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## Statistics and probability

### Course: Statistics and probability>Unit 9

Lesson 8: More on expected value

# Expected profit from lottery ticket

Sal multiplies outcomes by probabilities to find the expected value of a lottery ticket. Created by Sal Khan.

## Want to join the conversation?

• I could barely understand what Sal said at about P(small). However, it would be better if someone fully explained it by both ways (the shorter one as Sal did and the longer one). If I calculated the probability using the normal method, how would it be?
• P(grand prize) = 1/10 x 1/10 x 1/26 = 1/2600
P(only letter correct) = 9/10 x 9/10 x 1/26 = 81/2600
P(1 number and the letter right) = (1/10 x 9/10 x 2P2) = 18/100 x 1/26 = 18/2600
NOTE: 2P2 because you can get the first number right or the second number right. Order matters.

Expected value of grand prize = 1/2600 x \$10,405 = \$4
Expected value of smaller prize = (81/2600 + 18/2600) x 100 = \$3.81
Total expected value of prizes= \$7.81
cost = \$5
Expected profit = \$7.81 - \$5 = \$2.81
• When I was trying to calculate the probability of winning the small prize, I went about it a whole different way and I'm wondering if its correct. There are two different scenarios in which you win the small prize: getting both numbers wrong and getting the letter right, or getting one number wrong and getting the letter right.

So for the first scenario I did (8/9*)(8/9)*(1/26). For the second I did (1/9)*(8/9)*(1/26). I then added the values of both the answers. The answer was around 0.034. Is this a wrong way of calculating probability?
• Your intuition is partially correct. There are actually 3 scenarios in which you win the small prize: getting the left number right and the right number wrong, the left number wrong and right number right, or getting both numbers wrong - in all three cases you also have getting the letter right. Also please note there are 10 numbers not 9 (0-9).

So the answer looks like this: (1/26)*(1/10)*(9/10) + (1/26)*(9/10)*(1/10) + (1/26)*(9/10)*(9/10). If you do the maths that gives you the same answer as Sal's approach.

Hope you find it useful.
• Why does he distribute the "-5" into each probable case; wouldn't just tossing "-5" at the end of everything imply the same thing?
• You're absolutely right. Since all of the probabilities add to 1, this would work. The math comes out to this:

P(grand)(10405-5) + P(small)(100-5) + P(neither)(-5)
(1/2600)*(10405-5) + (1/26 - 1/2600)*(100-5) + (25/26)*(-5)
These terms can all be distributed like this:
(1/2600 * 10405) + (1/2600 * -5) + ((1/26 - 1/2600) * 100) + ((1/26 - 1/2600) * -5) + (25/26 * -5)
If we rearrange this, we get:
-5(1/2600) - 5(1/26 - 1/2600) - 5(25/26) + (1/2600 * 10405) + ((1/26 - 1/2600) * 100)
Because we have three terms which are divisible by -5, we can simplify it to this:
-5 * (1/2600 + (1/26 - 1/2600) + 25/26) + (1/2600 * 10405) + ((1/26 - 1/2600) * 100)
Then we can add up all the things in those parentheses next to the -5, and we get:
-5 * (1) + (1/2600 * 10405) + ((1/26 - 1/2600) * 100)
And of course, -5 * 1 = -5. Thus, our final result is just:
(1/2600 * 10405) + ((1/26 - 1/2600) * 100) - 5.

So as you can see, you were right. It's perfectly acceptable to just move the -5 to the end, and it all works out the same mathematically.

Great observation!
• How is 1/26 -1/2600 the probability of getting the small prize?
• It might help if you think of it this way:

What are the odds of getting the right letter? Well, there are 26 letters, and he guesses one of them, so the probability of him getting it right is 1/26.

However, if he gets the letter right, will he always win the small prize? No. If he gets the letter right, and he also gets the numbers right, then he won't win the small prize. Instead, he will win the grand prize. In other words, the only way to win the small prize is to get the letter right, and to NOT get the numbers right. So, what are the odds of him doing that?

Well, he has a 1/26 chance of getting the letter right. And he has a 1/2600 chance of getting both right. So the probability of him getting the letter right and NOT getting the numbers right is 1/26 - 1/2600.

Thus, this is his probability of winning the small prize. Hope this helped. Great question!
• why subtract 1/2600? plz explain?
• Form what I can gather, he subtracts the 1/2600 in order to factor out the P(grand). The reason for doing this, is that P(small) = (1/26 [chance of getting the letter correct, which implies you win regardless] - 1/2600 [the chance of getting the grand prize, since 1/26 as the first value, implies that you could also win the grand prize] )

Hope that helps somewhat!

-Sean
• I solved it in a simpler way & got the same answer.
I said, imagine Ahmed buys 2,600 tickets. It will cost him \$5*2,600= \$13,000. OK so that's his total cost. Now we expect him to win the big prize once so that's \$10,405. And he's going to win the small prize 100-1 times or 99*\$100= 9,900. The 2 prize monies total \$10,405+\$9,900= \$19,945. Now subtract the initial investment of \$13,000 & he has made \$7,305. Divide that by 2600 (the # of tix he bought)= \$2.81.
• it seems that what you're doing is somehow an "old-school" way of calculating probability without relying on a concrete concept of probablity

in other words, it does the same work as Sal's (with some tweak of factoring out the cost of 5 before calculating the earning) in a bit more graspable way for our human intuition. cause you're using the actual number of tickets as unit of probability

but when it comes to generalization, i believe the way of handling probability with 0 to 1 range has a lot greater potential and usage than doing the same with real number range like yours, 0 to 2600

in short, i really appreciate your intuition to figure out an intuitive way of solving this. but i also highly recommend you to embrace the concept and power of probability a bit deepr
(1 vote)
• Hello, I just wanted to clarify why the probability of getting a number right is 1/10 instead of 1/11?I think it is 1/11 because 0 is a part of the set of numbers that are used in the lottery tickets (when we count 0 in, we will have 11 numbers).Thanks!
• There are only 10 numbers to pick from: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9; therefore the probability of choosing a number correct is 1/10.
• At Sal calculates the probability of small to be (1/26-1/2600). Why isn't this probability equal to (9/10*9/10*1/26)?
Since, Probability of losing number = 9/10 and Probability of winning alphabet = 1/26.
• I did the problem like you say. Once you buy a ticket, the expected values are as follows:
Expected Value of \$5 payout = probability*value = 1 * (-\$5)
EV(grand prize) = P(x)*x = (1/10*1/10*1/26) * (10405) = 4.0019
EV(small prize) = (1/10*9/10*1/26) * 100 + (9/10*1/10*1/26) * 100 + (9/10*9/10*1/26) * 100 [ there are 2 ways to get 1 number, 1 way to get no numbers] = 0.34615 + 0.34615 + 3.1154
Total is 7.81 - 5 = 2.81