Expected value is a concept used in statistics and is very important in deciding how useful or harmful a given action will be. To calculate it, you need to understand each outcome of a situation and its probabilities, i.e. the chances of a particular case happening. This guide will help you through the process with a couple of example problems and teach you the concept of expected value.
Steps
Part 1 of 3: Elementary Problem
Step 1. Familiarize yourself with the problem
Before you think about the possible outcomes and probabilities involved in the problem, make sure you understand it. For example, consider a dice throwing game that costs $ 10 per spin. A six-sided die is rolled only once and your winnings depend on the side that comes up. If 6 comes out you get 30 euros; if 5 is rolled, you get 20, while you are the loser for any other number.
Step 2. Make the list of possible results
In this way you will have a useful list of possible outcomes of the game. In the example we have considered, there are six possibilities, which are: number 1 and you lose 10 euros, number 2 and you lose 10 euros, number 3 and you lose 10 euros, number 4 and you lose 10 euros, number 5 and you win 10 euros, number 6 and earn 20 euros.
Note that each outcome is 10 euros less than described above, as you still have to pay 10 euros for each play, regardless of the outcome
Step 3. Determine the probabilities for each outcome
In this case they are all the same for the six possible numbers. When you roll a six-sided die, the probability that a certain number will come up is 1 in 6. To make this value easy to write and calculate, you can transform it from a fraction (1/6) to a decimal using the calculator: 0, 167. Write the probability near each outcome, especially if you are solving a problem with different probabilities for each outcome.
- If you type 1/6 into your calculator, then you should get something like 0, 166667. It is worth rounding the number to 0, 167 to make the process easier. This is close to the correct result, so your calculations will still be accurate.
- If you want a really accurate result and you have a calculator that includes parentheses, you can type the value (1/6) in place of 0, 167 when proceeding with the formulas described here.
Step 4. Write down the value for each outcome
Multiply the amount of money related to each number on the dice by the probability that it will come out and you will find how many dollars contribute to the expected value. For example, the "prize" related to the number 1 is -10 euro (since you lose) and the possibility that this value will come out is 0, 167. For this reason the economic value linked to the number 1 is (-10) * (0, 167).
It is not necessary to calculate these values, for now, if you have a calculator that can handle multiple operations simultaneously. You will get a more precise solution if you insert the result in the entire equation later
Step 5. Add the various results together to find the expected value of the event
To always take the above example into account, the expected value of the dice game is: (-10 * 0, 167) + (-10 * 0, 167) + (-10 * 0, 167) + (-10 * 0, 167) + (10 * 0, 167) + (20 * 0, 167), that is - 1, 67 €. For this reason, when you play craps, you should expect to lose around € 1.67 in each round.
Step 6. Understand the implications of calculating the expected value
In the example we have just described, this indicates that you will have to expect to lose € 1.67 per game. This is an impossible result for any bet, since you can only lose 10 euros or earn 10 or 20. However, the expected value is a useful concept for predicting, in the long term, the average outcome of the game. You can also consider the expected value as the cost (or benefit) of the game: you should only decide to play if the fun is worth the price of € 1.67 per game.
The more the situation repeats itself, the more precise the expected value will be and it will approach the average of the outcomes. For example, you could play 5 times in a row and lose each time with an average expenditure of 10 euros. However, if you bet 1000 times or more, your average winnings should approach the expected value of -1.67 euros per play. This principle is called the "law of large numbers"
Part 2 of 3: Calculating the Expected Value in a Coin Toss
Step 1. Use this calculation to know the average number of coins you need to flip to find a specific resulting pattern
For example, you can use this technique to know how many times you have to flip a coin to get two "heads" in a row. The problem is slightly more complex than the previous one; for this reason reread the first part of the tutorial, if you are still unsure with the calculation of the expected value.
Step 2. We call "x" the value we are looking for
Suppose we want to find the number of times (on average) that a coin has to be flipped to get two "heads" consecutively. We will have to set up an equation that will help us find the solution that we will call "x". We will build the formula a little at a time, for now we have:
x = _
Step 3. Think about what would happen if the first throw was "tails"
When you flip a coin, half of the time, on your first toss you will get "tails". If this happens, then you will have "wasted" a roll, although your chances of getting two "heads" in a row have not changed at all. Just like just before the flip, you should expect to flip the coin a number of times before hitting heads twice. In other words, you should expect to do "x" rolls plus 1 (what you just did). In mathematical terms you can say that "in half of the cases you will have to flip the coin x times plus 1":
- x = (0, 5) (x + 1) + _
- We leave the space blank, as we will continue to add more data as we evaluate other situations.
- You can use fractions instead of decimal numbers if that's easier for you. Writing 0, 5 is equivalent to ½.
Step 4. Evaluate what will happen if you get “heads” on the first roll
There are 0, 5 (or ½) chances that on the first roll you get the side with the "head". This eventuality seems to bring you closer to your goal of getting two consecutive "heads", but can you quantify exactly how close you will be? The easiest way to do this is to think about the possible outcomes with the second roll:
- If on the second roll you get "tails", then you will end up again with two "wasted" rolls.
