Expected Value if One Outcome Is Roll Again

Contents:

1. What is Expected Value?
2. Formula
   • Basic Formula
   • Binomial Random Variable
   • Multiple Events
   • Expected Value for Continuous Random Variables
   • Expected Value Formula for an Capricious Role
3. Find an Expected Value by Hand
4. Observe an Expected Value in Excel
5. Find an Expected Value for a Discrete Random Variable
6. What is an Expected Value used for in Real Life?
7. St. Petersburg Paradox

What is Expected Value?

Expected value is exactly what you might think it means intuitively: the return yous can wait for some kind of activeness, similar how many questions you lot might become right if you guess on a multiple option exam.

Lookout this video for a quick caption of the expected value formulas:

Can't see the video? Click here.

For example, if you take a 20 question multiple-pick test with A,B,C,D every bit the answers, and you guess all "A", then you can wait to get 25% right (5 out of xx). The math behind this kind of expected value is:
The probability (P) of getting a question correct if you lot gauge: .25
The number of questions on the test (n)*: xx
P 10 north = .25 x xx = 5

*You might see this equally X instead.

This blazon of expected value is called an expected value for a binomial random variable. It's a binomial experiment because in that location are but ii possible outcomes: you get the answer right, or y'all get the answer wrong.

Formula

Basic Expected Value Formula

The bones expected value formula is the probability of an event multiplied by the amount of times the event happens:
(P(x) * due north).

The formula changes slightly according to what kinds of events are happening. For well-nigh simple events, you'll use either the Expected Value formula of a Binomial Random Variable or the Expected Value formula for Multiple Events.

Formula for the Expected Value of a Binomial Random Variable

The formula for the Expected Value for a binomial random variable is:
P(x) * 10.
X is the number of trials and P(x) is the probability of success. For example, if you toss a coin 10 times, the probability of getting a heads in each trial is 1/2 and then the expected value (the number of heads yous can expect to get in x coin tosses) is:
P(x) * X = .5 * 10 = v

Tip: Calculate the expected value of binomial random variables (including the expected value for multiple events) using this online expected value calculator.

Expected Value for Multiple Events

Of course, calculating expected value (EV) gets more complicated in real life. For example, You buy ane $10 raffle ticket for a new automobile valued at $15,000. Two m tickets are sold. What is the EV of your gain? The formula for computing the EV where there are multiple probabilities is:

E(X) = ΣX * P(10)
Where Σ is summation note.

The equation is basically the same, just here y'all are adding the sum of all the gains multiplied past their individual probabilities instead of just i probability.

Other Expected Value Formulas

The two formulas to a higher place are the ii about common forms of the expected value formulas that you lot'll see in AP Statistics or elementary statistics. However, in more rigorous or advanced statistics classes (like these), you might meet the expected value formulas for continuous random variables or for the expected value of an arbitrary function.
Expected Value Formula for an Arbitrary Function

Expected Value for Continuous Random Variables

The expected value of a random variable is simply the mean of the random variable. Y'all can calculate the EV of a continuous random variable using this formula:

Expected value formula for continuous random variables.

Where f(10) is the probability density function, which represents a function for the density curve.
The "∫" symbol is chosen an integral, and information technology is equivalent to finding the area under a curve.

Expected Value Formula for an Arbitrary Function

If an effect is represented by a function of a random variable (m(x)) and then that role is substituted into the EV for a continuous random variable formula to get:

Expected value formula for an arbitrary part.

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Calculate an Expected value in statistics by hand

Lookout man the video for an example:

How to find an Expected Value

Can't meet the video? Click here.

This section explains how to figure out the expected value for a single item (like purchasing a single raffle ticket) and what to do if you have multiple items. If you lot have a discrete random variable, read Expected value for a discrete random variable.

Example question: You buy one $10 raffle ticket for a new car valued at $fifteen,000. Two thousand tickets are sold. What is the expected value of your gain?

Step 1: Make a probability chart (encounter: How to construct a probability distribution). Put Gain(X) and Probability P(X) heading the rows and Win/Lose heading the columns.

Stride ii: Effigy out how much you lot could gain and lose. In our example, if nosotros won, nosotros'd exist upward $xv,000 (less the $10 price of the raffle ticket). If you lose, you'd be down $10.  Fill in the information (I'm using Excel here, and then the negative amounts are showing in ruddy).

Step 3: In the bottom row, put your odds of winning or losing. Seeing every bit 2,000 tickets were sold, you accept a i/2000 run a risk of winning. And you lot as well have a 1,999/2,000 probability chance of losing.

Footstep 4: Multiply the gains (10) in the height row by the Probabilities (P) in the lesser row.
$xiv,990 * 1/2000 = $7.495,
(-$ten)*(1,999/ii,000)= -$9.995

Step 5:Add the two values together:
$seven.495 + -$9.995 = -$two.5.

