# 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:** 2
**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 6, 2016, 4:10pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/21 "2016-10-06T16:10:01Z")
> [@chrisk](#): > > I would say that the wording is not only distracting, it’s ambiguous. We have no idea whether there was intention or not in removing the 6, so we can’t tell if it changes the odds or how. I suppose you could interpret it to mean “A die is removed, and that die happens to be a 6,” in which case the answer would be 1/6. But the more natural (IMHO) interpretation is that it’s part of the setup that a 6 specifically be removed. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [October 6, 2016, 4:10pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/22 "2016-10-06T16:10:49Z")
> [@x-ray\_vision](#): > > Right, but as has been mentioned, how is it that a 6 being removed is decided? > > We have: > > 1,6 > 2,6 > 3,6 > 4,6 > 5,6 > 6,6 > 6,5 > 6,4 > 6,3 > 6,2 > 6,1 > > Is on pair of dice randomly chosen to have a 6 removed?: 1 out of 11 pairs. > > Or is one die out of ever die showing a 6 in the running to be chose?: 1 out of 12 dice. I don’t think this makes sense… those 22 dice, 11 pairs, represent a possibility space. They’re not all on the table at the same time, as it were, and you can pick any of the 12 6’s to remove. You have to remove a 6 from each pair in the possibility space–assuming you’re looking for a 6 to remove. From 10 of the pairs, you have no choice under that assumption From the last pair, it doesn’t really matter what choice you make. ---
**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 6, 2016, 4:16pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/23 "2016-10-06T16:16:16Z")
I see your point about them not actually being on the table at the same time. Then I don’t understand what you meant by this: _We have no idea whether there was intention or not in removing the 6, so we can’t tell if it changes the odds or how._ ---
**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, 4:17pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/24 "2016-10-06T16:17:29Z")
This problem is going to produce endless fruitless debate, because, like the boy-girl problem, the specified conditions are ambiguous. The details of the randomization procedure that produced the roll with at least one 6 have not been specified, so there are at least two possible answers, depending on how the roll was produced. I would plead with everyone who is about to reply here with the answer that seems “obvious” to them to read [Wikipedia’s article on the boy-girl problem](https://en.wikipedia.org/wiki/Boy_or_Girl_paradox) first, to save a lot of pointless argumentation. Here’s one interpretation of the given problem: “Two dice are rolled numerous times. In the cases where a six appears on at least one of the dice, the observer is told that one die is a six and asked the probability that the other is a six.” In this case, 25/36 of the rolls will not produce a six and are ignored. Of the remaining cases, 1 in 11 will have two sixes, so the answer is 1/11. Here’s another interpretation of the problem: “Two dice are rolled once. It just so happens that at least one die shows a six. The observer is asked for the probability that the other is a six.” In this case the probability that the second die is a six is the same as the probability that any arbitrary thrown die is a six, 1/6. Clearly different people in this thread are already interpreting the problem in different ways, and therefore producing different results. –Mark ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [October 6, 2016, 4:18pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/25 "2016-10-06T16:18:30Z")
Yes, you have 11 possible combinations - but the case presented to you is that “one of the sixes is removed”. As the posters talking about red and green dice point out, there are in fact 2 cases of 6,6 - one where we remove the red and the green die is 6, and one where we remove the green and the red die is 6. so the odds at 2 in 12, not 1 in 11. It’s not the odds that 6,6 is rolled, it’s the odds that the case presented to you matches the criteria - at least one 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 6, 2016, 4:18pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/26 "2016-10-06T16:18:32Z")
> [@x-ray\_vision](#): > > Right. But how the choice is made to remove a 6 affects the answer to the question being asked. Is it decided by taking one randomly from one of the 11 pairs, or by taking one randomly from the 12 dice showing a 6? At the point the 6 is removed, there are only two dice, and they’ve already been rolled. At that point, if there’s any choice left about which 6 to remove, it doesn’t matter which it is—either way, the remaining die will show a 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 6, 2016, 4:27pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/27 "2016-10-06T16:27:28Z")
