So probability of getting $1$ ace (the one dot in dice) = $1/6$ i.e. in one out of six times we will get an ace.

But when we calculate the probability of getting at least one ace in six rolls, we get

$$= 1-\left(\frac{5}{6}\right)^6$$ $$= 0.665$$

I understand how the value is derived.

  1. But what is intuitive explanation for the same?

  2. Since it is so close to $68\%$, the percentage of population within $1$ standard deviation of normal distribution, does it have any relationship with normal distribution?

  • 4
    If someone said "If I flip this coin twice, I should get one heads and one tails", how would you respond? – Sam Sep 21 at 16:40
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    If you have a $1000$-sided die, the probability of getting no aces when you roll it $1000$ times is $$ \left( \frac{999}{1000} \right)^{1000} \approx 0.3676954 $$ In fact $$ \left( \frac{n-1} n \right)^n \to \frac 1 e \approx 0.367787944 \text{ as } n\to\infty. $$ – Michael Hardy Sep 21 at 18:01
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    @MichaelHardy Thanks Mr Hardy. The way things are interconnected in maths is very sacred experience. – q126y Sep 22 at 5:11

13 Answers 13

up vote 3 down vote accepted

The expected number of rolls until the first ace is $6$. Do you find this intuitive? If so, consider that there must then be lots of cases where you don't see an ace in the first $6$ rolls, or the expected waiting time would be less than $6.$

Think of it this way: roll the die a large number of times, and break the rolls up into groups of $6$. We expect that roughly $1/6$ of the die rolls will be aces, so on average the groups will contain a single ace. But some groups will have two aces. In order to have an average of 1 ace, for every group with two aces, there has to be another group with no aces. And of course for every group with three aces, we need two more groups with no ace. Etc.

When you realize how common multi-ace groups will be, it becomes obvious that there also has to be a lot of groups with no aces to balance them.

When you roll a die six times you shall expect to obtain one ace.

This is an average result, so the probability for obtaining zero aces should counterballance the probability for obtaining more than one aces.

Thus your intuition should anticipate that the probability for obtaining at least one ace is not too much more than one half.

  • 3
    "you shall expect to obtain one ace" -- this phrasing tends to mislead students. I don't expect this at all, seeing exactly one ace would be suprising! Saying "the average number of aces is one" is much clearer. – usul Sep 21 at 17:32
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    @usul: Well, "the average number of aces is one" is not quite the usual terminology, and usually, "average" refers to some sample population. So if you say "the average number of aces is one" and I try the experiment six times and get four aces, isn't the average number of aces 2/3, in my experiment? The term "expect", however, is a technical term which happens to have the correct meaning here. It's worth explaining what the word "expect" means, but I would be careful about replacing the technically correct word with a non-standard word or a term with unusual usage like "average". – Dietrich Epp Sep 21 at 18:01
  • @DietrichEpp In this context I would interpret the "average" to refer to the experiment itself, as in "in an average experiment the number of aces is one" – SamYonnou Sep 21 at 19:48
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    @DietrichEpp Yes, but I think usul's point was that the term "expected value" tends to mislead students, hence the different wording (meant to be used to explain the concept of "expected value" to people new to it, which the OP clearly is). – SamYonnou Sep 21 at 20:00
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    Part of what I'm saying is that it's easy for an expert to forget the difference a student might see between statements like "The expected value is 1 ace" and "you expect to see 1 ace". i.e. replacing the word average with 'mean' or 'expected' doesn't change my point much. – usul Sep 21 at 20:54

The intuitive explanation is that there are $6^6=46656$ different outcomes from rolling a dice 6 times, and $5^6=15625$ of these have no aces. $15625/46656 \approx 33.5\%$, so $33.5\%$ of outcomes have no aces and $66.5\%$ of outcomes have at least one ace. If the outcomes are equally likely then the probability of at least one aces is the same as the ratio of the number of favourable outcomes to the total number of outcomes i.e. $66.5\%$.

  • Thanks for the answer. It may be intuitive for you, but I am not able to get it. The answer is analytical to me. I mean if we have 1 in 6 chance of getting an ace, if we roll a dice 6 times, chance of getting an ace should be close to 1. – q126y Sep 21 at 14:10
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    @q126y: Perhaps you could tell us more about this intuition? Do you agree that it doesn't work for a coin? (The chance of getting at least one heads in two tosses is $\frac34$.) Do you find that counterintuitive, too? If not, why should it work better for a die? – joriki Sep 21 at 14:14
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    Perhaps your intuition is really related to the expected number of Aces which we will roll in 6 attempts -- that is 1?. And/or ... perhaps to challenge your own intuition, consider what will happen if we roll 12 times ... should the probability of rolling an Ace now be $\frac{12} 6$ i.e. 2? Obviously not! – BBO555 Sep 21 at 14:42
  • @q126y: There are 15625 cases to have no aces, that's no a small number, at least for me. My answer very similar to Mr. gandalf61 so I deleted mine. – Isana Yashiro Sep 21 at 15:16
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    67.5 should be 66.5 -- your probability knowledge is better than your arithmetic :) – Barmar Sep 21 at 17:56

If you write out all the combinations of rolling one die, you get a row of six numbers. The "border" of this list are the numbers $1$ and $6$. The number of aces is $1$, which is half the number of elements in the border. The length of the list is $s=6$ and the derivative of $s$ is $1$.

