For example, a census might say the average family in California has 2.4 children. The word average as it is used in this context has no physical meaning in reality since you cannot have a fraction of a child. It can be interpreted to mean that a typical family in California has either 2 or 3 children.
If you are familiar with basic probability theory and calculating expected values you can continue reading this post. If not I suggest you to read my previous post.
http://mikebibbygrigsby83.blogspot.com/2012/03/playing-mega-millions-lotto-please-read.html
In this post I will talk about what happens to the expected value when you purchase more than one lotto ticket for the Mega Millions jackpot. The results are surprising. Let me start by constructing a probability table for different values of 'n'. The letter 'n' will represent the number of tickets an individual may purchase without regard to the total number tickets purchased by other individuals.
Probability Table
----------------------------------------------------------------
n | Probability of Having a Winning Ticket
----------------------------------------------------------------
1 | 0.00000000569
2 | 0.00000001138
5 | 0.00000002846
10 | 0.00000005691
25 | 0.00000014228
50 | 0.00000028456
100 | 0.00000056911
250 | 0.00000142279
500 | 0.00000284557
1,000 | 0.00000569115
5,000 | 0.00002845573
10,000 | 0.00005691146
50,000 | 0.00028455730
100,000 | 0.00056911460
1,000,000 | 0.00569114597
----------------------------------------------------------------
When I calculated the probability of having a winning ticket for each value of 'n' I made the following assumptions about the tickets:
1.) All tickets have only one set of numbers. This is not the case in reality since you can purchase more than one play (and hence have more than one set of numbers) on a single ticket.
2.) All tickets purchased by an individual are unique. In other words you cannot have two tickets (or three, or four, etc.) where the first 5 numbers and the 6th 'mega' number are the same.
For example, consider the following two tickets with the following 6 numbers:
Both tickets have the same set of numbers and only the order of the first 5 numbers are different. Therefore, both tickets are the same and two or more tickets like this are not allowed for this particular scenario.
-----------------------
|Ticket 1 |
-----------------------
| mega |
|45 12 07 25 39 40 |
-----------------------
-----------------------
|Ticket 2 |
-----------------------
| mega |
|12 39 25 45 07 40 |
-----------------------
This table tells us that when you buy more tickets you increase your chances of winning. You don't need to do all these tedious calculations to figure that out :-P
Even if you buy 1,000 or 10,000 tickets your odds are no better than 1 in 175,000 or 1 in 17,500 respectively. Now, let's look at what happens to your expected net winnings. I assume that each ticket costs $1.
Expected Net Winnings
----------------------------------------------------------------
n | Expected Net Winnings ($)
----------------------------------------------------------------
1 | 2.07
2 | 1.07
5 | -1.93
10 | -6.93
25 | -21.93
50 | -46.93
100 | -96.93
250 | -246.93
500 | -496.93
1,000 | -996.93
5,000 | -4,996.78
10,000 | -9,996.36
50,000 | -49,982.70
100,000 | -99,940.02
1,000,000 | -994,305.78
----------------------------------------------------------------
According the table above your expected net winnings decrease as you buy more tickets. This would be a strong suggestion against buying more tickets even though your chances increase. I wouldn't buy a ton of tickets, because you are more likely to lose a lot of money if none of your tickets have the winning combination!
Let's look at the scenario from the perspective of the total number of plays purchased for the Mega Millions jackpot. In this case I will relax the two assumptions above and add one more assumption.
1.) People can purchase more than one play on a single ticket (i.e. you can have more than one combination of numbers on a single ticket).
2.) The plays themselves do not have to be unique. So it is possible for two or more people to have the winning combination.
3.) The combination of numbers selected for each play are independent of one another (i.e. the combination of numbers for one play does not effect the combination of numbers for another).
For this scenario I am still assuming that there are only two possible outcomes; either someone has a ticket with the winning combination or no one has a ticket with the winning combination. Therefore, someone either wins the jackpot or no one wins anything at all. And each play costs $1.
Given the three assumptions above we will use the binomial probability distribution to construct a table like those above. If you don't know what the binomial probability distribution is google it!
Let us first introduce the following parameters:
N - total number of plays purchased
p - probability of a play matching the winning combination
q - probability of a play not matching the winning combination
p = 1/175,711,536 = 0.00000000569
q = 1 - 1/175,711,536 = 0.9999999943
*Note: the values of 'p' and 'q' do not change and are fixed
Probabilities and Expected Net Winnings for all who Play the Mega Millions Lotto
----------------------------------------------------------------
N | Probabilities | Expected Net Winnings ($)
----------------------------------------------------------------
1 | 0.00000000569 | 2.07
2 | 0.00000001138 | 4.15
5 | 0.00000002846 | 10.37
10 | 0.00000005691 | 20.73
25 | 0.00000014228 | 51.83
50 | 0.00000028456 | 103.66
100 | 0.00000056911 | 207.32
250 | 0.00000142279 | 518.30
500 | 0.00000284557 | 1,036.61
1,000 | 0.00000569113 | 2,073.22
5,000 | 0.00002845573 | 10,366.02
10,000 | 0.00005691146 | 20,731.88
50,000 | 0.00028451682 | 103,653.31
100,000 | 0.00056895268 | 207,291.34
1,000,000 | 0.00567498209 | 2,070,165.31
10,000,000 | 0.05532229244 | 20,427,260.84
100,000,000 | 0.43397362257 | 177,743,118.44
540,000,000 | 0.95372802684 | 490,026,268.99
1,000,000,000 | 0.99662427778 | 534,801,387.78
----------------------------------------------------------------
The probabilities in this table represent the chance that someone has a ticket with the winning combination given that the total number of plays purchased is 'N'. This allows the possibility that more than one person has a winning ticket. The figures in the second column represent the expected net winnings for all who play the lotto. We are not concerned with individual players here! As we saw in the previous table the expected net winnings for an individual become negative once they purchase more than three plays.
The table suggests that as the total number of plays purchased increases the likelihood that someone will win the jackpot. The expected net winnings for all players also increases as more tickets are purchased. Furthermore, the probability that someone has a winning ticket approaches 1.00 and the expected net winnings approaches the jackpot of $540 million. This means that it is very likely that some will win the big jackpot tonight. I hope it's me!
Thanks again for reading part 2 of my post about the Mega Millions Lotto. I spent the entire night on this and my previous blog (probably more time than I should have).
Even though I know a lot of probability theory I should say that I make no guarantees as to the accuracy of the numbers and the methods I used to obtain those numbers. I am only human.
Upon scrutinizing these results if you should find any errors in my calculations or methods, please let me know. I can always revise these blogs or post future blogs with more accurate information.
Upon scrutinizing these results if you should find any errors in my calculations or methods, please let me know. I can always revise these blogs or post future blogs with more accurate information.
Thanks again :-)
In my previous post and in the first few paragraphs of this post I mention the phrases "purchase one ticket", "purchase a single ticket" or "purchase a ticket". This is a little ambiguous.
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