That's information technology!

Note on multiple items: for example, what if you purchase a $10 ticket, 200 tickets are sold, and besides as a auto, y'all have runner up prizes of a CD player and luggage gear up?

Perform the steps exactly as to a higher place. Brand a probability chart except you'll have more items:

Then multiply/add the probabilities as in step 4: 14,990*(i/200) + 100 * (one/200) + 200 * (one/200) +  -$10 * (197/200).

You'll note at present that because you lot have iii prizes, you have 3 chances of winning, so your risk of losing decreases to 197/200.

Note on the formula: The actual formula for expected gain is East(Ten)=∑X*P(X) (this is likewise 1 of the AP Statistics formulas).  What this is maxim (in English language) is "The expected value is the sum of all the gains multiplied by their individual probabilities."

Like the explanation? Check out the Practically Adulterous Statistics Handbook, which has hundreds more than stride-by-step explanations, just like this one!

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Notice an Expected Value in Excel

Step i: Type your values into 2 columns in Excel ("10" in one cavalcade and "f(x)" in the side by side.
Step 2: Click an empty cell.
Step three: Type =SUMPRODUCT(A2:A6,B2:B6) into the jail cell where A2:A6 is the actual location of your x variables and f(x) is the actual location of your f(x) variables.
Step four: Press Enter.

That's it!
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Find an Expected Value for a Discrete Random Variable

Y'all tin think of an expected value every bit a mean, or boilerplate, for a probability distribution. A detached random variable is a random variable that can only take on a certain number of values. For example, if you were rolling a die, information technology tin simply accept the set of numbers {1,2,3,four,5,half dozen}. The expected value formula for a discrete random variable is:

Basically, all the formula is telling you to practise is find the mean by adding the probabilities. The mean and the expected value are then closely related they are basically the same thing. You'll demand to exercise this slightly differently depending on if you have a prepare of values, a set of probabilities, or a formula.

Expected Value Discrete Random Variable (given a listing).

Case trouble #1: The weights (X) of patients at a clinic (in pounds), are: 108, 110, 123, 134, 135, 145, 167, 187, 199. Assume one of the patients is chosen at random. What is the EV?

Step 1: Discover the mean. The mean is:
108 + 110 + 123 + 134 + 135 + 145 + 167 + 187 + 199 = 145.333.
That's it!

Expected Value Detached Random Variable (given "10").

Case problem #ii. You toss a fair coin three times. X is the number of heads which announced. What is the EV?
Step one: Figure out the possible values for X. For a three money toss, you lot could get anywhere from 0 to 3 heads. So your values for 10 are 0, one, 2 and 3.

Step 2: Figure out your probability of getting each value of X. You may need to use a sample space (The sample infinite for this problem is: {HHH TTT TTH THT HTT HHT HTH THH}). The probabilities are: ane/8 for 0 heads, 3/8 for i head, three/viii for ii heads, and i/eight for iii heads.

Step iii: Multiply your X values in Pace 1 by the probabilities from step two.
E(X) = 0(1/8) + 1(three/8) + 2(3/viii) + 3(ane/8) = iii/2.

The EV is 3/2.

Expected Value Discrete Random Variable (given a formula, f(x)).

Example problem #3. You toss a coin until a tail comes up. The probability density function is f(x) = ½^x. What is the EV?
Step 1: Insert your "10" values into the kickoff few values for the formula, ane by one. For this item formula, y'all'll get:
one/ii⁰ + one/two¹ + 1/2² + 1/2³ + i/2⁴ + 1/2⁵.

Step 2: Add up the values from Footstep one:
= 1 + one/two + i/iv + one/8 + ane/16 + one/32 = 1.96875.

Annotation: What you lot are looking for here is a number that the serial converges on (i.due east. a set number that the values are heading towards). In this case, the values are headed towards two, and then that is your EV.

Tip: You can but use the expected value detached random variable formula if your function converges admittedly. In other words, the office must stop at a item value. If information technology doesn't converge, then in that location is no EV.

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What is Expected Value in Statistics used for in Existent Life?

Expected values for binomial random variables (i.e. where you have two variables) are probably the simplest type of expected values. In existent life, you're likely to encounter more circuitous expected values that have more than than 2 possibilities. For example, you might buy a scratch off lottery ticket with prizes of $thousand, $10 and $1. Y'all might desire to know what the payoff is going to be if you lot go alee and spend $ane, $5 or even $25.