If you write out all the combinations of rolling two dice, you get a $6\times 6$ table. The rolls with aces are the top row and the left column, which is half the border of the table. In an $s\times s$ square, half the border is $2s$ which is the derivative of $s^2.$ But this counts one roll twice, so we have to subtract $1$ for each intersection of border edges. There's only one $={2\choose2}$ intersection , so we have $2s-1 = 11$ rolls with aces.

If you make a 3-D array of the all the combinations of rolling 3 dice, you get a $6\times 6\times 6$ cube of ordered triples. The rolls with aces are the top face, the left face and the front face, so again, half the border, which turns out to be $3s^2$. But again we have to subtract the intersections. This time there are three $={3\choose 2}$ intersections so we subtract three $s\times 1$ edges. Oh, but we need to add back in all triple intersections. There is one $={3\choose 3}$, so I end up with $3s^2 - 3s+3$ rolls with aces.

Continue in this fashion till you get to $6$ dimensions. Half the border of your cube will be the derivative of $s^6$, that is $6s^5$. The ratio $\frac{6s^5}{s^6} = 1$ for $s=6$, which is what your intuition wanted. It's the subtraction of the six 5-dimensional faces (and the addition of the sixty 4-dimensional faces times ${6\choose 2}$ and the subtraction of ${6\choose 3}$ times 160 cubic faces etc.) that makes the numerator much smaller.

  • "The ratio $\frac{6s^5}{s^6}=1$ for s=6, which is what your intuition wanted." As tempting as it sounds to have such a mathematical basis for my intuition, but it was more like okay we have the chance of 1 in 6, therefore, we should get it once in six rolls. In you working, 1 equals due to double counting. Though for both method, the wrong answer that one concludes is 1. But they are unrelated, according to my understanding. Or are they related somehow? – q126y Sep 22 at 10:05

Consider the following thought experiment:

Take $N$ balls in $N$ containers; pull them out, then assign each ball a new container randomly, without regard to where the previous balls went; pick a container at random. What is the probability that this container is filled? Close to one? That would mean that all of the containers hold a ball, and none contain two.

But there are $n^n$ ways to shuffle the balls, but only $n!$ ways to shuffle them in a way that leaves all containers filled. In fact, the probability that a random shuffle leaves all containers filled is $\prod_{x=0}^n \frac x n = \frac 1 n \cdot \frac 2 n \cdot \frac 3 n \cdot \dots$. As $n$ grows, this probability quickly drops to zero. In all other cases, there's at least one container empty.

This already tells us that the chance of getting an ace in six rolls should be less than $5/6.$ The actual value is close to $4/6.$ Not bad for such a weak argument.

In fact, the probability of getting a probability $\frac 1 n$ event in $n$ tries converges to $1-\frac 1 e = 0.6321$ (ref).

While the answers so far give a good and correct answers, I'll offer my thoughts on the thinking process you may go through to gain more intuition. I would say the previous answers are obvious to most with a probability background, but I remember it took me several math courses before I could think of problems intuitively right away. Everyone starts on different levels and takes time to learn, but that doesn't mean you can't reach the intuition and appreciation for probability.

To gain intuition about why it wouldn't be close to 1, think about what happens when you roll a die many times. Your certainty that an ace shows up is never 1. Even if you roll the dice a trillion times it would still be 0.999999999... likely. While this may seem obvious, it's always important to remember that a finite number of rolls never hits 1 so you shouldn't just expect it to be close to 1 after only 6 rolls.

Another thing is the high chance that some faces will repeat themselves in the 6 rolls you do. If the faces didn't repeat themselves, only then would you have 100% certainty. Think about what happens when you take 6 balls out of a bag. If you put the ball back in the bag each time, there is a good chance you take out the same ball twice, and each time you do this, it limits your chance of getting the ball you are looking for. It just so happens that accounting for all of these replacement possibilities leads to the numbers you derived.

Generally when thinking about probability intuitively, you have to think about what are all the possible combinations. You shouldn't let large numbers intimidate your intuition because taking 6^6 is a big number. Factorials and exponential numbers are fundamental to probability, but we can use ratios to simplify the results instead of counting all possibilities in our heads.

A good approach is to think, what are all the ways in which I don't get what I am lookin for (ie, 1 ace). Think about the chance of getting multiple aces, multiple 1's and no ace, multiple 2's and no ace. When you add all of these up you get a much larger value than 0.01 (if you expected 99% probably). In fact, you get thousands, as mentioned in the other comments.

