Probability of rolling a consecutive $1, 2, 3, 4, 5, 6$ on a die

Question

Questions

I was wondering what the probability of rolling a consecutive $1 ,2, 3, 4, 5, 6$ on a dice is?

For realism, is there any way to calculate an 'extra' factor, such as someone kicking the table (as in reality this does happen and does effect where the die will land)? Can this 'extra' probability also be calculated using math?

Answer 1

I'm assuming you mean a fair, six-sided die, although this can be generalized.

In general, the probability of multiple independent (read: unrelated) events happening is the product of the probabilities of each individual event. For example, the question of "what is the probability of rolling a consecutive 1 ,2, 3, 4, 5, and 6" boils down to "what is the probability we roll a 1 and then what is the probability we roll a 2 and then what is the probability we roll a 3... and so on.

Here, this means that this boils down to $P(1) \cdot P(2) \cdot P(3) \cdot P(4) \cdot P(5) \cdot P(6)$

On a six-sided die, the probability of rolling any individual number is 1/6. Thus, this is

$\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}$

If your extra event is kicking the table, however, there are some complications to this. If you know this kicking the table is designed to skew the probabilities of each die roll (say, someone "accidentally" bumps the table sometimes when a certain number is rolled), you have to deal with conditional probabilities (the probability of an outcome occurring given that some other outcome has happened)

If you have a large number of measurements, you could (in most cases) use the Law of Large Numbers, which says that as you increase your number of measurements you get close to the expected value for that measurement (i.e., if you repeat this experiment tons of times, the proportion of time the outcome occurs is the "true" probability of the event). This is known as a frequentist approach, which assumes you know nothing about things ahead of time. In practice, you would probably use what's known as a Bayesian approach, which assumes you have some expectation on the outcome already (possibly $\frac{1}{6}$ chance of a given event here) and incorporates that information along with the results you actually observe to come up with an estimate on the probability. Both of these approaches converge the more measurements you take, though.

For a more comprehensive explanation of using the Bayesian approach (but also with some sections targeted at an audience with a knowledge of discrete math), I go a bit more in depth at https://dem1995.github.io/machine-learning/curriculum/probability/probability.html#independence-and-the-product-rule

Answer 2

There are two questions you've asked and so I'll split my answer into two to address them both separately.

Question 1: Can we determine the probability of rolling a $1,2,3,4,5,6$ consecutively?

Yes, we can. Assuming the dice is fair then we have a $\frac{1}6$ chance of landing on each of the 6 numbers. In the situation where we roll the dice exactly 6 times then since each of the 6 rolls are independent of one another, we have the probability of this occurring as being $\frac{1}{6^6}=\frac{1}{46656}$. This can be adapted without too much added complexity in the case where we have more than 6 rolls.

Question 2: Can we determine the effect of some extra factor on the distribution of probabilities?

Yes we can. I will, however, point out the fact that if we rolled a dice a large number of times and someone kicked the table every time, then we would still expect there to be about $\frac{1}6$ of each number, as the way the dice is rolled will be different every time and so will the force that the the table is kicked with.

However, let's assume, for the sake of argument, that kicking the table changes the probabilities of landing on a $1$ to $0.3$, landing on a $2$ to $0.1$, landing on a $3$ to $0.2$, landing on a $4$ to $0.1$, landing on a $5$ to $0.2$, and landing on a $6$ to $0.1$. Now the new probability of rolling those 6 numbers consecutively will be $(0.3)(0.1)(0.2)(0.1)(0.2)(0.1)=0.000012$.

The problem is that you might be interested in knowing how we found out these probabilities for each number. There are two ways we can look at this issue - either through a Bayesian or Frequentist approach.

The Bayesian approach would be to keep rolling the dice a very large number of times (possibly millions) and then record what percentage landed on each number and assign these as the probabilities for each number.

The frequentist approach may involve studying the amount of force applied to the table in order to work out the objective probabilities.

Sometimes, it is easy to work out the exact probability, but in the real world, we often need an approximate approach - which is what the Bayesian school of thought offers.

Answer 3

In my view, the aforementioned answer (which is the typical answer to this question) is not actually answering the question “What is the probability of getting one of each number in a roll of six dice?”.

Rather, it is answering the question “What is the probability of die A being a 1, die B being a 2, die C being a 3…” and so on. In other words, each die has a specific, individual target value defined ex ante.

The answer to the question “What is the probability of getting one of each number in a roll of six dice?” must instead be

6/6 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6

The reason being that the first die that stops (die A) can be any number; the next die to stop (B) can be any number except the same as A; the third die to stop (C) can be any number except the same as A or B; etc.

The probability of this happening is considerably higher than 1/46656, which is why we do experience it occasionally.