## What is 32 plus 3?

What is $32+3$? What is $35-32$? This was a question that apparently stumped a clerk in a Macy’s Department Store. My wife recently had to return a wallet back to Macy’s that was purchased for $32. It was early in the morning and the cash drawer did not have singles. So the clerk did not have the exact amount of cash for$32. However, the cash drawer had $30 in cash plus a$5 bill. My wife offered a solution: taking $35 from the clerk and returning$3 to the clerk. The clerk, a woman possibly in her early 30s, did not believe that this is a correct transaction. After my wife explained to her a few times, the clerk reluctantly agreed to this exchange.

This is how the math works in this situation. A customer returned an item back to Macy’s that was worth $32 plus$3 in cash. The total value returned to Macy’s was $35. It was only fair that the clerk gave$35 back to the customer. This is $32+3=35$.

We can also look at this as a repurchase. Macy’s Department Store was to purchase an item ($32 in value) from my wife and pay$35 in cash. Then my wife gave the clerk $3 in change. This is $35-32=3$. This is a telling example. Knowing how to do $32+3=35$ in a math assignment at school, at least for some people, does not mean they know how to use arithemtic to make decisions in real life situations. Some people call this having a number sense. Having a strong number sense is the first step in achieving a larger set of skills commonly called quantitative literacy, which usually includes critical thinking, problem formulation, and written and oral communication. After I heard this story from my wife, I could not help but wanting to give quantitative literacy a plug. Having a better number sense certainly would help this clerk become more effective at her job. Quantitative literacy can also help open doors. There is a whole host of many exciting, highly in demand and well paying careers that require math and science courses in particular and a comfort level with quantitative reasoning in general. Some of these careers only require training at the community college level (an array of jobs in health sciences and computer technology comes to mind). Being proficient quantitatively will give one the confidence and ability to pass the necesary course work to pursue these career options. How to become more proficient quantitatively? There is no magic bullet. It will take time and effort. There are books that one can read and courses to take. I would also recommend to start with one’s own situation. Start paying attention to the aspects in one’s life that involve quantitative concepts (e.g. work, buying of consummer goods, personal finance, household buget). In the case of the clerk at the Macy’s store, learning how to make changes correctly is a good place to start. Learning the concept of interest (with respect to saving as well as borrowing) and learning how credit card works are a couple of worthwhile examples. Working with quantitative examples in one’s personal life is a good first step. Advertisements ## Taking another look at the California Lottery What are the odds of winning the lottery? If the lottery is a 49/6 game (i.e. choosing 6 numbers out of 49 numbers), the odds are 1 in 14 million (one in 13,983,816 to be precise). I would like to show you that for the California Lottery, the odds of winning a jackpot of$1 million or more are 1 in 36 million. Let me show you how I arrive at this estimate.

Statistics about winning tickets are available from the official website of the California Lottery. But one has to do some digging to get the data (I searched at the Lucky Retailer Search). There are 58 counties in California. I simply searched for the 58 counties one by one. Only 28 of the counties had winning tickets. Since the inception of the California Lottery 25 years ago, there were 247 winning tickets as of November 1, 2010.

Here’s the summary information. All of these 247 tickets paid out $1 million or more. The largest jackpot was$110 million. The earliest winning ticket was on 3/21/1987, bought from a retailer in Imperial County. The most recent one was on 10/9/2010, bought from a retailer in Ventura County. The sum of all the winning amounts for these 247 tickets was $4,535,519,264 (about$4.5 billion). Thus each winning ticket prize was, on average, $18,362,426 (about$18 million).

So there are about 250 winning tickets that paid $1 million or more in the 25 years of history of the California Lottery. On average, there were about 10 winners a year. If you do not think that the odds are infinitesimally small, read on. By law, the California Lottery has to pay out at least 50% of the revenue as winning. The total winning amount for these 247 tickets was$4.5 billion. This implies that the $4.5 billion in winnings was paid out from the sales of$9 billion worth of tickets (equivalently 9 billion tickets since the ticket price was $1). So out of 9 billion tickets bought, there were about 250 winners. Thus the odds of winning are 250 in 9 billion or 1 in 0.036 billion (9/250=0.036). The odds of 1 in 36 million followed from the following translation. $\text{1 billion = 1,000,000,000 (1 followed by 9 zeros)}$ $\text{0.036 billion} = 0.036 \times 1,000,000,000=36,000,000 \text{ (36 million)}$ Of course, the California Lottery will never tell you that the odds of winning the big prizes are 1 in 36 million. One has to dig to find the information like I did. In fact, in the same page where I did the digging, they claim that “Since 1985, the California Lottery has distributed more than$27 Billion in winnings (including annuitized) with more than 2,842,467,062 winning tickets sold!

