Problem Statement
The CA Super Lotto is a sequence of five randomly-drawn numbers between 1 and 47, plus a mega number randomly drawn between 1 and 27. The first five numbers do not repeat, but the sixth number may repeat. To win, you have to match all 6 numbers. The first five numbers don’t need to be in any particular order, but the mega number must be the sixth draw. Each lottery ticket costs $1. If you win, you are awarded $8,000,000 To solve the CA Super Lotto problem, we must find the total number of combinations, the probability of winning, and the expected winnings.
The CA Super Lotto is a sequence of five randomly-drawn numbers between 1 and 47, plus a mega number randomly drawn between 1 and 27. The first five numbers do not repeat, but the sixth number may repeat. To win, you have to match all 6 numbers. The first five numbers don’t need to be in any particular order, but the mega number must be the sixth draw. Each lottery ticket costs $1. If you win, you are awarded $8,000,000 To solve the CA Super Lotto problem, we must find the total number of combinations, the probability of winning, and the expected winnings.
Process and Solution
My initial attempt to solve the problem was loaded with bumps and mistakes. One part of the process stands out to me the most. I was having a hard time grasping why the numerator of the probability was a five instead of a one. Suddenly, it "clicked", and I immediately understood. It was like an epiphany. |
1) To find the total number of combinations, let’s look at the sample space for the first draw:
47
The sample space for the first draw is 47 because the first draw is made out of 47 numbers. Now, let’s move on to the second draw:
46
The sample space of the second draw has decreased to 46. But why? The reason is because one number has already been drawn from the 47 original, leaving 46 numbers to be drawn from in the second round. Now we will check the sample space of the third draw:
45
Again, the sample space has been reduced by one because yet another number has been drawn. This pattern continues for the next two draws. Finally, let us examine the sample space of the mega number draw:
27
The last number is drawn out of 27. Therefore, the last sample space is 27.
Now that we have the denominators, or for each draw, we can find the total denominator. To do this, we’ll multiply all of them together, like so:
47 x 46 x 45 x 44 x 43 x 27 = 4,969,962,360
The total number of combinations or “sample space” of winning the CA Super Lotto is 4,969,962,360.
47
The sample space for the first draw is 47 because the first draw is made out of 47 numbers. Now, let’s move on to the second draw:
46
The sample space of the second draw has decreased to 46. But why? The reason is because one number has already been drawn from the 47 original, leaving 46 numbers to be drawn from in the second round. Now we will check the sample space of the third draw:
45
Again, the sample space has been reduced by one because yet another number has been drawn. This pattern continues for the next two draws. Finally, let us examine the sample space of the mega number draw:
27
The last number is drawn out of 27. Therefore, the last sample space is 27.
Now that we have the denominators, or for each draw, we can find the total denominator. To do this, we’ll multiply all of them together, like so:
47 x 46 x 45 x 44 x 43 x 27 = 4,969,962,360
The total number of combinations or “sample space” of winning the CA Super Lotto is 4,969,962,360.
2) Next, we will find the total probability of winning.
For the first draw, the probability is:
5/47
The probability of winning the first draw is 5/47 because the first number can match in any of the first five draws, and the first draw is made from 47 numbers. Now we will check the probability of winning the second draw:
4/46
You may notice that the numerator has decreased by one. The reason for this is because one number has already been drawn, so there are now only four chances for the number to match. Here is probability of winning the third draw:
3/45
This pattern continues for the next two draws. However, the last draw is a bit different:
1/27
The probability of winning the final draw is 1/27 because the mega number must also be the sixth draw. This leaves only one opportunity for the number to match.
Now that we have the probabilities of winning each draw, we can find the probability of winning the entire lottery. This is how it’s done:
5/47 x 4/46 x 3/45 x 2/44 x 1/43 x 1/27= 120/4,969,962,360
The probability of winning the CA Super Lotto is 120/4,969,962,360
For the first draw, the probability is:
5/47
The probability of winning the first draw is 5/47 because the first number can match in any of the first five draws, and the first draw is made from 47 numbers. Now we will check the probability of winning the second draw:
4/46
You may notice that the numerator has decreased by one. The reason for this is because one number has already been drawn, so there are now only four chances for the number to match. Here is probability of winning the third draw:
3/45
This pattern continues for the next two draws. However, the last draw is a bit different:
1/27
The probability of winning the final draw is 1/27 because the mega number must also be the sixth draw. This leaves only one opportunity for the number to match.
Now that we have the probabilities of winning each draw, we can find the probability of winning the entire lottery. This is how it’s done:
5/47 x 4/46 x 3/45 x 2/44 x 1/43 x 1/27= 120/4,969,962,360
The probability of winning the CA Super Lotto is 120/4,969,962,360
3) Finally, we must determine the expected winnings of playing the CA Super Lotto.
