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I understand how we get 3.5 as the expected value for rolling a fair 6-sided die. But intuitively, I can expect each face with equal chance of 1/6.

So shouldn't the expected value of rolling a die be either of the number between 1-6 with equal probability?

In other words, when asked the question 'what's the expected value of throwing a fair 6-sided die?', one should answer 'oh, it can be anything between 1-6 with equal chance'. Instead it's 3.5.
Intuitively in real world, can someone explain how 3.5 is the value I should expect on throwing a die?
Again I don't want the formula or the derivation for the expectation.

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    $\begingroup$ en.wikipedia.org/wiki/Expected_value#History $\endgroup$ – Alex Mar 3 '17 at 1:27
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    $\begingroup$ You can think it's like average. $\endgroup$ – SmallChess Mar 3 '17 at 1:28
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    $\begingroup$ @Tim Okay, but I still don't understand why we call expected value as expected value. I'm just curious if I'm missing some intuition here. Even a story of how/why we started looking at expectation would also help I believe $\endgroup$ – Nithish Inpursuit Ofhappiness Mar 3 '17 at 1:29
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    $\begingroup$ The expected value of rolling a die isn't the number you expect to get; it's the amount of money you expect to get if you know that you are going to be paid the number that comes up. If you expect to go home with less than \$3.50 then you are being too pessimistic; if you expect to go home with more than \$3.50 then you are expecting too much. Even though you know you can't end up with exactly \$3.50, you wouldn't pay \$3.51 for a chance to roll. $\endgroup$ – Flounderer Mar 3 '17 at 1:51
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    $\begingroup$ @Flounderer By that argument, nobody would pay for lottery tickets ... but they seem to sell quite well $\endgroup$ – Glen_b Mar 3 '17 at 3:11
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Imagine that you are in Paris in 1654 and you and your friend are observing a gambling game based on sequential rolling of a six sided dice. Now, gambling is highly illegal and busts by the gendarme are quite frequent, and to be caught at a table with stacks of livre is to almost surely guarantee a lengthy stint in the Chateau d'If.

To get around this you and your friend have a gentleman's agreement on a bet made between two of you prior to the last die roll. He agrees to pay you five livre if you observe two sixes in the next five rolls of dice, and you agree to pay him the same amount if two ones are rolled, with no other action if these combinations do not come up.

Now, the last die roll is a six so you are on the edge of your seat, figuratively. At this moment, heavily armed guardsmen burst into the den and arrests everyone at the table, and the crowd disperses.

Your friend believes that the bet made between the two of you is now invalidated. However, you believe that he should pay you some amount as one six had already been rolled. What is a fair way of settling this dispute between the two of you?

(This is my interpretation of the origins of the expected value as presented in here and discussed in greater detail here)

Let's answer this question of fair value in a non rigorous way. The amount your friend should pay you can be calculated in the following manner. Consider all possible rolls of four dice. Some sets of rolls (namely those containing at least one six) will result in your friend paying out the agreed amount. However, on other sets (namely, those not containing a single six) will result in you receiving no money. How do you balance the possibility of these two types of rolls happening? Simple, average out the amount you would have been paid over ALL possible rolls.

However, your friend, (quite unlikely), can still win his bet! You have to consider the number of times two ones will be rolled in the remaining four dice, and average out the amount you will pay him over the number of all possible rolls of four dice. This is the fair amount you should pay your friend for his bet. Thus the amount you end up getting is the amount your friend should pay you, minus what you should pay your friend.

This is why we call it the "expected value". It is the average amount you expect to receive if you are able to simulate an event happening in multiple simultaneous universes.

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  • $\begingroup$ I was just going to submit my answer, but your explanation is much better! $\endgroup$ – Flounderer Mar 3 '17 at 1:44
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Excellent question. It's more subtle than it seems at first. It has to do with the random event and random variable (number, value). Your confusion stems from mixing together these two related but distinct concepts.

Let's start with an event. From the way you formulated your question, it appears that you consider the outcome of a dice throw an event. It's random, so you may get one of its six sides with equal chance, as you wrote. It makes a perfect sense.

What is expected value of this experiment? The expectations are defined for random variables (values) not events. For you the numbers 1 to 6 on the dice are simply the ways to distinguish its sides (in the context of your question's formulation). Imagine you instead used letters: A, B, C, D, E, and F. Replace the numbers with letters and repeat your question as follows:

In other words, when asked the question 'what's the expected value of throwing a fair 6-sided die?', one should answer 'oh, it can be anything between A and F with equal chance'

Now try to come up with an expected value. It's not defined!

The expectations show up when you define the random values, such as 1 to 6. You map the values to the event space, for instance, you define that side A is 1, side B is 2 etc. Now you have 6 numbers and can calculate the expectation, which happens to be 3.5.