Let's look at the case of three rolls. Suppose you have 216 people who are each rolling dice three times. After one roll, one sixth of them, or 36, will get an ace, leave 180 people with no aces. The next roll, 36 out of all the people will get an ace, but only 30 of them will be among the 180 people who didn't get an ace the first time. On the third roll, there will be only 25 people who are getting an ace for the first time. The more times you roll the die, the fewer people will be getting an ace for the first time, so rolling the die has diminishing returns as far as how many people get at least one ace.

If you look at the probability of getting no aces (.335), it's close to $e^{-1}$ = .36787. The reason for this becomes more clear if you look at the log of the probabilities. The first time you roll the die, the probability of getting no aces is just $\frac56$, and the log of that is $\ln(\frac56)$. To get no aces after two rolls, you have to get no aces on either roll, so the probabilities get multiplied: ${\frac56}^2$, and the logs get added: $2\ln(\frac56)$. After $n$ rolls, you have $n\ln(\frac56)$. If the number of faces on the die is $f$, then we have $n\ln(\frac{f-1}f)=n\ln(1-\frac{1}f)$. Going further requires some calculus: $\ln(1-\frac{1}f)$ is approximately $\ln(1)$ minus the derivative of $\ln$ times $\frac1f$. The derivative of $\ln(x)$ is $\frac1x$, and since $x$ is 1, the derivative is 1. And $\ln(1)$ is zero. Thus, $\ln(1-\frac{1}f)$ is approximately $-\frac{1}f$, and $n\ln(1-\frac{1}f)$ is approximately $\frac{-n}f$. Since this is the log of the probability, the probability is $e^{\frac{-n}f}$. If $n=f$, then we have $e^{-1}$. As f gets larger, this approximation gets closer. For instance, the probability of getting no aces after rolling a 100-sided die 100 times is .36603.

What's the chance that at least one of the first $3$ rolls will be an ace? Well, each roll individually has a $1/6$ chance of being an ace, so the chance is no more than $1/6 + 1/6 + 1/6 = 3/6$. But, in fact, the chances don't quite "add up" like that, so the chance of at least one of the first $3$ rolls being an ace is actually less than $3/6$.

What this means is that the first $3$ rolls have at least a $50\%$ chance of failing to contain an ace at all. Likewise, the last $3$ rolls also have at least a $50\%$ chance of failing to contain an ace. Take these two facts together, and we can conclude that the entire set of $6$ rolls has at least a $25\%$ chance of failing to contain an ace.

As a matter of fact, the set of $6$ rolls has about a $37\%$ chance of failing to contain an ace, so this estimate of "at least $25\%$" is really pretty good.

Because 5/6(83.3333%) are good chances of survival, and even if you have to survive 6 times in a row they are not terrible.

If you were throwing a coin (and had 50% chance of survival) then it would be highly unlikely you could survive 6 coin tosses.

Your intuition seems to be that since there are 6 faces, you expect to see about one of each face in 6 rolls (you only asked about the ace in the question, but you could just have easily asked about one of the other faces and the result would be the same). But this is actually a very unlikely occurrence, because each roll is independent of previous rolls. There are 66 = 46656 different sequences of 6 rolls, but only 6! = 720 of them have one of each face -- that's only about 1.5% of the time.

If you roll the die a thousands of times, you do expect the number of occurrences of each face to be about equal, each of them about 1/6 of the total. And the more times you roll, the more likely it is that there will be an ace somewhere in there; if you rolled 100 times and never got an ace, it would be reasonable to be suspicious about it being a fair die. But when you have a small number of rolls, like only 6, significant variation from the expected distribution is likely.

  • The approximate relationship you spotted is not particularly there: it would have been even closer using $1-\left(\frac34\right)^4 \approx 0.6836$ but that too is a coincidence

  • Your normal probability of being within one standard deviation is $\Phi(1)-\Phi(-1) \approx 0.6827$ using the standard normal cumulative distribution function does not have a nice closed form

  • Your $1-\left(\frac56\right)^6 \approx 0.6651$ might be seen as an example of $1-\left(\frac{n-1}{n}\right)^n$, which for increasing $n$ would tends towards $1-\frac1e \approx 0.6321$, getting further away from your normal distribution calculation

  • It is possible to find an artificial relationship using the standard normal density function at one standard deviation: $1 - 2\pi \left(\phi(1)\right)^2 = 1-\frac1e$ too, but this expression has no natural interpretation

The " philosophical" point is that the die does not have memory : when you throw it for the sixth time, it doesn't know what happened before, and does not care whether the ace already has appeared or not. Sixth draw is same as first. Yet the die is also "fair": it does not have preferences. And it is the combination of memoriless and fairness that produce "randomness".
And randomness has only a "regularity" within statistics, as explained in the other answers.

However, that the concept is difficult to grasp intuitively is demonstrated by all the people following the lotto numbers .. in delay.

In this respect, consider the following situation. You launched the die four times without aces when a naughty boy steals your die and make a couple of draws getting two aces and returns the die to you: what are the chances you are left now ?

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