# 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:** 6
**Author:** ![SigMan](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@SigMan](https://boards.straightdope.com/u/SigMan)\ **Post date:** [October 6, 2016, 10:22pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/101 "2016-10-06T22:22:04Z")
If you roll two dice for one die to show a six won’t that be odds of 1 in 3? ---
**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, 10:27pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/102 "2016-10-06T22:27:05Z")
> [@octopus](#): > > Yeah but in a chart of events where a die is being a revealed and a die is being concealed there are 12 permutations which I listed above will do so again. > > Revealed Red result Green Result > > Red 6 1 > Red 6 2 > Red 6 3 > Red 6 4 > Red 6 5 > Red 6 6 > Green 1 6 > Green 2 6 > Green 3 6 > Green 4 6 > Green 5 6 > Green 6 6 > > Out of the 12 possibilities of a particular die being revealed and a particular die being concealed there are only 2 that have two 6s. 2/12 = 1/6. > > If you do this as a real life experiment with two people. You will see that 1/6 of the time when at least one six is rolled that the concealed die is a 6. Do it with real dice and two people. You’re counting 6, 6 twice, which I initially thought had to be done. See the following: [http://www.math.hawaii.edu/~ramsey/Probability/TwoDice.html](http://www.math.hawaii.edu/~ramsey/Probability/TwoDice.html) As borschevsky is alluding to in the question he asked you, you will get a 1 and a 6 twice as often as a 6 and a 6. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [October 6, 2016, 10:44pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/103 "2016-10-06T22:44:54Z")
> [@Chronos](#): > > I’ll join in with the “problem is ambiguous” crowd, here. There are multiple possible experiments that are all consistent with the problem statement, and which have different answers. Here are two examples: > > 1: I roll a pair of dice. I then choose one die at random, look at the number on the die, and announce “At least one die is a \_\_\_”, where the blank is the number I looked at. On this particular iteration of the experiment, the die I looked at was a 6, and I announce as such. What is the probability that the other die is a 6? > > Answer: The probability that the other die is a 6 is 1 in 6. > > 2: I roll a pair of dice. I then examine both dice, and if it is the case that at least one of them shows a 6, I announce “At least one die is a 6”, and otherwise I say nothing. On this particular iteration of the experiment, I saw at least one 6, and so I announce as such. What is the probability that both dice are 6? > > Answer: The probability that both dice are 6 is 1 in 11. I think you’re right that the answer you get depends very much on how the situation is framed. Lay out 36 pairs of dice in a grid showing all 36 possible results. Of the 6 pairs in which the die on the right side of the pair is six, the left die will also be 6 once (one chance in 6). Of the 6 pairs in which the die on the left side of the pair is six, the right die will also be 6 once (one chance in 6). Of the 11 pairs in which at least one of the dice in the pair is 6, the other die will be 6 in one case (1 in 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 6, 2016, 10:49pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/104 "2016-10-06T22:49:09Z")
> [@Thudlow\_Boink](#): > > When the dice are _thrown_, there are 11 possible outcomes, all equally likely. > > In the case of (6, 6), there are two possible ways that one of the dice could be revealed, but that doesn’t make that outcome more likely to have come up in the first place. Take two real dice and see if you come up with 1/11. Ok, I just rolled two dice for a good chunk of time. I discarded all rolls that did not include a 6 and I stopped after 20 double sixes. I had 132 rolls that included at least one 6. Here are the ratios for the rolls. 4 22 5.5 5 26 5.2 6 29 4.833333333 7 30 4.285714286 8 35 4.375 9 41 4.555555556 10 85 8.5 11 86 7.818181818 12 96 8 13 98 7.538461538 14 99 7.071428571 15 111 7.4 16 112 7 17 117 6.882352941 18 118 6.555555556 19 129 6.789473684 20 132 6.6 ---
**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, 10:55pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/105 "2016-10-06T22:55:45Z")
I’m afraid you’ll have to explain to me what those numbers represent. ---
**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, 10:56pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/106 "2016-10-06T22:56:06Z")
octopus, then you think with a roll of two dice, you have an equal chance of rolling two sixes as you do rolling a 1 and a 6? Do you agree that the probability of rolling a 1 and a 6 is 1/18? ---
**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, 11:15pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/107 "2016-10-06T23:15:38Z")
> [@x-ray\_vision](#): > > octopus, then you think with a roll of two dice, you have an equal chance of rolling two sixes as you do rolling a 1 and a 6? > > Do you agree that the probability of rolling a 1 and a 6 is 1/18? Of course it is. You have the chance of rolling the pair 1,6 or 6,1 assuming you can tell the rolls apart with some mechanism. Now the chance of the ordered pair 1,6 or 6,1 is actually 1/36. But notice that a 1,6 is the same chance as a 6,6 if you can discern the dice. The sum of 1 and 6 is 7 and that is a 1/6 chance of occurring. So what are we looking for sums or particular probabilities of particular events. In that example he is summing events. If 1,6 is considered exactly equivalent to 6,1 than yes those 2 have twice the probability of 6,6. Let’s make the math easier and