I’m blaming my confusion on LSLGuy. 😃 > [@LSLGuy](#): > > IMO the above part is right. Then you go wrong here:How many different ways can we remove a six? Let’s assume we have a green and a red die so we can tell them apart. As you said, the possibilities expressed in green, red order are: > > 1,6 > 2,6 > 3,6 > 4,6 > 5,6 > 6,6 > 6,5 > 6,4 > 6,3 > 6,2 > 6,1 > > If the green die is a 6 we’re looking at one of the bottom 6 possibilities. 1/6th of which have another 6, the red die. > > If the red die is a 6 we’re looking at one of the top 6 possibilities. 1/6th of which have another 6, the green die. > > Result: it doesn’t matter which die is a 6. Red, green, or both. After you remove one of the sixes the chance of the other being a 6 is 1/6. He has every combination twice because order matters, but he doesn’t for the 6,6 combination. The order of 6s matters too. So it should be: 1,6 2,6 3,6 4,6 5,6 6,6 6,6 6,5 6,4 6,3 6,2 6,1 Answer: 1 out of 6 ---
**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 6, 2016, 4:32pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/28 "2016-10-06T16:32:16Z")
Nope. Anything involving two 6,6 pairs is wrong from the git-go. ---
**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, 4:39pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/29 "2016-10-06T16:39:49Z")
In the boy/girl paradox, if _at least_ one child is a boy, and we _remove_ that boy, what are the odds of having two boys left over … the answer is zero. Or are we asking, after we remove the boy, what are the odds of the one left being a boy … that’s 1/2. We don’t care about the odds of having at least one 6, that’s 11/36. A least one 6 is a certainty, we pick the die up and examine the remain die in isolation, the odds of a 6 is 1/6. ---
**Author:** ![Ignotus](https://avatars.discourse-cdn.com/v4/letter/i/ecc23a/32.png) [@Ignotus](https://boards.straightdope.com/u/Ignotus)\ **Post date:** [October 6, 2016, 4:41pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/30 "2016-10-06T16:41:19Z")
> [@markn\_1](#): > > This problem is going to produce endless fruitless debate, because, like the boy-girl problem, the specified conditions are ambiguous. The details of the randomization procedure that produced the roll with at least one 6 have not been specified, so there are at least two possible answers, depending on how the roll was produced. I would plead with everyone who is about to reply here with the answer that seems “obvious” to them to read [Wikipedia’s article on the boy-girl problem](https://en.wikipedia.org/wiki/Boy_or_Girl_paradox) first, to save a lot of pointless argumentation. > > Here’s one interpretation of the given problem: > > “Two dice are rolled numerous times. In the cases where a six appears on at least one of the dice, the observer is told that one die is a six and asked the probability that the other is a six.” > In this case, 25/36 of the rolls will not produce a six and are ignored. Of the remaining cases, 1 in 11 will have two sixes, so the answer is 1/11. > > Here’s another interpretation of the problem: > “Two dice are rolled once. It just so happens that at least one die shows a six. The observer is asked for the probability that the other is a six.” > > In this case the probability that the second die is a six is the same as the probability that any arbitrary thrown die is a six, 1/6. > > Clearly different people in this thread are already interpreting the problem in different ways, and therefore producing different results. > > –Mark Is the observer aware of _us_ observing _him_? If so, does that knowledge somewhat skew his judgment? Is my cat dead or alive? ---
**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, 5:07pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/31 "2016-10-06T17:07:13Z")
100 valid trials: 1 = 11 2 = 21 3 = 14 4 = 15 5 = 25 6 = 14 Fuzzy data … tends to the 1/6 prediction … need more trials I would say. ---
**Author:** ![OldGuy](https://avatars.discourse-cdn.com/v4/letter/o/3bc359/32.png) [@OldGuy](https://boards.straightdope.com/u/OldGuy)\ **Post date:** [October 6, 2016, 5:07pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/32 "2016-10-06T17:07:27Z")
> [@Thudlow\_Boink](#): > > As in the case of the infamous Monty Hall problem, one way of settling this would be experimentally. > … > Repeat this experiment many times, and then calculate the number of YESes you wrote down, dividing by the total number of trials. I tend to avoid these discussions now in the Dope, as some posters are never satisfied with probability proofs or the discussion gets sidetracked on the exact meaning of things like “at random”. But this is good advice. If you don’t like someone else’s calculation do a simple simulation. Doing this not only gives you your answer, it makes you think precisely what you mean by any of the instructions because you have to program it into the simulation in a precise way that the computer understands. ---
**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, 5:14pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/33 "2016-10-06T17:14:57Z")
I ran a simulation 100K times. I ignored any roll without at least one six. For the remaining rolls, here are the counts: ```auto **Die occurrences %** 1 5428 17.7% 2 5627 18.4% 3 5533 18.1% 4 5536 18.1% 5 5690 18.6% 6 2801 9.1% ``` So assuming the interpretation that the question is asking about the odds over multiple rolls, the answer is clearly 1/11. I understand **markn+'s** point about the answer depending on how you interpret it, but the second interpretation of “on this **one** roll where one die just happened to be 6, what are the odds the other one is a 6” doesn’t work for me. The whole point of probabilities and odds is assuming what happens over multiple events. Pretending that we’re talking only about this single roll isn’t a probability. It smacks too much of “either it’s a 6 or it’s not, so the odds are 50/50.” ---