They claim that there were 2,842,467,062 (2.8 billion) winning tickets since 1985. How does this number squared with the 247 tickets that I found? I wrote about this point in a previous post called Shining a light on the California Lottery. Except for 247 tickets, these tickets paid out small prizes (on average just under $10). Their information is correct but can give the impression that there are many millionaires running around (could be as many as half the world’s population)! The usual refrain of many lotto players is that you have to buy a ticket in order to win. Winning is desirable for sure. In the case of mega lotto jackpot such as the games of Mega Million and SuperLottoPlus in the California Lottery, you have to buy millions of tickets before you have a realistic chance of winning (could very well be 36 million tickets). If you treat the game of lottery as a money making opportunity or a way to become an instant millionaire, you better count the cost. Some play the lottery for its entertainment value and the excitement. If you spend a small sum every week buying tickets for the huge jackpots, the entertainment value is about the only benefit you will receive from playing. ## The stealth price increases at the grocery store One particular snack containing almonds, cashews and cranberries that I buy at a local grocery store is now in a reduced size of 14 ounces per pack (previously a 16-ounce pack). I am getting 2 ounces less with the same price. This is a stealth price increase. In this tough economic time, this must be a favorite way for manufacturers to raise prices. Breakfast cereal boxes seem to be getting smaller. It will be an interesting exercise to find out what the percentage price increases are for these items with reduced sizes. For example, what is the percentage price increase for the snack pack that I buy (going from 16-ounce to 14-ounce)? Since the amount I pay at the cash register is the same whether it is 16-ounce or 14-ounce, I must look at the unit price (price per ounce). The old unit price is$0.28 (=$4.49/16) and the new unit price is$0.32 (=$4.49/14). Thus they are charging me 4 cents more per ounce by giving me two ounces less almonds and cashews. The percentage increase in price is 14.29% (0.04/0.28=0.1429). As a percentage, this is a hefty price increase! The following is a general formula for finding percentage price increase. $\displaystyle (1) \ \ \ \ \ \ \ \ \ \ \ \text{Percentage Price Increase}=\frac{\text{Dollar Price Increase}}{\text{Old Dollar Price}}$ There is another way to compute the percentage increase. Observe that 0.32/0.28=1.1429. Thus the ratio of the new price to the old price contains the information of the percentage increase. All we need to do is to subtract 1 from this ratio. We have the following formula, which is equivalent to the above formula. $\displaystyle (2) \ \ \ \ \ \ \ \ \ \ \ \text{Percentage Price Increase}=\frac{\text{New Price}}{\text{Old Price}}-1$ It turns out that in the situation of price increase by reducing weight while charging the same price, calculating the percentage increase does not require the price amount. In our example, the percentage price increase does not depend on the price being$4.49. All we need to do is to compute the ratio of the old weight to the reduced weight. This ratio contains the information of the percentage price increase. Simply subtract 1 from this ratio, we obtain the percentage increase. In our example, 16/14=1.1429.

$\displaystyle (3) \ \ \ \ \ \ \ \ \ \ \ \text{Percentage Price Increase}=\frac{\text{Old Weight}}{\text{Reduced Weight}}-1$