- If the second roll were "heads", then you would have achieved your goal!
Step 5. Learn how to calculate the probabilities of two events happening
We know that a roll has 0.5 chances of showing the head side, but what are the odds of two consecutive rolls giving the same result? To find them, multiply the probabilities of each side together. In this case: 0, 5 x 0, 5 = 0, 25. This value also indicates the chances of getting heads and then tails, as both have a 50% chance of showing up.
Read this tutorial that explains how to multiply the decimal numbers together, if you do not know how to perform the operation 0, 5 x 0, 5
Step 6. Add the result for the "heads followed by tails" case into the equation
Now that we know the probabilities of this outcome, we can extend the equation. There are 0.25 (or ¼) odds of flipping the coin twice without getting a useful result. Using the same logic as before, when we assumed that a "cross" would come out on the first roll, we will still need a number of "x" rolls to get the desired case, plus the two we have already "wasted". By transforming this concept into mathematical language we will have: (0, 25) (x + 2) which we add to the equation:
x = (0, 5) (x + 1) + (0, 25) (x + 2) + _
Step 7. Now let's add the "head, head" case to the formula
When you get two consecutive head-side throws, then you've achieved your goal. You got what you wanted in just two rolls. As we saw earlier, the chances of this happening are exactly 0.25, so if that's the case, let's add (0.25) (2). Our equation is now complete and is:
- x = (0, 5) (x + 1) + (0, 25) (x + 2) + (0, 25) (2).
- If you fear that you have not thought about all the possible outcomes of the launches, then there is an easy way to check the completeness of the formula. The first number in each "fragment" of the equation represents the probabilities of an event occurring. The sum of these numbers must always be equal to 1. In our case: 0, 5 + 0, 25 + 0, 25 = 1, so the equation is complete.
Step 8. Simplify the equation
Try to make it easier by doing multiplication. Remember that if you notice data in brackets like (0, 5) (x + 1), then you have to multiply each term of the second bracket by 0, 5 and you will get 0, 5x + (0, 5) (1) that is 0, 5x + 0, 5. Continue like this for all the fragments of the equation and then combine them together in the simplest way possible:
- x = 0.5x + (0.5) (1) + 0.25x + (0.25) (2) + (0.25) (2).
- x = 0.5x + 0.5 + 0.25x + 0.5 + 0.5.
- x = 0.75x + 1.5.
Step 9. Solve the equation for x
Just like in any other equation, your aim is to find the value of x by isolating the unknown on one side of the equal sign. Remember that the meaning of x is "the average number of throws to be performed to get two consecutive heads". When you have found the value of x, you will also have the solution to the problem.
- x = 0.75x + 1.5.
- x - 0.75x = 0.75x + 1.5 - 0.75x.
- 0.25x = 1.5.
- (0, 25x) / (0, 25) = (1, 5) / (0, 25)
- x = 6.
- On average, you'll have to expect to flip six times the dime before getting two heads in a row.
Part 3 of 3: Understanding the Concept
Step 1. Understand the meaning of the concept of expected value
It is not necessarily the most likely outcome to be achieved. After all, sometimes an expected value is downright impossible - for example, it could be as low as € 5 in a game that only offers € 10 prizes. This figure expresses how much value you should give to the event. In the case of a game whose expected value is greater than $ 5, you should only play if you believe the time and effort is worth $ 5. If another game has an expected value of $ 20, then you should only play if the fun you get is worth $ 20 lost.
Step 2. Understand the concept of independent events
In everyday life, many people think they have a lucky day only when good things happen and might expect that such a day holds many pleasant surprises. On the other hand, people believe that on an unfortunate day the worst has already happened and that one cannot have a worse fate than this, at least for the moment. From a mathematical point of view, this is not an acceptable thought. If you throw a regular coin, there is always a 1 in 2 chance of having heads or tails. It doesn't matter if at the end of 20 throws you only get heads, tails or a mix of these outcomes: the next throw will always have a 50% chance. Each launch is completely "independent" from the previous ones and is not affected by them.
The belief that you have had a lucky or unfortunate series of tosses (or other random and independent events) or that you have ended your bad luck and that from now on you will only have lucky outcomes is called the bettor's fallacy. It was defined this way after noticing the tendency of people to make risky or crazy decisions while betting when they feel they have a "lucky streak" or that luck "is ready to roll"
Step 3. Understand the law of large numbers
Perhaps you might think that expected value is a useless concept, as it rarely seems to tell you the outcome of an event. If you calculate the expected value of roulette and get -1 € and then play three games, most of the time you may find yourself losing 10 euros, earning 60 or other amounts. The "law of large numbers" explains why the expected value is much more useful than you think: the more games you play, the closer your results come to the expected value (the average result). When you consider a large number of events, then the total result is most likely close to the expected value.
Advice
- For situations in which there may be different outcomes, you can create an excel sheet on the computer to proceed with the calculation of the expected value of the outcomes and their probabilities.
- The example calculations in this tutorial, which took euros into account, are valid for any other currency.
