Lottery Tickets and Gambling

1. Lottery Tickets and Gambling

2. Lottery Spending in the United States • It is estimated that Americans spend $50 billion each year on lottery tickets. • This averages out to $160 for every man, woman and child. • This is over $200 for each adult. • According to one academic study (this # is debated)… • Households earning less than $13,000 a year, spend 9% of income on lottery tickets. • That is almost $1,200 a year. • Chances of winning • 2/3rds of lottery receipts are returned in prize money. • 1/3 is not returned as prize money. • The average person loses 1/3 of their money. • A few people win a lot of money, nearly everyone else loses much more than 1/3. • Lottery Games should not be used for investment purposes!

3. Odds of Winning • The classic lottery would choose 3 balls each with a different digit. • If your $1 ticket matched those digits, you would win $500. • You have a ticket numbered 216 • What is your chance of winning? • It depends….. • There are different ways the balls could be chosen, so let’s evaluate some different scenarios. • Each ball is chosen independently from its own bin of 10 balls • Each ball is chosen from the same bin of 10 balls • Match one ball to win • The numbers 2-1-6 win in any order

4. Probability • To understand the lottery, it is essential to understand probability. • Probability is a measure or estimation of how likely something is to happen or how likely it is that something is true. • Probabilities are typically measured in %. • They range between • 0% - (There is no chance something will happen. It will not happen) • 100% - (The thing absolutely will happen) • Probability is often measured as … number of successes number of possibilities

5. Each ball is chosen independently from its own bin • In this scenario, each digit is chosen “independently”. • Independent – One event does not affect another event. • Imagine 3 bins, each holding 10 balls numbered 0-9. • The lottery commission will chose 1 ball from each bin. • The chance of you getting a 2 on the first ball is 1 out of 10 or 10% • One way to succeed out of 10 possible balls • The chance of you getting a 1 on the 2nd ball is also 10%. • The chance of you getting a 6 on the 3rd ball is also 10%. • All three things need to happen to win, so you multiply the odds • You need a 2 and a 1 and a 6. • You need all 3 things to occur, so your chance of winning is …. • .1 x .1 x .1 = .001 or 0.1% or 1 out of 1000.

6. Each ball is chosen from the same bin of 10 balls • In this scenario there are only 10 balls total. • The odds of the first ball being a 2 are still 1 out of 10 or 10%. • The odds of the 2nd ball being a 1 are 1 out of 9 (There are only 9 balls left). • The odds of the 3rd ball being a 6 are 1 out of 8 (There are only 8 balls left). • In this scenario the odds of winning are 1/10 x 1/9 x 1/8 • = 1/720 = .00139 = 0.139% • This is often referred to in probability as “probability without replacement” • In this situation, each ball is not “independent”. The odds of each ball getting chosen change each time another ball is pulled.

7. And vs Or • Let’s pretend that in a different lottery game you could win your dollar back if any of your three numbers (2,1,6) matches the first number they pick. • Meaning you win if they choose a 2 or a 1 or a 6. • This is known as a “union”. • Each of your numbers has a 10% chance. • In a “union” (and “or” situation), you can add the probabilities together. • Your chances are 10%+10%+10% = 30%.

8. More Unions • In the previous situation only one ball was being chosen and you had 3 ways to win. • Imagine that the lottery chooses 3 balls. You win if any of the 3 matches any of the 3 on your ticket (2 or 1 or 6). • On the first ball, your odds of winning are 30%. 3 ways to win/ 10 possibilities. • The odds of the 2nd ball winning are also 30%. It seems the odds of either happening would be 30%+30% = 60%, but they aren’t • You win if you match a ball from the first bin. You win if you match a ball from the 2nd bin, but you still only won 1 time if you match a ball from both bins. • To calculate this scenario correctly, you add the probability of winning from Bin A or Bin B then subtract the probability of both happening. • .3 + .3 – (.3x.3) = .51 or 51% • From the 3rd bin your odds of getting a winning ball are again 30%. • .3 + .51 – (.51 x .3) = .657 or 65.7% • P (A or B) = P(A) + P(B) – P(A and B)

9. Assumption of Independence • If there are 2 events happening, they are “independent” if one event happening does not affect the odds of the other event happening. • Example: • We know in a deck of cards the odds of getting a particular suit are 1/13 • We know the odds of getting a specific numbered card are 1/4 • If you are dealt 2 cards …. • The odds of getting a club and then a 7 are 1/4 x 1/13 = 1/52 • The odds of getting a club and then another club are 1/4x 12/51= 12/204 • In the first scenario the individual probabilities are multiplied because those are independent events (getting a club does not change the odds of getting a 7) • In the 2nd scenario the odds of getting a club the 2nd time have changed.

