what advantage is gained by buying multiple squares in a super bowl board

Question

If people were to play a gambling game, and there were 100 randomly numbered squares on a board.  All of these squares cost exactly 1$.  Then 1 number would be randomly drawn, and whoever had the matching number would win the entire 100$ prize.  If there is no limit on the number of squares that can be purchased, what advantage is gained by purchasing multiple squares.

 

The probability of getting your number drawn would be exactly related to the number of squares owned.  2 squares = 2%  (3=3%, 4=4%, etc)  However, you also had to pay exactly 1$ more for purchasing the 2nd square (or more).  

 

It seems like every time you purchase a square you are increasing your risk by the exact same ratio.  

 

Any help would be appreciated.

Thanks!

 

 

Answer 1

EV = [P(Winning) * $100] + [P(Losing) * Amount Spent]

 

Let x be the number of tickets bought

 

[(x/100) * $100] + [(100-x)/100 * -x]

x + [(100/100) - (x/100)]*-x

x + [1-(x/100)]*-x

x - x + x^2/100

x^2/100

 

I went ahead and graphed the results of each amount of tickets bought as a visual:

https://s15.postimg.org/mfb7pdeiz/Tickets_Bought_Plot.png

 

As you can see, there is actually an exponential relationship between tickets bought and expected value.  As you buy more tickets, your EV is decreasing proportional to the amount of money you are spending.  So while your odds of winnings are increasing, your downside risk is increasing at a higher rate (this is because your upside is always $100; it doesn't increase relative to the extra risk you are incurring).

 

So while your instinct was correct that buying more tickets is not optimal, it's actually worse than you thought.