California Super Lotto! Rules are choosing any 5 numbers between 1-47 that do not repeat. Then picking a mega number between 1-27 that can repeat from the initial 5 numbers. Lastly to win you have to match all 5 numbers and the mega number.
We had to answer these 3 questions.....
How many different number combinations are possible for a CA Super Lotto ticket?
What is the probability of winning the CA Super Lotto?
If you match all 6 numbers, you win $8,000,000. It costs $1 to play. What are your expected winnings?
When I first read this problem, before even learning about probability I guessed the probability would be 1 in a million, because that's what you usually always hear for events where your chance is slim. No calculations where made. Little did I know.....
So I first started by thinking about how to find all the total combinations, I knew I had to use some sort of multiplication rule. But what do I multiply? I thought about the first 5 numbers on the ticket. The total numbers were 1- 47. 47 to the fifth power was my first guess. But then I realized numbers couldn't be repeated, and every time a number was written down. our sample space shrunk by one. Therefore, I would multiply 47 x 46 x 45 x 44 x 43. The product, 184,072,680. But that is not counting the mega number. The combos for the mega number is 27 because its a single number with any number between 1-27. We multiplied 27 x 184,072,680 to receive our overall number of probabilities, 4,969,962,360.
Next problem was finding the probability of winning the lotto. So we needed to distinguish what our sample space was and what our "success" numbers were. 47 x 46 x 45 x 44 x 43. These were our denominators. Taking in consideration that the numbers for the lotto could be out of order as long as they're matching to win ,the first numerator would be 5. 5/47 is the chance you get the first number right, and it decreases and decreases each number. the numerators were then 5 x 4 x 3 x 2 x 1. And adding the mega number chance at the end, 1/27. Multiplying all this together we get 120/4,969,962,360. Those are pretty great chances, right? (Not.) Comparing this to my original guess( 1 in a million) I realized I was way off, it was much more slim than that. Converting this to percentage form revealed the chances of winning , that is .000000024%.
Last question, what are the expected winnings. To solve this we used the formula to calculate the expectation of a random value (x) . E(x)=payout if you win(probability of winning)+(-money you bet) (probability of losing). We plugged the numbers in and got 96,000,000/4,969,962,360-4,009,962,240/4,969,962,360. Then subtracted this and calculated the percent. -.80. The expected winnings for each dollar bet are 80 cents.