10. Assumption of Independence - Unions • Imagine our lottery game with 10 balls numbered 0-9. • Also imagine that the balls 0, 2 and 4 are red and the rest are white • If we pull one ball what are the odds of getting a getting an even and red ball. • We can simply count and we know there are 3 even, red balls. 30% is correct. • If we try our formula ½ are even and 3/10 are red. • ½ x 3/10 =3/20 • The formula doesn’t work because the 2 events are not independent • If we imagine pulling 2 balls with replacement we create independent events and the odds would be 3/20 • http://www.ted.com/talks/peter_donnelly_shows_how_stats_fool_juries.html

11. Expected Value • Expected Value – The weighted average of all possible values of X • Expected Value – The theoretical long term average value after repeating a trial an infinite number of times. • Expected Value = (Probability of gain or loss) x (Value of gain or loss) • Example – Flip a coin, if it is heads you win $10, if it is a tail you lose $8 • Expected Value = ($10 + -$8)/ 2 = $1 • Expected Value = .5 x $10 + .5 x (-$8) = $1 • http://www.ted.com/talks/dan_gilbert_researches_happiness.html

12. Expected Value • Expected Value = probability of success x payout • The expected value is just an average. It may not even be possible to win that exact amount of money. • Example – Six people each pay $1 and pick a number between 1-6. • A six sided die is rolled and whoever has that number wins all the money. • The expected return is … 1/6 x $6 = $1 • None of the players will return home with $1. • 5 players will have nothing and 1 player will have $6, but the expected return is still $1. It is the average return expected.

13. Lottery Expected Value • In our first example – • 3 balls chosen independently, must match each one in order • The payout is $500 • We’ve already determined the odds of winning are 1 out of 1000 • Expected value = .001 x $500 = $0.50 • If a lottery ticket only cost $0.50 or less then it isn’t a bad bet. However since lottery tickets of this kind cost $1, you have an expected loss of $0.50 each time you play. • Think of it this way… • If you were to buy 1,000,000 lottery tickets • You would likely win 1m x .001 = 1000 times • Each win is $500, so you would spend $1m to get back $500,000. • Expected Return = .001 x $500 x 1 million = $500,000 • You spent $1m to get back $500,000. You lost $500,000 which is $0.50 per ticket.

14. 2,1,6 in any order • In this scenario, you have a bin with 10 balls. 3 balls will be chosen without replacement. You will win if you choose 2, 1, 6 in any order. • Let’s think about how many ways there are to win. • You could get a 2, then a 1, then a 6. • You could get a 2, then a 6, then a 1. • You could get a 1, then a 2, then a 6. • You could get a 1, then a 6, then a 2. • You could get a 6, then a 1, then a 2. • You could get a 6, then a 2, then a 1. • There are 6 ways to win. • How many total possibilities are there when choosing 3 balls out of 10? • How many ways are there to choose 3 balls from a group of 10? • (0, 1, 2), (0, 1, 3), (0, 1, 4) …... (9, 8, 5), (9, 8, 6), (9, 8, 7) • If you actually show all of these possibilities you will have 720 ways to choose 3 balls from a bin of 10 balls.

15. 2,1,6 in any order • This is known as a permutation. • Permutation – A rearrangement of a certain number of items that are chosen without replacement. • To find the value of a permutation Pkn= n! / (n-k)! • n is the total number of items in a population • k is the number of items selected • In our example of selecting 3 items from 10 total • P310 = 10! / (10-3)! • P310 = (10x9x8x7…..x1) / (7x6x5x4….x1) • P310 = 10 x 9 x 8 = 720 • There were 6 = (3!) ways to win, so the odds of winning are 6/720 or 1/120

16. Probability Review Probability = number of successes number of possibilities • P (A and B) = P(A) x P(B) • P (A or B) = P(A) + P(B) – P(A and B) • Probability Tree • Expected Value = probability of success x payout