use the coin example. Throw out all double tails. Use a penny and a nickel. What are the outcomes Penny Nickel H T T H H H How many possible ways can I show one head? I can’t show one head with double tails so that isn’t even in the denominator. It’s not counted. Correct? So how many ways can I show you a head? A) I can show you Penny’s head if Penny comes up head and Nickel comes up tails. Correct? That is one count. B) I can show you Penny’s head if Penny comes up head Nickel comes up head. That is two. C) I can show you Nickel’s head if Nickel comes up head and Penny comes up tails. That is three. D) I can show you Nickel’s head if Nickel comes up head and Penny comes up head. That is four. Ah - ha ha ha. Four ways to show you a head in which at least one head was flipped. These are four permutations of a state in which one coin is revealed and one is concealed with the constraint that there will only be a revelation if at least one is a head. Correct? Now how many of those permutations satisfy the second case where the concealed coin is also a head? Two events B and D. 2/4 is 1/2 which is what you’d expect with a fair coin. ---
**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, 11:17pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/108 "2016-10-06T23:17:13Z")
> [@octopus](#): > > Of course it is. Cool. Now please Google **probability of rolling snake eyes**. ---
**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, 11:20pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/109 "2016-10-06T23:20:01Z")
> [@Skammer](#): > > ETA: Thudlow Boink is also correct. If I said “the red die is a six. What are the odds that the green die is a six?” the answer is 1/6. But that’s not the question. Yes it is, it’s just more detailed. If the red die is a 6, the odds of a green 6 are 1/6. If the green die is a 6, the odds of a red 6 is 1/6. That covers all possibilities, and in every case, it is 1/6. It doesn’t mater if the dice are colored – it’s 1/6 in every possible case. ---
**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, 11:22pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/110 "2016-10-06T23:22:05Z")
> [@octopus](#): > > If 1,6 is considered exactly equivalent to 6,1 than yes those 2 have twice the probability of 6,6. Then why are you claiming this?: > [@](#): > > Red 6 1 > Red 6 2 > Red 6 3 > Red 6 4 > Red 6 5 > Red 6 6 > Green 1 6 > Green 2 6 > Green 3 6 > Green 4 6 > Green 5 6 > Green 6 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, 11:23pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/111 "2016-10-06T23:23:21Z")
What I’m enjoying now are the folks writing simulations and doing manual dice throwing trials. The ambiguity is in the problem statement. Once you decide on your preferred interpretation of the problem statement then your result (1/11 or 1/6) is foreordained. And can be proven with a few short lines of small words. So any effort spent to simulate is foolish. It’s merely proving the trivial conclusion you already (perhaps unwittingly) assumed. ---
**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, 11:25pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/112 "2016-10-06T23:25:08Z")
> [@x-ray\_vision](#): > > Cool. Now please Google **probability of rolling snake eyes**. Snake eyes is 1/36. 1,1 is one ordered pair out of 36. Simple. I’m not the one confusing what the actual constraints of the question are. Shoot let’s say we only count the probability of the die roll after we rolled a 6. Instead of revealing a 6 and asking we are only counting 6s after we rolled a 6. What’s the probability of that? Why would the probability of the die change? It only has 6 possible states regardless of what was rolled before. ---
**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, 11:30pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/113 "2016-10-06T23:30:01Z")
> [@octopus](#): > > Snake eyes is 1/36. I’m confused. You agree the probabilty of rolling a 1 and a 6 is 1/18, and that rolling a 6 and a 6 is 1/36, yet you count 6,6 twice here: > [@](#): > > Red 6 1 > Red 6 2 > Red 6 3 > Red 6 4 > Red 6 5 > Red 6 6 > Green 1 6 > Green 2 6 > Green 3 6 > Green 4 6 > Green 5 6 > Green 6 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, 11:32pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/114 "2016-10-06T23:32:33Z")
Do you agree with the following?: [http://www.math.hawaii.edu/~ramsey/Probability/TwoDice.html](http://www.math.hawaii.edu/~ramsey/Probability/TwoDice.html) There are 36 equal rolls of two dice. ---
**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, 11:33pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/115 "2016-10-06T23:33:54Z")
> [@Chronos](#): > > Answer: The probability that both dice are 6 is 1 in 11. 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. > [@Thudlow\_Boink](#): > > If nothing that happened before matters, then what determines which die is the “other die,” the one that you are shown? It doesn’t matter which is the “other die,” since _this is not a conditional situation._ All conditions (no 6’s found, etc.) are eliminated. All that matters is the odds of ONE toss of ONE die, with the possibilities of 1…6. And just in case, running my program for 10,000,000 iterations confirms a 1:6 ratio, within 3 decimal places. I doubt if running it longer will change much, so it’s the premise that counts here. I stand by my [post #89,](http://boards.straightdope.com/sdmb/showpost.php?p=19680760&postcount=89) more than ever. Anyone want 10 million data points to look at? ---