Formulas (1) and (2) are general formulas that work for all situations. Formula (3) can only be used for calculating percentage increases as a result of reducing the weight of the product (while charging the same dollar amount). Formula (3) works because the price of the product (e.g. $4.49 in our example) is canceled out in the derivation of Formula (2). Formula (3) can be handy for displaying scenarios (e.g. a manufacturer may want to determine the percentage increases for different sizes of reduction in weight). The following matrix displays the percentage increases for various sizes of reduction of a 16-ounces (or any other unit of measurement for weight). Reducing a 16-unit package: $\displaystyle \begin{pmatrix} \text{Weight Reduction}&\text{Percentage Increase} \\{\text{ }}&\text{ }&\text{ } \\\text{1}&\text{6.67 percent} \\\text{2}&\text{14.29 percent} \\\text{3}&\text{23.08 percent} \\\text{4}&\text{33.33 percent} \\\text{5}&\text{45.45 percent} \\\text{6}&\text{60.00 percent} \\\text{7}&\text{77.78 percent} \\\text{8}&\text{100.00 percent}\end{pmatrix}$ Obviously, the more the reduction in weight, the higher the percentage increase in price. I can see that a company could very well look at a matrix such as the one above to see what they can get away with. Reducing a 16-ounce package by one ounce only gives a 6.67% increase, probably not high enough from a manufacturer’s perspective. By reducing too much, the increase will no longer be stealth. So they probably would like to target a reduction that is not so conspicuous and yet one that can achieve a significant percent increase. In the case of our example of the snack packs of almonds and cashews, reducing the package by 2 ounces may not be too noticeable while acheiving a double digit percent price increase. That was a winner for the manufacturer! If a manufacturer is bold enough to reduce the 16-ounce package to an 8-ounce package while charging the same price, the price increase would be 100%! You would to pay twice as much to get the same 16-ounce of the product! I am sure no manufacturer in their right mind would go for this. One final observation. For a fixed amount of reduction in weight (say 2 ounces), the larger the original weight, the smaller the percentage price increase. For example, if the original weight is 24 ounces, a reduction of 2 ounces would only acheive a 9.09% increase. See the following matrix. Each percentage is calculated using Formula (3). For example, for a reduction of 2 ounces from a 24-ounce package, the percentage increase in price is 24/22 – 1 = 0.0909. Reducing a 24-unit package: $\displaystyle \begin{pmatrix} \text{Weight Reduction}&\text{Percentage Increase} \\{\text{ }}&\text{ }&\text{ } \\\text{1}&\text{4.35 percent} \\\text{2}&\text{9.09 percent} \\\text{3}&\text{14.29 percent} \\\text{4}&\text{20.00 percent} \\\text{5}&\text{26.32 percent} \\\text{6}&\text{33.33 percent} \\\text{7}&\text{41.18 percent} \\\text{8}&\text{50.00 percent}\end{pmatrix}$ ## Shining a light on the California Lottery In the back of my mind, winning the lottery means becoming a millionaire (or better). Thus I find the following statements found in the website of the California Lottery very interesting. Since 1985, the California Lottery has distributed more than$27 Billion in winnings (including annuitized) with more than 2,842,467,062 winning tickets sold!

The amount of $27 billion is a lot of money. This amount ought to be reassuring to anyone dreaming of winning big (think sitting in a beach chair holding a martini in a posh beach resort in the Caribbean). The number of winning tickets 2,842,467,062 is a huge number too. If I keep playing, could I be joining this huge legion of winners? I noticed something else. Why is the total amount of winnings stated in a nice round number while the total number of winning tickets is not? Note that both figures are not meant to be exact amounts (more than$27 billion and more than 2,842,467,062 winning tickets).

Why not express the number of winning tickets in a nice round number too? Could it be that this is intentional? Could it be that the display of the number 2,842,467,062 is designed to be fantasy inducing? I do not know for sure. But I have my suspicion. Understanding how to read large numbers will clear things up.

One million is 1,000,000 (one followed by 6 zeros). Note that one million is 1000 times of 1000. Putting it another way, if you receive $1000 from each of 1000 people, you become a millionaire. One billion is 1000 times of one million (1,000,000,000 or one followed by 9 zeros). If you have$1 billion in wealth and you only spend $1 million a year, it will take you 1000 years to deplete your wealth! Of course, in this scenario, we are not taking the time value of money into account (but that is another story). So the number of winning tickets for the California Lottery since 1985 is 2,842,467,062 (about 2.8 billion tickets). Interestingly, this means that the average winning amount per ticket is slightly under$10! This implies that most of the 2.8 billion tickets are for small prizes (way smaller than $1 million). So how many lottery prizes of$1 million or more were won by players in the California Lottery in its 25-year existence? Fortunately, the data are available in the official website of the California Lottery, just that they are not conveniently summarized. I had to search for them county by county (there are 58 counties in California).

