Counter-Intuitive Probability: The Snake Eyes Riddle



You close your eyes and roll a pair of dice. The casino dealer announces that at least one of the rolls is showing a 1.What is the probability that both dice are showing a 1? Watch the video for the solution. (Snake eyes is a term for rolling a pair of 1s. Note: if you do not roll a 1, the dealer asks you to roll again.)

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Fahad Hameed

Fahad Hashmi is one of the known Software Engineer and blogger likes to blog about design resources. He is passionate about collecting the awe-inspiring design tools, to help designers.He blogs only for Designers & Photographers.

41 thoughts on “Counter-Intuitive Probability: The Snake Eyes Riddle

  • September 25, 2017 at 4:23 pm
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    What is the difference between e.g. (1,4) roll and (4,1) ? One of the dice is showing 1 and one is showing 4, from my perspective it doesn't matter which is which! It doesn't have to be consecutive roll, 1 after something or vice versa…

    Reply
  • September 25, 2017 at 4:23 pm
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    This is BS. It's the same thing as putting the one dice with the number one facing up and then rolling the other. Having one of the values known, the only value that matters is the second one.

    If you still don't believe me, try it in real life. Have 1 dice to have the value 1, roll the other. Approximately 1/6 of the time it will show 2 ones.

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  • September 25, 2017 at 4:23 pm
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    Keeping the A and B labels. Interestingly enough, If we are only told that A is a 1, (if and when) then we can be assured that B has a 1/6 chance of also being a 1. Is that right? For whatever reason this is very hard to grasp. I understand the table you made. I also understand that your result is correct.
    But let's say if we only include the sample of when, "A" is a 1 ("one"), regardless of B, but also recording B. Then I'd expect that there is a 1/6 chance of "B" being 1 ("one") therefore a 1/6 for both being a 1. I know that these are inherently different questions, but as I said it's very abstract and it's very hard to see the difference.

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  • September 25, 2017 at 4:23 pm
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    There's a mistake in the dealers declaration- if he says 'if one of the dice is a one' then your calculation is correct.

    However if he were to say, after looking at the outcome, 'one of the dice is one' then the the probability of the other being a one is 1/6 (since all possibilities where neither die is a one is effectively removed upon the dealer's declaration)

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  • September 25, 2017 at 4:23 pm
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    So then if at the beginning the dealer says that at least one of the dices will be 1 which is the probability to have both dices showing 1? Looks like he is not very useful with sharing the dice information in the video scenario isn't he? Or maybe the solving method is wrong…

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  • September 25, 2017 at 4:23 pm
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    Can someone explain me why is the solution 1/11 and not 1/36? if there are 36 possible combinations, and only 1 out of them is (1,1)?

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  • September 25, 2017 at 4:23 pm
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    no there's a 1 in 6 for each dice the result of the first die does not influence the result of the 2nd die

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  • September 25, 2017 at 4:23 pm
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    The flaw here is it expects that the game is roll the dice, if a one appears, then you bet. If a one also appears, you lose. It's just a shitty game. No one would play that game outside a mathematical thought experiment. Therefore, we reject it in favor f the more simple "One's a one. What's the other?" which for each time it's posed is 1/6.

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  • September 25, 2017 at 4:23 pm
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    You're wrong, it's 1/6, he already said one dice was one so it's 1/6 that the other would be one taoo

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  • September 25, 2017 at 4:23 pm
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    Great riddle! It's like the opposite of the Monty Hall Problem. We know the probability of throwing snake eyes is 1/36, but the we are given the information that one die is a 1, so we disregard it and work out the probability of a single die being a 1, which is 1/6. And also wrong 🙂 I'm going to try The same thing with 10 coins.

    Reply
  • September 25, 2017 at 4:23 pm
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    The way he sets up this problem is untenable compared to how he solves it.To prove this consider rolling 1,000,000 dice at a casino and the dealer tells you at least 999,999 are 1s…. What is the probability that all 1,000,000 dice are 1s?Using his solution the chance is astronomical. But the true answer is still 1/6 because the dealer has gotten rid of all other possibilities.

