Probability and the "out of" thing"

Question

I have quite an odd question:

I am not able to fully understand the concept of "out of". If I roll a dice once, from a total of $6$ possible outcomes, I'll get 1. Why does that mean a fraction $1\over 6$ = approx $16.67 \%$ and why does that mean that on average one out of $6$ rolls, I will get for example "$1$" on dice.

Where does the fraction say that for every $6$ rolls, I'll get on average one roll I wanted to get.

Why does $5$ out of $7$ mean $5\over 7$, why does that mean that it's on average $5$ per every $7$ people? Because when I want to get $5\over 7$ of something, I divide something into $7$ parts and get $5$. Is that the second look at this matter, that it can be seen like, for example: for every $7$ (divide some number by $7$ to find out how many $7$s are there) and then multiply by $5$, because for every seven that is included in the number it will be $5$.

Are my thoughts correct? How is the correct way of seeing these things?

Thanks for help in advance.

dont answer like Maths 90-page long thesis, I just want an answer that is en explanation in your own words. What I struggle is probably the fractions, what does out of mean and why... and you explain everything but this.

Answer 1

Here I define some of the basic terms used in introductory probability.

An experiment is a task/action with measurable/perceivable distinct outcomes. The set of all outcomes is referred to as the sample space. An event is a subset of the sample space.

For example, the experiment might be "roll two dice and compute their sum." The sample space will be $\{2,3,4,5,6,7,8,9,10,11,12\}$. An event such as "the sum is greater than or equal to 10" would refer to the subset of the sample space $\{10,11,12\}$.

The probability of an event occurring is the ratio of times that if we repeat the experiment (independently), we expect the outcome to be one of the outcomes in the event in question.

Note, we require a few properties of a probability distribution as a result of this:

• $Pr(\emptyset)=0$
• $Pr(S)=1$
• If $E\cap F=\emptyset$ then $Pr(E\cup F) = Pr(E)+Pr(F)$
• $0\leq Pr(E)\leq 1$ for all $E$

(in words of measure theory, probability acts as a measure such that $Pr(\Omega)=1$)

It is worth noting that several set theory and counting tools are directly applicable to probability as well such as the principle of inclusion exclusion.

In the special case that all outcomes in the sample space are equiprobable, in other words equally likely to occur, we say that the sample space is unbiased. In this special case, for sample space $S$ and event $E$, we have the following:

$$Pr(E)=\frac{|E|}{|S|}$$

where $|E|$ denotes the number of elements in the set $E$.

In the question of dice rolling, the sample space $\{2,3,4,5,6,7,8,9,10,11,12\}$ is not equiprobable, so we may not use the formula above. We may however use instead the sample space being all ways to roll two differently colored dice and describe our event as a subset of that instead. In this case, we have the sample space $\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),\dots,(6,4),(6,5),(6,6)\}$ with sample space size $|S|=36$. Our event corresponds to the subset $\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\}$, so our probability $Pr(\text{two dice rolled sum to at least 10}) = Pr(E) = \frac{|E|}{|S|}=\frac{6}{36}=\frac{1}{6}$.

Note: If we had incorrectly used the sample space $\{2,3,\dots,12\}$, and used the formula, it would have given us an answer of $\frac{3}{11}$ which is incorrect! Event size divided by sample space size is only usable if the sample space is unbiased!

For a smaller example, let our experiment be rolling a single die, and our outcome we are curious to find the probability of be "is even." Our sample space is $\{1,2,3,4,5,6\}$ and our event is $\{2,4,6\}$. The probability is then $Pr(\text{is even}) = \frac{|E|}{|S|}=\frac{3}{6}=\frac{1}{2}$.

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Without going into too much detail about expected value, we define the expected value of a discrete random variable $X$ as

$$\mathbb{E}[X]=\sum_{x\in S} x Pr(X=x)$$

This gives us a way of talking about the "average" outcome. For example, in rolling one fair six-sided die, the expected value of the outcome will be $3.5$. Of course, that doesn't mean that we will ever roll a $3.5$ exactly (it isn't even a side on the die), but it means that averaging out the results over "several" attempts, the average will approach $3.5$. In the case of either "success" or "failure" it is easier as we can represent a success as a $1$ and a failure as a $0$ in terms of value.

If we were to run an experiment with probability of success $p$ a total of $n$ times, you will find that the expected number of successes is $np$ (provable from definitions).

In this sense, yes indeed, a probability of $Pr(\text{rolling a six})=\frac{1}{6}$ implies that we expect that in six rolls, one $6$ will occur on average. It also implies that in one hundred rolls, $16.\overline{6}$ sixes will occur on average. Herein lies the usefulness of referring to things as a "percentage" (per: for each, cent: hundred). An event having probability $Pr(E)=72\%$ implies that if we were to run the experiment $100$ times, we expect a success $72$ of those times. It is not so much encoded in the number itself, as it is within the theory that these numbers can be interpreted in this way.

