# The meaning of expected value for discrete random variable in dice experiments

Does the expected value always speak to a payoff? Or can the expected value be thought of independent of payoffs?

I don't understand when we say a fair die has an expected value of 3.5. Does that mean we should expect to roll a 3 or 4 most of the time, or that if we had to pay for each roll and the payout was 1 dollar for 1, 2 dollars for 2, etc., that you shouldn't pay more than $3.5 per roll because in the long run that would be fair (meaning we'd break even). • It's just historic terminology for the average over possibilities. As with 3.5 for throwing a die with 6 sides, the "expectation" does not even have to be a feasible value. Jul 13, 2016 at 18:48 • Value here doesn't imply monetary value. The word "payoff" seems to suggest that you are thinking of money. Even throwing dice or choosing playing cards need have no monetary implication. Jul 13, 2016 at 18:57 • Expected value relates to the expectation of some random variable (a formally-defined random variable is always numeric, but not necessarily monetary). Jul 13, 2016 at 23:47 • @Glen_b could you check my example 2? Am I right the expectation does not have any physical meaning such case? Jul 14, 2016 at 14:46 • @hxd1011 this is already addressed in my previous comment. Your$Y$does not formally meet the requirements of a random variable in the standard sense. For example see the distinction drawn here and here between a random variable and a random element. Sometimes people may call such a variable a random variable but the most common definition would require us to map outcomes to elements of$\mathbb{R}$to get a random variable. Jul 14, 2016 at 18:07 ## 3 Answers There are several words that can have different meanings in statistics and other fields or every day life in your question. For instance fair in the context of throwing a die means that all sides have the same probability of occurring. The same word has a different meaning in game theory, where fair price means a price you'd pay for a lottery which eliminates the profit from the payoffs. The word expected in statistics means this: $$E[x]=\sum_{i=1}^6p_ix_i=\frac{1}{6}\1+\frac{1}{6}\2+\dots+\frac{1}{6}\6$$ where$i\in{1,2,3,4,5,6}$- an index of the side,$p_i$the probability to get this side, and$x_i$- the outcome of a trial (payoff). If the payoff was in dollars and the die is fair you could easily see how this is equal to$3.5.

The outcome (payoff) that happens most often is called a mode. In the case of a fair die, all outcomes are equally probable, so the mode is not an interesting measure.

UPDATE: So, if you think of a fair dice as a tool which generates payoff randomly depending on which side was tossed, then yes, $3.5 is a fair price you'd pay for this tool assuming there's no time value of money. There are complications, into which you can dig in if interested, things like ST. Petersburg paradox etc. UPDATE 2: @hxd1011 asked whether there's a physical meaning to that referring to his Example 2 with the possible outcome to be animals. Right, the expected value is of the values of a random variable. The random variables in statistics are defined as some - usually real - values that are linked to events from the event space. Do not mix them with indices. For instance, in your example 2 let's denote the events with indices$j={1,2,3}$, then we can enumerate all possible events:$\omega_1=\text{cat},\omega_2=\text{dog},\omega_3=\text{pig}$. Let's say we have the associated probabilities$p_1=p_2=p_3=1/3$. If you did not define the random variable$x_i$yet, then there's no point in talking about the expected value at all. Expected value of what? Of index$j$? It doesn't have any physical meaning as you wrote. Let's now define a random variable$xon the probability space as follows (suppose we have some hypothetical game that if you guess correctly you will get some money): \begin{align} x&=\10,\quad\text{if }\omega=\text{cat} \\ x&=\$20,\quad\text{if }\omega=\text{dog} \\ x&=\$30,\quad\text{if }\omega=\text{pig} \end{align} Now we can talk about the expected value of $E[x]$. We can calculate it easily using the equation above, it's $\$20$of course. • could you check my example 2? does the expectation have some physical meaning for that? Jul 14, 2016 at 14:45 • @hxd1011, I updated my answer. sorry for confusion Jul 14, 2016 at 17:28 • +1 for really nice answer that point out "random variables are defined as some real values that are linked to events from the event space." Jul 14, 2016 at 17:36 The expected value corresponds to the value you should expect to obtain, on average, if you repeat the action MANY times. In your example of the fair dice, if you assume that all sides are equally likely (probability 1/6 each) and if you were to do try many times, on average you would get the value: 1 * 1/6 + 2 * 1/6 + 3 * 1/6 4 * 1/6 + 5* 1/6 + 6 * 1/6 = 3.5 But, if the dice was not fair (assume for example that it never gives you a 1, but the 6 is twice as likely as usual), then the expected value would be: 1 * 0 + 2 * 1/6 + 3 * 1/6 4 * 1/6 + 5* 1/6 + 6 * 2/6 = 4.3 Note that the expected value does not need to correspond with any actual (integer) value you can obtain in one particular realization. Your confusion is coming from the definition of discrete random variables. I will use two examples to illustrate. Example 1. Assume you can receive$X$mails a day,$X$is a discrete number that can be$0,1,2,\cdots$. Assume you collected 1 week data, and it is$(1,2,3,4,5,6,7)$. So, the expected value / average is$4$. So you know the physical meaning is on average you will receive$4$mails a day. This is the case that the expected value has intuitive physical meaning. Example 2. Assume there is an animal in the room, but you do not if it is a cat or dog or pig. Let's use$\Omega \in \{\text{cat,dog,pig}\}$to represent the possible outcome. Note, As @Glen_b, @whuber, and @Aksakal mentioned, we cannot think$\Omega$is a random variable, because iti s not a real number. So, we need to use some mapping to build random variables. Let us use$Y \in \{1,2,3\}$to be a discrete random variable and represent$\{\text{cat,dog,pig}\}$, and you collect$7$data points, they are$(1,1,1,1,1,1,2)$so the expected value is$1.142$. Here is where your confusion comes from: what does this number mean? This number does not have physical meaning, i.e., which animal it is. To conclude, you confusion comes from the definition of discrete random variable and the expectation. The expectation does always have concrete physical meaning. • There appears to be a confusion here between an outcome and a random variable. In your example, despite the suggestive notation,$Y\$ is an outcome, not a random variable.
– whuber
Jul 14, 2016 at 18:46
• @whuber thanks for the comment, you are wright some of my concept is not clear. I also learned form Glen_b, that difference between random variable and random elements. Jul 14, 2016 at 19:07
• If you are interested, I attempted to address these issues in a non-mathematical but rigorous way in the thread on what is meant by a random variable.
– whuber
Jul 14, 2016 at 19:31