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  • September 25, 2017 at 4:23 pm
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    ok-since you like casino duce games(craps here)

    what is the probability of ,say a 5 ,showing up on 3 standard fair dice.
    this game is tai sai

    Reply
  • September 25, 2017 at 4:23 pm
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    Hmmm, what he just said doesn't make sense "What is the probability that both rolls show a 1". At least the description says "both die show 1". As the dealer didn't announce which die showed the 1, then the odds are going to be lower that 1 in 6.

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  • September 25, 2017 at 4:23 pm
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    Flawed logic.

    Using the logic you explain here would mean there is a 1/11 chance of any of each of the 6 numbers showing up on the second unkown die. Since there are 6 possibilities, this would give a 6/11 chance of one of those 6 numbers showing up leaving a 5/11 chance of none of those numbers showing up. Which is impossible.

    The flaw is that we want to know the odds of it being a 1 AFTER already finding out that one of them is a 1. This means we are now only calculating the odds for one die landing on one particular number from six equal possibilities i.e. a 1/6 probability. Another way to think of it is that there are 36 possible combinations on the two dice. Six of them are doubles. That's a 1/6 chance of getting a double. So in all cases, regardless of which number the dealer announces, the probability of the other displaying the same number is 1/6.

    If we were working out the odds beforehand of the second die being a 1 in the case that the dealer announces that one of them is a 1 and after having already subtracted the odds of neither of the dice being a 1 then you would come out with a 1/11 chance. Or if we asked "What are the odds of the second die also being a 1?" before the dealer announces that one of them is a 1 then we obviously have a 1/36 chance.

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  • September 25, 2017 at 4:23 pm
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    The problem with this answer is that you treat both dice as seperate entities. Or in probability with grabbing and laying back.cbut the order in which the dice are is still not important. You will always in this have 1 dice showing a one and it does not matter if it is dice one or two. So you always 1/1 have 1 dice showing that 1. Then there are 6 possibilities for the other dice. Without the order of the dice being important and counting them as seperate gives you 1/6. When you do use the order as important you get to 1/11

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  • September 25, 2017 at 4:23 pm
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    I was way off. I had the idea of the equation being 1/x*y with X being the number of sides of the die, and Y being the number of die rolled.

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  • September 25, 2017 at 4:23 pm
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    The answer is 1/6. If the dealer announces that one of them is a 1, then they have become independent outcomes. The outcome of 1 does not affect the outcome of the other.

    The graph showing the outcomes is analyzed wrong, if one of the dice is a 1, then you would eliminate either the column or the row, as each represents a dice roll. But we already have 1 dice roll, so the answer is 1/6.

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  • September 25, 2017 at 4:23 pm
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    It's called conditional probability. As both dice are independent, we do not count one die being 1 and the other being 6 as any different than one being 6 and the other being 1. When we don't know either dice, there is a 1/36 possibility of snake eyes, but as soon as we learn the outcome of one dice, there are only 6 states the second one could assume. The right answer should be 1/6.

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  • September 25, 2017 at 4:23 pm
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    Flawed logic. There are 11 possibilities if 1,2is not equal to 2,1. However in a casino the dice are not numbered or alphabetised. 1,2 = 2,1 We know the outcome of one roll (doesn't matter which) the only variable is the roll of one D6.

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  • September 25, 2017 at 4:23 pm
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    Wait… since this is all a probability, would the chance of getting snake eyes 1/36 since theres 36 different ways the dice could end up?
    (1/6 * 1/6)

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  • September 25, 2017 at 4:23 pm
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    You still haven't fixed this video? It's completely wrong but it's still here.

    Reply
  • September 25, 2017 at 4:23 pm
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    Now, I don't have a very great track record in math, and I tend to study paradoxes and philosophy more than mathematical puzzles, but something about this puzzle seems off in my mind. I am not saying that the narrator is wrong and that I am right, so correct me in my comparisons and logical steps if I am mistaken.

    So, when the two dice are rolled with my eyes closed, I am assured that one die of the pair is a one. Now, I need to calculate the odds that both dice are ones. Rather than calculate the odds of both as seen in the video, couldn't I just logically say "because I know that at least one of these two dice are displaying a one, I should just calculate the odds of rolling a one on a standard six sided die?

