There is a local drawing that involves being selected out of an estimated 6000 entries, and then correctly selecting 1 of 3 numbers in order to win. The numbers have are actually cards in a deck that have been selected once per week. At this point only 3 cards remain and in order to win, you must first be selected and also have picked the "correct" card. That correct card is already predetermined (queen of hearts).
My question: What are my odds of winning by purchasing 300 entries (estimated 6000 total entries for this contest) and does it make any statistical difference if I play all my chances on 1 number, or spread 100 entries each across all 3 numbers (300/3)?
Lastly, at what number of entries would I be statically be increasing my odd by a significant percentage?
Thanks!
Best Answer
You first have to win the first drawing. It is not fair to assume there are exactly $6000$ tickets sold if you are going to buy very many. A more reasonable model is that there are $6000$ tickets bought by others. If you buy $n$ tickets, your chance of winning the first draw is $\frac n{n+6000}$. For $300$ tickets, you haven't increased the denominator much and your chance here is just under $5\%$. If you bought $6000$ tickets your chance of winning the first draw would not be $100\%$ (which it would be if there were $6000$ sold) but $50\%$ You can plot $\frac n{n+6000}$ to see how the probability increases.
For the second draw, it doesn't matter whether you pick the same card for all your tickets or spread them around. One particular one of your tickets has won the first draw and you have one chance in three it matches the card. Of course, if you know the correct card (you said it was predetermined), you should buy all your tickets with that.