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    $\begingroup$ Thank you Aksakal. Your answer together with Alex's make perfect sense to me now! Yours points out how my question is valid and what assumption (misconception) I was making, and Alex's provides more elucidated information on what you concisely wrote in your last para. $\endgroup$ – Nithish Inpursuit Ofhappiness Mar 3 '17 at 2:43
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"Each of the values equally likely", or "some value most likely" is the definition of mode, not expected value.

Imagine we are playing a coin-tossing game. Each time I toss heads, I give you 1\$, each time I toss tails, you give me 1\$. How much money would you expect to win or loose in the long run? Amounts are equal, probabilities of throwing them are equal, expected value is zero.

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The expected value is called so because if you average all dice rolls you expect to get this expected value in the long run. The expected value is not related to any single dice roll.

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From an historical point of view, the concept seemed to appear in different countries, so I would consider the use of this word as a convenient convergence between similar concepts across languages.

My starting point was the excellent Earliest Uses of Symbols in Probability and Statistics:

Expectation. A large script E was used for the expectation in W. A. Whitworth's well-known textbook Choice and Chance (fifth edition) of 1901 but neither the symbol nor the calculus of expectations became established in the English literature until much later. For example, Rietz Mathematical Statistics (1927) used the symbol E and commented that "the expected value of the variable is a concept that has been much used by various continental European writers..." For the continental European writers E signified "Erwartung" or "espérance (editor's note: mathématique)."

The term is sometimes "attributed to" Huyghens, which is discussed in Huygens Foundations Of Probability:

It is generally accepted that Huygens based probability on expectation. The term "expectation," however, stems from Van-Schooten's Latin translation of Huygens' treatise. A literal translation of Huygens' Dutch text shows more clearly what Huygens actually meant and how he proceeded.

Additional details with respect to Fermat, Pascal can be found in Expectation and the early probabilists.

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Interestingly, the more general concept than expected value is location. Thus, the concept of expected value has subtle implications that are somewhat confusing.

It is reasonable to question what it means to have 3.5 as anything to do with an anticipated outcome for a die. The answer is that although the average value of rolled dice outcomes is 3.5, that the expected value concept only signifies mean or average value, and is only an expectation for a limited class of functions, that specific to the question here does not include die roll outcomes. To put it another way, although the average roll outcome is 3.5, So what? True enough, one can invent a context (in some alternate universe), where an average value has meaning, but, die outcomes $\leq 3$ pays $\$1$, and outcomes $\geq 4$ loses $1, works as well as an average, with the advantage of actually having outcomes in this universe.

The reason for the inordinately restricted association between the term "expected value" and "mean value" appears to be historical rather than semantically correct, or even particularly cogent. That is, the context in which a calculated expected value is consistent the expectation of a location characterizing behaviour in a data set is limited to only certain distributions of data, and not others.

That this is historical is supported by the notion of statistical moments. It is widely acknowledged that the first proof of the central limit theorem up to modern standards of rigor was given by Chebyshev in 1887. His argument introduced the method of moments.. Now the first moment of $f$ was, for Chebyshev, the mean value of a Borel set. The concept of a mean value being thus an expected value for the normal distribution, that is, the density function to which the central limit theorem applies is thus tracable to Chebyshev 1887. Such is the strength of the central limit theorem that it became a parenthetical expression to associate expected value with a mean value, as opposed to a more general measure of location.

But what about data distributions that are not normal for which other measures are more stable and/or more representative of that data? For example, the mid-range value or average extreme value of data from a uniform distribution is more accurate and stable, i.e., precise and converges faster than the mean or median of that distribution. For log-normal distributions, e.g., (much of the treatment of) income data, the anti-log of the mean of the logarithm of data (A.K.A. geometric mean, e.g., moderate income data), rather than the data mean (e.g., mean income) itself, would be more indicative of what an individual thinking (or anticipated datum) to become inserted into that data might have as an anticipated outcome. That this is well-known is illustrated by the phrase, "I am anticipating a 5-figure salary." Here is an example of this for actual incomes. Another example, Pareto distributions, also used for income calculations, (see 80/20 law, and high income data) often have an undefined expected value (infinite first moment of $\alpha \beta ^{\alpha } t^{-\alpha -1}$ when $\alpha \leq 1$), such that for such distributions, it would be a mistake to anticipate an outcome to be an expected value. In that case, see Pareto distribution, the median, geometric mean, and harmonic mean are better measures of location, not only because the $\alpha \leq 1$ requirement is removed, but also because they are less variable even when $\alpha \gt 1$. Further information is found here in Clauset A, Shalizi CR, Newman ME. Power-law distributions in empirical data. SIAM review. 2009;51:661-703, and here.

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