    To give an example of what I am thinking of, imagine that you are rolling a single die. You roll a 3. Then, on your second roll, you roll another 3. Then, on your third roll, you roll another 3. If you calculate your odds of rolling that die and getting a 3, it is still one in 6. There was a 1/216 chance of rolling 3 3's in a row, but rolling a 3 on a six sided die is still 1 in 6.

    Again, I am no mathematician. This isn't some grand way of saying that I am right or the video is wrong. I am simply looking to confirm or deny this line of thought with a bit of help.

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  • September 25, 2017 at 4:23 pm
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    No, the dice are independent of each other. It's 1 in 6 for either independently and since we already know one of them is a 1, this has absolutely no effect on the other die and it has a 1 in 6 probability.

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  • September 25, 2017 at 4:23 pm
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    Hey… Wait a minute there… The guy told you AT LEAST ONE DICE IT'S 1, he didn't tell you which one. For exemple, the FIRST dice could 1 one and the SECOND dice could be any other number, including 1… As well as the SECOND dice could be 1 and the FIRST dice could any any other number, including 1…

    This means it could be 2 of 12 possibilities.
    1-1, 1-2, 1-3, 1-4, 1-5, 1-6
    1-1, 2,1, 3-1, 4-1, 5,1 6-1

    This means 1-1 could appers 2 times.

    Reply
  • September 25, 2017 at 4:23 pm
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    The real problem here is the way the pairs of dice are being generated is ambiguous. You can't say that "you know one of the dice is a 1" (the dealer calls it out) unless you start the game with a die that has all 1s, or reroll the dice until a 1 is rolled. Your simulation is doing the latter, which is what messes with the probabilistic outcome and leads to all the confusion. It's a bad problem, and the assumptions we make to begin with affect the outcome. It's the same with the frog riddle – when you say "you know one of the frogs is male", does that mean that we generate a bunch of pairs of frogs and throw out the roughly 25% that don't have a male? Or does that mean that all pairs we generate have a guaranteed male, and none are thrown out? Again, the probabilities are skewed depending on how the pairs are generated – I wrote a simple program that simulates this both ways and with the first method roughly 66% of the remaining pairs had a female while with the second method roughly 50% had a female. Now modifying it to test this dice problem I get the same results – if I do what you did I get roughly 9% but if I guarantee that one of the dice is a 1, I get roughly 16.66%. There's no way to guarantee that one of the dice at random will be a 1, unless we keep rerolling until we get a 1. In the frog riddle, we can't guarantee that one of the frogs at random will be male unless we keep generating frog pairs until we get one with a male. That's why this is a bad problem – it presents itself as a real-life possible situation and glosses over the anomaly in the description; solvers are left to make assumptions and argue about what the problem was asking. Ironically, the argument that is generated is a probabilistic certainty given this problem, if we assume that roughly half will assume each interpretation of what is being asked.

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  • September 25, 2017 at 4:23 pm
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    I could be wrong, but if you use identical dices and roll them at the same time, wouldn't it change the outcome of your experiment?
    It has been a while since I had stochastics class, but I think I remember something like this.

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  • September 25, 2017 at 4:23 pm
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    1/6

    Вероятность выпадения 1 на одном кубике = 16
    Вероятность выпадения 1 на одном из двух кубиков = 1136( что близко к 13, а 13 является суммой двух 16), что больше 16, а на одном из 3-х ещё больше, на одном из 6 вероятность будет 100%
    Вероятность выпадения 1 на двух кубиках = 136

    Вероятность выпадения 1 на двух из трёх кубиках =
    на трёх из трёх = 216

    Вероятность выпадения 1 на 5 из шести кубиках =

    Reply
  • September 25, 2017 at 4:23 pm
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    If we consider that the opposite side where "1" is located, is always "6", then if the top face of the die is not 6, we always have a "1" in other showing faces, that is if we calculate the probability of the top face not having "6", we are calculating the other showing faces with "1" on them, so Pr(6 not on top) = 1 – 1/6 = 5/6, then we have 2 dice and because their outcomes are independent to each other, so the probability of the 2 die showing "1" is (5/6)*(5/6).

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  • September 25, 2017 at 4:23 pm
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    But we know at least one is a one. So in the excel evidence, every one of them had to have 1. I didn't make it but it would be probably different than we got.

    Reply

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