Let's say your school is raffling off a season pass to a local theme park, and that value is $200. If the schoolhouse sells one one thousand $ten tickets, every person who buys the ticket volition lose $nine.80, expect for the person who wins the flavor pass. That's a losing proposition for you (although the school volition rake it in). You lot might want to save your money! Here's the math backside it:

1. The value of winning the season ticket is $199 (you don't get the $ten back that yous spent on the ticket.
2. The odds that you win the season laissez passer are ane out of 1000.
3. Multiply (i) past (2) to get: $199 * 0.001 = 0.199. Set this number bated for a moment.
4. The odds that y'all lose are 999 out of chiliad. In other words, your odds of ending up minus ten dollars are 999/1000. Multiplying -$10, you get -nine.999.
5. Adding (3) and (4) gives united states the expected value: 0.199 + -9.999 = -9.80.

Here'due south that scenario in a tabular array:

St Petersburg Paradox

What is the Saint petersburg Paradox?

The Leningrad Paradox has been stumping mathematicians for centuries. It's near a betting game you can always win. But despite that fact, people aren't willing to pay much money to play it. It's called the Petrograd Paradox considering of where information technology appeared in print: in the 1738 Commentaries of the Regal Academy of Science of Petrograd.

The Paradox is this: In that location'due south a simple betting game you can play where your winnings are always going to be bigger than the amount of money y'all bet. Imagine buying a scratch off lottery ticket where the expected value (i.e. the amount you can look to win) is ever higher than the amount you pay for the ticket. You could buy a ticket for $1, $10, or a 1000000 dollars. You volition always come ahead. Would you lot play?

Bold the game isn't rigged, yous probably should play. Merely the paradox is that most people wouldn't be willing to bet on a game like this for more than a few dollars. Then, why is that? At that place are a couple of possible explanations:

1. People aren't rational. They aren't willing to risk their money fifty-fifty for a sure bet.
2. There has to exist something wrong with the game's odds. Surely the odds of winning can't always be that good, tin can they?

The short answer is, people are rational (for the most function), they are willing to office with their money (for the near part). And, there is admittedly zilch incorrect with the game. If you're confused at this bespeak — that is why it'due south called a paradox.

The St. petersburg Paradox Game.

How much would y'all bet if you could e'er win?

The original paradox wasn't about lottery tickets (they didn't exist in 1738). It was about a coin toss game. Suppose yous were asked past a friend to play a coin toss game for $2. Assume the coin is fair (i.e. information technology isn't weighted). Yous toss the coin until the first tails comes upwardly, at which time you would earn $two^northward and the game would end. In other words, if tails come up upward on the first toss, you would win $2¹ = $2. If tails comes up on the tertiary toss, you would win $2³ = $eight. And if you had a run and tails showed upwards on the 20th toss, you lot would win $ii²⁰ = $one,048,576.

If you lot figure out the expected value (the expected payoff) for this game, your potential winnings are infinite. For example, on the first flip, you have a l% risk of winning $2. Plus y'all get to toss the coin again, so yous also accept a 25% chance of winning $4, plus a 12.5% run a risk of winning $8 and so on. If y'all bet over and over again, your expected payoff (gain) is $1 each time you lot play, as shown by the following table.

P(due north)	Prize	Expected payoff
1	1/two	$two	$1
2	1/4	$iv	$1
3	i/eight	$8	$1
4	1/xvi	$16	$1
5	1/32	$32	$one
half dozen	1/64	$64	$one
7	1/128	$128	$1
8	ane/256	$256	$1
ix	1/512	$512	$1
10	ane/1024	$1024	$1

Y'all tin't maybe lose money. Nonetheless, despite the expected value being infinitely large, most people wouldn't be willing to fork out more a few bucks to play the game.

The Leningrad paradox has been debated by mathematicians for near three centuries. Why won't people risk a lot of money if the odds are certainly in their favor? Every bit of nonetheless, no ane has found a satisfactory answer to the paradox. As Michael Clark states: "[The St. petersburg Paradox] seems to exist 1 of those paradoxes which we accept to swallow." A couple of solutions, which have been presented and yet have failed to offer a satisfactory respond:

• Limited utility (suggested by Jacob Bernoulli). Basically, the more we have of something, the less satisfied we are with information technology. You can utilize this to candy; You lot're probable to be satisfied with ane bag, but afterwards vi or vii bags, you lot're probable to not desire whatsoever more than. Yet, you tin can't utilize this to money. Everyone wants more coin, correct?
• Hazard aversion. The average person might consider putting a few thousand dollars in the stock market. But they wouldn't exist willing to run a risk their entire life savings. You can't apply this dominion to the St. Petersburg Paradox game considering in that location is no take chances.

Next: Powerball Expected Value

References

Clark, Michael, 2002, "The Leningrad Paradox", in Paradoxes from A to Z, London: Routledge, pp. 174–177.
Papoulis, A. "Expected Value; Dispersion; Moments." §5-iv in Probability, Random Variables, and Stochastic Processes, 2nd ed. New York: McGraw-Loma, pp. 139-152, 1984.

Related manufactures:
Online expected value calculator.

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