Expected winnings can be calculated using the following equation:
E(x) = i(summation)n = xipi+xnpn
This can be restated as:
Expected winnings = (value of event)(probability of event)+(value of alternate event)(probability of alternate event)
In this situation, the first event is winning the lottery. The alternate event is losing the lottery. Given these two things, here is how the equation is written:
E(x) = $8,000,000(120/4,969,962,360)$ -1(4,969,962,240/4,969,962,360)
As you can see, the value applied to each event is the amount of money gained. The value of the first event (winning the lottery) is $8,000,000 because you gain $8,000,00 if you win the lottery. The value of the second event (losing the lottery) is -1 because you gain nothing and a lottery ticket costs $1. The fractions in the parenthesis are the probabilities of each event.
E(x) = $8,000,000(120/4,969,962,360) $ -1(4,969,962,240/4,969,962,360)
Now let us simplify the equation by multiplying the value and probability of each event:
E(x) = $960,000,000/4,969,962,360 $-4,969,962,240/4,969,962,360
To further simplify, we’ll turn those fractions into decimals:
E(x) = $0.193160416611284 -$0.999999975854948
Finally, we need to subtract:
E(x) = $-0.806839559243664
The expected winnings of playing the CA super lotto is $-0.806839559243664. This means that for every dollar you spend on a lottery ticket, you can expect to lose 80.6839559243664% of your money. This ratio can be used to determine the amount of money lost/gained over a long period of time. For example, say you purchased a million lottery tickets, your expected winning would be:
$-0.806839559243664 x $1,000,000 = $-806839.559244
If you spent a million bucks on lottery tickets, you are almost guaranteed to lose $-806839.559244 of that money.
Expected winnings can be calculated using the following equation:
E(x) = i(summation)n = xipi+xnpn
This can be restated as:
Expected winnings = (value of event)(probability of event)+(value of alternate event)(probability of alternate event)
In this situation, the first event is winning the lottery. The alternate event is losing the lottery. Given these two things, here is how the equation is written:
E(x) = $8,000,000(120/4,969,962,360)$ -1(4,969,962,240/4,969,962,360)
As you can see, the value applied to each event is the amount of money gained. The value of the first event (winning the lottery) is $8,000,000 because you gain $8,000,00 if you win the lottery. The value of the second event (losing the lottery) is -1 because you gain nothing and a lottery ticket costs $1. The fractions in the parenthesis are the probabilities of each event.
E(x) = $8,000,000(120/4,969,962,360) $ -1(4,969,962,240/4,969,962,360)
Now let us simplify the equation by multiplying the value and probability of each event:
E(x) = $960,000,000/4,969,962,360 $-4,969,962,240/4,969,962,360
To further simplify, we’ll turn those fractions into decimals:
E(x) = $0.193160416611284 -$0.999999975854948
Finally, we need to subtract:
E(x) = $-0.806839559243664
The expected winnings of playing the CA super lotto is $-0.806839559243664. This means that for every dollar you spend on a lottery ticket, you can expect to lose 80.6839559243664% of your money. This ratio can be used to determine the amount of money lost/gained over a long period of time. For example, say you purchased a million lottery tickets, your expected winning would be:
$-0.806839559243664 x $1,000,000 = $-806839.559244
If you spent a million bucks on lottery tickets, you are almost guaranteed to lose $-806839.559244 of that money.
Problem Evaluation
Like most other math problems, I enjoyed this one when I understood the math. Not grasping a math concept is frustrating to me, but like I said, it is always very satisfying to finally understand.
Self Evaluation
I am satisfied with the DP write-up that I have a made for this math problem. I think that my past write-ups for math were too complicated. This time, I simplified my words to make my writing easier to understand.
Peer Edits
My peer editors were Sydney and Ana. Sydney told me to: "clarify what the 6th number is", "explain more about why the last draw is 27", and "double-check the fractions and the equation in the probability of winning". Ana told me to: "try adding pictures of the expected winnings to make it clearer", "add evaluations", and "add the launch where we played the lottery".
Like most other math problems, I enjoyed this one when I understood the math. Not grasping a math concept is frustrating to me, but like I said, it is always very satisfying to finally understand.
Self Evaluation
I am satisfied with the DP write-up that I have a made for this math problem. I think that my past write-ups for math were too complicated. This time, I simplified my words to make my writing easier to understand.
Peer Edits
My peer editors were Sydney and Ana. Sydney told me to: "clarify what the 6th number is", "explain more about why the last draw is 27", and "double-check the fractions and the equation in the probability of winning". Ana told me to: "try adding pictures of the expected winnings to make it clearer", "add evaluations", and "add the launch where we played the lottery".