, 7:07pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/56 "2016-10-06T19:07:11Z")
Okay, thanks for all the discussion. The way both I interpreted the question, as the OP: 1. Two (normal, fair, six-sided) dice are rolled. The result can be one of any 36 possible combinations. You don’t know the results. 2. You are told that at least one of the dice is a six. Could be either die or both. 3. You are shown one of the dice, which is the (or a) six. What are the chances the other one is a six? Thanks to Troutman and Shagnasty and whomever else ran the simulations and confirmed my 1/11 response based on those assumptions. ---
**Author:** ![Cartoonacy](https://avatars.discourse-cdn.com/v4/letter/c/848f3c/32.png) [@Cartoonacy](https://boards.straightdope.com/u/Cartoonacy)\ **Post date:** [October 6, 2016, 7:09pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/57 "2016-10-06T19:09:05Z")
> [@octopus](#): > > If it’s only asked when one is a 6? Then the answer is 1/6. Independent each have 1/6 chance. > > But lets do the red die green die list of possibilities. > > R G > 1 6 > 2 6 > 3 6 > 4 6 > 5 6 > 6 6 > 6 1 > 6 2 > 6 3 > 6 4 > 6 5 > 6 6 > > There are 12 different combinations in which a 6 can be given. Six that are green and six that are red 6s. That’s why you can count the 6 6 twice because you are given a different die. You are given either the red or the green so it’s two events. Ok you have twelve possible and equally likely events. Only two still have a 6 left over. 2/12 gives 1/6 which is what should be expected treating dice as individual events and only doing this in the trials you have at least one 6 rolled. In other words, you’re saying that the possibilities are: R G REMOVED 1 6 Green 2 6 Green 3 6 Green 4 6 Green 5 6 Green 6 6 Green 6 6 Red 6 5 Red 6 4 Red 6 3 Red 6 2 Red 6 1 Red Hmmm – not the way I’d have worked it out, but I’ll go along with it. ---
**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, 7:10pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/58 "2016-10-06T19:10:45Z")
> [@TroutMan](#): > > This is exactly what I did - if **neither** die is a six, the roll is discarded. The answer is 1/11. And I bet the simulation has this flaw. Counting the double six event wrong. The case in which two sixes are rolled needs to be counted twice because the way that event is handled has two permutations. Permutation one is when die 1 is removed and shown. Permutation two is when die 2 is removed and shown. Those are not the same event. If one die is red and one die is green there are 6 events where you show a green 6. There are 6 events you show a red 6. These are the 12 events possible. Only two events have a 6 remaining and those are show red 6 while holding a green 6 or show green 6 while holding a red 6. These are distinct events and need to be counted as such. That is 2/12. Or 1/6. 1/11 can only occur when you treat different events as one event. Showing the red is not the same as showing the green. ---
**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 6, 2016, 7:10pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/59 "2016-10-06T19:10:47Z")
> [@DrDeth](#): > > If one die is always a six, then what is the %? 1 in 6. One die that always rolls a six is not a normal, fair die. ---
**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 6, 2016, 7:12pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/60 "2016-10-06T19:12:42Z")
**octopus** , you are wrong. The event where the red die is 6 and the green die is six is the SAME EVENT as the green die being six and the red die being six. You can only count it once. If you play out every one of the 36 possible combinations of two dice you only get (6,6) once. [Previous page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=2) [Next page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=4)