**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, 11:36pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/116 "2016-10-06T23:36:42Z")
> [@octopus](#): > > Shoot let’s say we only count the probability of the die roll after we rolled a 6. Instead of revealing a 6 and asking we are only counting 6s after we rolled a 6. What’s the probability of that? Why would the probability of the die change? It only has 6 possible states regardless of what was rolled before. Different problems. > [@](#): > > 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? > Gardner initially gave the answers 1/2 and 1/3, respectively, but later acknowledged that the second question was ambiguous.[3] Its answer could be 1/2, depending on how you found out that one child was a boy. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Bar-Hillel and Falk,[4] and Nickerson.[5] > **[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:** ![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, 11:38pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/117 "2016-10-06T23:38:55Z")
> [@LSLGuy](#): > > What I’m enjoying now are the folks writing simulations and doing manual dice throwing trials. > > The ambiguity is in the problem statement. Once you decide on your preferred interpretation of the problem statement then your result (1/11 or 1/6) is foreordained. And can be proven with a few short lines of small words. > > So any effort spent to simulate is foolish. It’s merely proving the trivial conclusion you already (perhaps unwittingly) assumed. You’d like to think so. But some are still arguing about questions unrelated to the ambiguity of the problem statement, like red vs green dice and ordered pairs. For those, a simulation can be handy to demonstrate why 1/11 is correct according to one interpretation. ---
**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, 11:41pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/118 "2016-10-06T23:41:42Z")
> [@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. > > It doesn’t matter which is the “other die,” since _this is not a conditional situation._ All conditions (no 6’s found, etc.) are eliminated. All that matters is the odds of ONE toss of ONE die, with the possibilities of 1..6. > > And just in case, running my program for 10,000,000 iterations confirms a 1:6 ratio, within 3 decimal places. I doubt if running it longer will change much, so it’s the premise that counts here. > > I stand by my [post #89,](http://boards.straightdope.com/sdmb/showpost.php?p=19680760&postcount=89) more than ever. Anyone want 10 million data points to look at? Once again you’ve assumed one interpretation of the problem. And then spent effort tautologically proving your own assumption. Yes, Virginia, the odds of a 1:6 outcome really are 1:6. FWIW, I prefer your interpretation of the problem too. But it’s just that: one interpretation of the subtly (and deliberately) ambiguous sentences in the original problem statement. The fact the multiple choice answers have choices which are correct for each interpretation and for the common confusion of [is R6-G6 different from G6-R6?] pretty well proves the problem was carefully deliberately \*designed \*to produce these incompatible lines of thinking. ---
**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, 11:45pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/119 "2016-10-06T23:45:52Z")
> [@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… 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. ---
**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, 11:46pm UTC](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763/120 "2016-10-06T23:46:00Z")
> [@x-ray\_vision](#): > > Then why are you claiming this?: Because what is being revealed and what is being concealed are different states. Look at it closely. There are 11 different pairs. 1,6 2,6 3,6 4,6 5,6 6,6 6,1 6,2 6,3 6,4 6,5 That’s 11 things right? 6,6 is 1/11 correct? That is out of this set of ordered pairs 6/6 occurs 1 time out of 11. Makes sense right? But we have 11 ordered pairs that could exist. The other 25 are discarded. Each of these 11 have at least one 6. If you reveal the 6 you see what the other number could be. But look! There are two different 6s in the 6,6 example that could be revealed. The left or the right, the first or second, the red or green, etc. So there are actually 12 different sixes that could be revealed. And out of that 12 only 2 of the revealed have another 6 concealed and that is reveal red 6 conceal green 6 and reveal green 6 conceal red 6. That’s why the denominator is not 11. Because we aren’t dealing with a set of 11 ordered pairs. We are dealing with revealing a \*particular \*6 from a set of 11 ordered pairs. But the question isn’t what is the probability of one of these to occur. The question is what is the probability if I am shown a 6 that the other number is a 6. Well with colored dice there are 12 combinations that can occur because what matters is what is the state of the dice. Are they revealed, concealed? What number is on them. So if I can reveal a green 6 with 6 equally likely permutations on a red die that’s 6 events. If I can reveal a red 6 with 6 equally likely permutations on a green die that’s 6 different events. Yes or no? Is a revealing a green 6 the same as revealing a red 6? No. isn’t. If I show you a green 6 what are states the red die can be in? 1,2,3,4,5,6. Correct? Why would one number be favored over another. If I show you a red 6 what are the states the green die can be in? 1,2,3,4,5,6. Correct? Why should 6 be a lessor probability than the 1,2,3,4 or 5? There’s no logical reason to treat 6 differently. [Previous page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=5) [Next page](https://boards.straightdope.com/t/odds-problem-roll-a-pair-of-dice/767763.md?page=7)