Answer 2

The classical definition of probability is

"The probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes".

If the number of trials (n) goes to infinity the probability is of event A is

$$P(A)=\lim_{n \to \infty}\frac{N(A)}{n}$$

$N(A)$ is the number of favorable outcomes. In your case, for instance, 5 of 7 peoples are females. Therefore $\frac{5}{7}$ it is just the ratio of number of females (F) and number of females and men (M):

$P(A)=\frac{F}{F+M}$.

And for a sample size of n the expected value of females is $$E(x)=P(A)\cdot n=\frac{5}{7}\cdot n$$

This is only the expected value. That means on average $\frac{5}{7}$ of the sample are womans. It is not sure that the number of females is 15 ($=21\cdot \frac {5}{7}=3\cdot \frac {5}{1}$) if the sample size is $21$. It has only the highest probability. The number of females can be from $0$ to $21$. But you can say, that for each single drawing of a person the chance is $\frac{5}{7}$ that this person is a female, if your sampling is with replacement.

Answer 3

The (long) answers are correct, but the OP wants a shorter one, so I will try.

When you roll a die, there are 6 possible things that could happen. If you assume the die is fair (not weighted, no rounded corners) then each of the six possibilities is equally likely. If you are betting on a "5" then you win with just one out of the six possibilities. "one out of six" is the fraction 1/6. If you are betting that the number will be even then you win in three of the six cases, so the probability is 3/6, or 1/2.

I think your problem starts with an idea of fractions that's just a little too narrow. You don't always think of 4/7 as dividing a pizza into 7 pieces and taking 4 of them. You might be thinking about 70 people in a room - 40 women and 30 men. Then 4/7 of the people are women. If you pick someone at random the probability that you pick a woman is 4/7.

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I hope those paragraphs help. But probability is subtle, so I will write a little more about two possible misunderstandings.

The "equally likely" is important. If you think about trying to roll a total of 3 with two dice you can't just list the possible totals (2 through 12) and say that the probability of a total of 3 is 1/11. When you roll the two dice there are $6 \times 6 = 36$ (equally likely) things that can happen (for example, 4 on the first die and 2 on the second die). Of those 36, just two give a total of 3, so the probability of a total of 3 is 2/36, or 1/18. (@JMorawetz discusses this example in detail in his answer.)

For a single roll of one die the assumption that the die is fair means that in the long run you will see a "5" (for example) about 1/6 of the time - just the fraction you get by counting the possibilities. Making sense of "about" and "in the long run" is a hard problem. It took mathematicians centuries to figure it out.

Answer 4

Where does the fraction say that for every $6$ rolls, I'll get on average one roll I wanted to get.

I don't think it's helpful to focus on it being $6$ rolls exactly, or to think in terms of a single roll. In probability, we have the law of large numbers which, in this case, means that the more rolls there are, the more the ratio of $1$s rolled to the total rolls will be to the number $\frac{1}{6}$.

In other words, if you make $600$ rolls of the dice, the number of $1$s you're going to roll is very likely to be very close to $100$. (As will the number of $2$s, $3$s, etc.) Now, this isn't all the $\frac{1}{6}$ probability means, but it's one of its more intuitive consequences.

Answer 5

I think it may help to understand what we are looking for when we ask for the probability of some event occurring. Given a large number of tests $N$, the probability of a particular event $E$ occurring should be some number $p$ such that $p\times N$ produces the expected number of events $E$. We are looking for the number of times a certain outcome will occur out of the total number of outcomes. When we have 100 events, probability asks, "How many are we interested in out of those 100 events?"

For example, when rolling a die, the probability of rolling a 1 is $\frac{1}{6}\approx16.67\%$. This means that if we roll the die $6$ times, we can expect to roll a 1 about $p\times N=\frac{1}{6}\times6=1$ time. Similarly, if we roll the die $300$ times, we can expect to roll a 1 about $p\times N=\frac{1}{6}\times300=50$ times.

The reason why the probability of rolling a 1 is $1$ out of $6$, or $\frac{1}{6}$, has to do with the definition of probability. Let's ask the same question as before. "How many events are we interested in out of the total number of possible events?" Well, there are $6$ possible outcomes (we can roll a 1, 2, 3, 4, 5 or 6), and we are looking for only $1$ of those outcomes (rolling a 1). Therefore, the probability of rolling a 1 is exactly $1$ (what we're interested in) out of $6$ (the total), or $\frac{1}{6}$. This is the same as saying, "For every 6 events, we can expect about 1 of them to be what we're looking for."