I searched the winning tickets by county and I found a total of 247 winning tickets, all in the amount of $1 million or more. These 247 tickets amount to$4,535,519,264 (or $4.5 billion). So more than$22.5 billion (=27-4.5) in winnings are for smaller prizes (e.g. a few hundreds to tens of thousands in dollars).

Out of 25 years of history in the California Lottery, there are only about 250 winning tickets with $1 million or more in winning. On average, there are about 10 such winning tickets a year. So winning a small prize may have good odds (about 2.8 billion instances of small winning so far). But winning a huge jackpot in the California Lottery, one that you normally think of as setting you up for life, had only happened 250 times so far. The California Lottery is in the business of selling dreams. It seems that fuzzy numbers help keep dreams alive. Interestingly, a large number such as 2,842,467,062 was transformed into a fuzzy number by not rounding it. Even if I did not dig up numbers from the official website, I can still get a sense that there is only a small number of winning tickets worth$1 million (or more). We can compare 2,842,467,062 with the sizes of the population in California, the United States and China.

The number 2,842,467,062 is almost 77 times the population of California (36.9 million in 2009), and is over 9 times the population of the United States (307 million in 2009). The population of China is 1.3 billion (in mid 2008). The number 2,842,467,062 is over twice the population of China.

Imagine that the number of millionaires created as a result of playing the California Lottery is twice the China population! If true, California would truly be a “Golden State”!

What can one get from buying a $1 lottery ticket? Not sure what one can get other than a chance to fantasize (a form of cheap entertainment I suppose). The fact speaks for itself. Of the tens of billions of tickets bought in the 25-year history of the California Lottery, there were only 247 winning tickets that paid out$1 million or more.

## The lottery of drunk driving fatality

Several years ago, I read of a news account of a mother driving with her infant son in a minivan. A drunk driver was driving on the same freeway but in the wrong direction and collided with the minivan. The mother died upon impact. Amazingly the infant survived. News accounts like this one are always heart wrenching.

I always think of dying from a crash involving an alcohol-impaired driver is a lottery. It is a negative lottery for sure since no one would want to win it. Los Angeles Angels pitcher Nick Adenhart and two of his friends were the unfortunate “winners” of this negative lottery in 2009.

