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Course: Statistics and probability > Unit 9
Lesson 8: More on expected value- Term life insurance and death probability
- Getting data from expected value
- Expected profit from lottery ticket
- Expected value while fishing
- Comparing insurance with expected value
- Expected value with empirical probabilities
- Expected value with calculated probabilities
- Making decisions with expected values
- Law of large numbers
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Expected profit from lottery ticket
Sal multiplies outcomes by probabilities to find the expected value of a lottery ticket. Created by Sal Khan.
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I could barely understand what Sal said at4:31about 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?(26 votes)
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(46 votes)
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?(6 votes)- 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.(21 votes)
Why does he distribute the "-5" into each probable case; wouldn't just tossing "-5" at the end of everything imply the same thing?(12 votes)
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!(7 votes)
- How is 1/26 -1/2600 the probability of getting the small prize?(6 votes)
- 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!(15 votes)
why subtract 1/2600? plz explain?(8 votes)
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(8 votes)
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.(9 votes)
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!(2 votes)
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.(10 votes)
At4:34Sal 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.(2 votes)- 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(2 votes)
Why is the outcome of the number $2.81? Why is it an odd number and not rounded to 0? Why did the outcome be $2.81 anyways, and not him either winning the grand, the small, or nothing?(2 votes)- The expected value is used to show you whether you will have profit if you play the game. It makes no sense when you the game once because $2.81 never come out. But according to the theoretical probability, if you play the game for 2600 times, you will likely get 1 grand prize and 99 small prized and you will have to pay 2600x5$, the profit will be 7305$ = 2.81$ x 2600.(1 vote)
- Does the order of the numbers matter ? Meaning if 04R considered a winning ticket, is 40R also considered a winning ticket and if yes would that change the expected value ? Thanks.(2 votes)
The order of the numbers matters in this problem.
If it didn't matter, we would get a different expected value.(1 vote)
4:31
Video transcript
Voiceover:Ahmed is playing a lottery game where he must pick two
numbers from zero to nine and then one letter out of the
26 letter English alphabet. He may choose the same number both times. If his ticket matches the two numbers and one letter drawn in order, he wins the grand prize
and receives $10,405. If just his letter matches but one or both of his numbers do not match, he wins the small price of $100. Under any other outcome, he
loses and receives nothing. The game costs him $5 to play. Under any other outcome he
loses and receives nothing. He has chosen the ticket 04R. Assuming he's paying the $5 to play and he picks the ticket 04R. Let's say we define a random variable X and let's say that this random variable is the net profit from
playing this lottery game. What is the expected from ... I guess we could even say the expected from the net profit from playing 04R, so Ahmed's particular
ticket right over here. Let's just say X is the random variable, is the net profit from
playing this ticket. What I want to think about in this video is what is the expected value of that? What is the expected net
profit from playing 04R? I encourage you to pause the video and think through it on your own. Let's think about what expected value is. It's the probability of
each of those outcomes times the net profit from those outcomes. There's the probability
of the grand prize. I can write that, let me
do that in that red color. There is the probability
of getting the grand prize and what would times his net
payoff from the grand prize. What would that be? Well he gets $10,405 but
that's not his net payoff or his net profit I should say. His net profit is what he gets
minus what he paid to play. He paid $5 to play. That's that, plus the probability of getting the small
price times the pay off of the small price which
is going to be $100 or times the net profit I guess
if you get the small price. You get a payoff of a 100 minus you have to pay $5 to play and then finally you have
the probability of neither. You're essentially not winning and in that situation,
what is the net profit? Well in that situation your
net profit is negative five. You paid $5 and you got nothing in return. To figure out the expected value, you just have to figure
out these probabilities. What's the probability of the grand prize? I'll do that over here,
probability of grand prize. Well the probability that he
gets the first letter right is one in 10, there's 10 digits there. Probability he gets
the second letter right is one in 10, these are all independent and probability he gets the letter right, there's 26 equally likely letters that might be in the actual one so he has a one in 26
chance of that one as well. The probability of the
grand prize is one in 2600. This is one in 2600. Now what's the probability
of getting the small price? Well let's see, he has a one in 26 chance. The small prize is
getting the letter right but not getting both of the numbers right. He has a one in 26 chance
of getting the letter right but we're not done here
just with the one in 26 because this one in 26, this includes all the scenarios where he gets the letter right, including the scenarios where
he wins the grand prize, where he gets the letter and
he gets the two numbers right. We need to do is we need to
subtract out the situation, the probability of
getting the two numbers, getting the letter and
the two numbers right and we already know what that is, it's one in 2600. It's one and 26 minus one and 2600. The reason why I have to
subtract out at this 2600 is he has one in 26 chance
of getting this letter right. That includes the scenario
where he gets everything right but the small prize is only
where you get the letter and one or none of these. If you get both of these then you're at the grand prize case. You essentially have to
subtract out the probability that you won the grand prize, if you got all three of them to figure out the probability
of the small prize. Now what's the probability
of essentially losing? The probability of neither. Well it's just kind of
that's everything else. It would be one minus these probabilities right over here. It would be one minus the probability of the small prize. The probability of the small minus the probability of the grand, these are the possible outcomes so they have to add up to one or a 100%. This is one less probability small minus probability of large or I'll say grand prize. Let's fill this in. The probability of the small
one, this right over ... I'm using that red too much. This right over here is one in 26 minus one in 2600 and then this right over
here is one minus the small which is one in 26 minus one in 2600 minus one in 2600. This simplifies to let's see, this is one minus one over 26 plus one in 2600 plus
or minus one in 2600. These cancel and you're left
with one minus one in 26. Why does this make sense? The way you get nothing is
if you get the letter wrong. You have a one in 26 chance
of getting the letter right and then you're going to be
in one of these two categories or you have a one minus one 26 which is equal to 25 of 26. You have a 25 26 chance of
getting the letter wrong in which case you get nothing, in which case you completely lose. Let's just get our calculator
out and calculate this and we'll round to the nearest penny here. Let's see, it is going to be one 2600. One divided by 2600 times let's see, 10,405 minus five is going to be 10,400, times 10,400, that's your net profit when you win the grand prize and then you're going to
have plus one divided by 26 minus one divided by 2600 times your net profit for the small price is a 100 minus five which is 95, and then finally plus 25 26. 25 divided by 26, actually I'll
put parenthesis around here just to make it consistent. 25 divided by 26 times that net payoff. When you got nothing, well
you have to pay out $5 and you got nothing in
return, times negative five. Actually I don't know if
it's going to recognize that as times so I'll just
write times negative five and let me delete that and
we deserve a drum roll now. We get a expected net profit of playing as $2.81 if we round up to the nearest penny. This is all going to be equal to $2.81. This is actually a very
unusual lottery game where you have a positive
expected net profit as a player. Usually the purpose on
operating the lottery, the state, or the casino, whoever it is, they're the ones who have
the expected net profit and then the player has
the expected net loss but this actually would
make rational sense to play which is not the case
with most lottery games and if by playing you actually
expect a $2.81 net profit.