**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 6, 2016, 5:15pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/34 "2016-10-06T17:15:42Z")
> [@LSLGuy](#): > > Nope. Anything involving two 6,6 pairs is wrong from the git-go. Then the answer is 1/11. And someone is willing to bet on this with actual dice rolls. > It seems like the OP’s problem is theoretically the same. ---
**Author:** ![CharmaChameleon](https://avatars.discourse-cdn.com/v4/letter/c/9d8465/32.png) [@CharmaChameleon](https://boards.straightdope.com/u/CharmaChameleon)\ **Post date:** [October 6, 2016, 5:20pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/35 "2016-10-06T17:20:24Z")
> [@Ignotus](#): > > Is the observer aware of _us_ observing _him_? If so, does that knowledge somewhat skew his judgment? > > Is my cat dead or alive? It’s almost impossible not to ask, “Is that Schrödinger’s cat?” (But, I refrain.) Just as the “observer” in the problem cannot change the outcome of the initial roll by any action on his/her part, we, as “meta-observers”, cannot “meta-change” any “meta-outcome” in the problem space, as all of that is inviolable from change caused by any event residing in internal or external (base or “meta”) space. **(The dice are rolled, and cannot be un-rolled.)** ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [October 6, 2016, 5:21pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/36 "2016-10-06T17:21:12Z")
> [@LSLGuy](#): > > Nope. Anything involving two 6,6 pairs is wrong from the git-go. Yes, except that: (To complicate things still further 😃 ) If you assume that either die is removed randomly, then each of the 11 cases is twinned, like so: ( **1** , 6) (1, **6** ) ( **2** , 6) (2, **6** ) ( **3** , 6) (3, **6** ) ( **4** , 6) (4, **6** ) ( **5** , 6) (5, **6** ) ( **6** , 1) (6, **1** ) ( **6** , 2) (6, **2** ) ( **6** , 3) (6, **3** ) ( **6** , 4) (6, **4** ) ( **6** , 5) (6, **5** ) ( **6** , 6) (6, **6** ) Where the bolded number is kept on the table. Out of those 22 cases, we can remove 10 on the grounds that a number other than 6 is removed, leaving: ( **1** , 6) ( **2** , 6) ( **3** , 6) ( **4** , 6) ( **5** , 6) (6, **1** ) (6, **2** ) (6, **3** ) (6, **4** ) (6, **5** ) ( **6** , 6) (6, **6** ) So, out of 12 remaining cases, in 2 there’s a 6 still on the table, so the odds are 2/12 or 1/6. ---
**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, 5:24pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/37 "2016-10-06T17:24:23Z")
> [@markn\_1](#): > > This problem is going to produce endless fruitless debate, because, like the boy-girl problem, the specified conditions are ambiguous. The details of the randomization procedure that produced the roll with at least one 6 have not been specified, so there are at least two possible answers, depending on how the roll was produced. I would plead with everyone who is about to reply here with the answer that seems “obvious” to them to read [Wikipedia’s article on the boy-girl problem](https://en.wikipedia.org/wiki/Boy_or_Girl_paradox) first, to save a lot of pointless argumentation. Yes and here is the relevant quote from Martin Gardner “\*Gardner argued that a “failure to specify the randomizing procedure” could lead readers to interpret the question in two distinct ways:…” \* (spoken in the voice of the Captain) _What we got here is failure to_ specify the randomizing procedure. ---
**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, 5:26pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/38 "2016-10-06T17:26:50Z")
> [@TroutMan](#): > > I ran a simulation 100K times. I ignored any roll without at least one six. For the remaining rolls, here are the counts: … [snip] I just realized I’m throwing the Black Death Dice that came with my +1 (to damage) two-handed sword … [smile] … `detect magic` doesn’t work on it. **Troutman** is throwing fair dice, the Law of Large Numbers can be applied here which gives 1/11 odds. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [October 6, 2016, 5:35pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/39 "2016-10-06T17:35:35Z")
> [@DrDeth](#): > > Yes and here is the relevant quote from Martin Gardner “\*Gardner argued that a “failure to specify the randomizing procedure” could lead readers to interpret the question in two distinct ways:…” > \* > > (spoken in the voice of the Captain) _What we got here is failure to_ specify the randomizing procedure. _Golf clap for combining Martin Gardner and Cool Hand Luke in one post._ 😃 ---
**Author:** ![Shagnasty](https://avatars.discourse-cdn.com/v4/letter/s/9dc877/32.png) [@Shagnasty](https://boards.straightdope.com/u/Shagnasty)\ **Post date:** [October 6, 2016, 5:53pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/40 "2016-10-06T17:53:59Z")
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. [Previous page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=1) [Next page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=3)