What are the odds of winning this negative lottery? How often is this negative lottery “won”? Let’s compare the number of drunk driving deaths with the number of winning Lotto tickets that pay out $1 million or more. The comparison is done on an annual basis and as a rate per certain number of minutes. It turns out that winning this negative lottery is far more frequent than winning the Lotto. Driving under the influence of alcohol is a serious problem. The numerical data presented here bear that out. Drivers are considered to be alcohol-impaired when their blood alcohol concentration (BAC) is 0.08 grams per deciliter (g/dL) or higher. Thus any fatality occurring in a crash involving a driver with a BAC of 0.08 or higher is considered to be an alcohol-impaired driving fatality. According to a report from the Center for Disease Control and Prevention (CDC), one alcohol-impaired driving fatality occurs every 45 minutes (see note 1 at the end of the post). This rate of occurrence is derived from the fact that in 2008, 11,773 people were killed in alcohol-impaired driving crashes. What is the total number of Lotto winners in a year? Here, we only focus on the winning prizes of$1 million or more. Would the number of Lotto winners approaches 11,773 in a year? I do not think so. I hunted for data from the website of the California Lotto, I found that as of today’s date, there are only 247 winning tickets that pay out $1 million or more since the inception of the California state lotto in 1985. So in 25 years, there are only about 250 winners in the state of California. Thus on average there are only about 10 “million dollar plus” Lotto winners a year in California. Multiplying across the 50 states in the United States, the total number of “million dollar plus” Lotto winners should be no more than 500 in a year. This should be a pretty conservative estimate since California is the largest state in the country (with the largest population). Not all 50 states in the country have Lotto. Some states do not have their own Lotto and are part of a multi-state Lotto. So 500 is a good upper bound on the total number of “million dollar plus” Lotto winners in a year. With 500 winners a year, there is one “million dollar plus” Lotto winner every 1050 minutes (see note 2). Or one winner in every 17.5 hours (see note 2). Thus alcohol-impaired driving deaths occur at least 23 times more frequently (one death per 45 minutes vs. one per at least 1050 minutes). Of the 11,773 alcohol-impaired driving deaths in 2008, how many of them were the drunk drivers? According to a report from National Highway Traffic Safety Administration (NHTSA), an agency within the Department of Transportation, we have the following breakdown of the drunk driving fatality in 2008. $\displaystyle \begin{pmatrix} \text{Role}&\text{Number}&\text{Percent of Total} \\{\text{ }}&\text{ }&\text{ } \\\text{Driver With BAC=0.08+}&8,027&\text{68 percent} \\\text{Passenger Riding w/Driver With BAC=0.08+}&1,875&\text{16 percent} \\\text{Occupants of Other vehicles}&1,179&\text{10 percent} \\\text{Nonoccupants}&692&\text{6 percent} \\{\text{ }}&\text{ }&\text{ } \\{\text{Total Fatalities}}&11,773&\text{100 percent}\end{pmatrix}$ Most of the deaths were self-inflicted (68%). According to the same report, 216 of the 1,875 deaths in the above table were children. The group of 2,087 (=216+1,179+692) were truly victims and their deaths were senseless. Angels pictcher Nick Adenhart and the mother mentioned at the beginning belong to this group. They did not do anything wrong and were just in the wrong place at the wrong time. If we just focus on this group of children riding with the drunk drivers and the occupants of other vehicles and nonoccupants, the rate of occurence is still pretty high (one fatality every 252 minutes; see note 3). That comes up to be about one fatality every 4.2 hours (see note 3). Focusing on this group alone, the rate of drunk driving fatalities is over 4 times higher than winning the Lotto (one death per 252 minutes vs. one winner per at least 1050 minutes). Eliminating the category of fatality represented by these 2,087 senseless deaths would drive down the rate of fatality and would go a long way to address the problem of drunk driving. Though this would be only modest improvement, it would be a step in the right direction. ************* Note 1 To derive the rate of one death per 45 minutes, we need to calculate the total number of minutes in a year. There are 365 x 24 x 60 = 525,600 minutes in a year. Then divide 525,600 by 11,773 to obtain 44.64 minutes = 45 minuties. We can get a perspective of this calculation by looking at an example of taking an exam. For example, if you have two hours (120 minutes) to take an exam and the exam has 10 problems, then on average you have 12 minutes to work one problem. Thus if you can work one problem per 12 minutes, you can expect to finish the exam in the allotted time. Back to the alcohol impaired driving situation, there are 525,600 minutes in a year and there are 11,773 events. Thus on average there are 45 minutes allotted for each event. $\displaystyle \frac{365 \times 24 \times 60}{11,773}=44.64 = 45 \text{ minutes}$ Note 2 $\displaystyle \frac{365 \times 24 \times 60}{500}=1051.2 = 1050 \text{ minutes}$ $\displaystyle \frac{1050}{60}= 17.5 \text{ hours}$ Note 3 $\displaystyle \frac{365 \times 24 \times 60}{2087}= 252 \text{ minutes}$ $\displaystyle \frac{252}{60}= 4.2 \text{ hours}$ ## An American in Paris We in the United States still use the English system of measurements while most of the rest of the world uses the metric system. For example, we use pounds and ounces in food packaging. Our motoring experience is measured in miles per gallon (MPG) and miles per hour (MPH). The distances shown in road signs are in miles. Suppose you are an American driving on a French highway toward Paris and you see a road indicating a distance of 270 kilometers to Paris. How many miles are in 270 kilometers? From my road race running experience in the past, I learned that a 5K road race is 3.1 miles long. This information provides a quick and dirty way of making the conversion from kilometers to miles and vice versa. To convert from kilometers to miles, the key is to determine the number of multiples of five in the distance with kilometer as unit. For example, 10K would be 3.1 x 2 = 6.2 miles, 15K would be 3.1 x 3 = 9.3 miles and so on. In other words, multiple of 5K would be the same multiple of 3.1 in miles. The distance of 270 kilometers is 54 times of 5K. Thus 270K = 54 x 3.1 = 167.4 miles. The above is a speed limit sign from Australia. Note that 80 = 16 x 5. Thus 80 K = 16 x 3.1 = 49.6 miles. So the speed limit of 80 K/h is about 50 MPH. To convert from miles to kilometers, we do the opposite, that is, determine the number of multiples of 3.1 (or 3) in the distance with mile as unit. For example, the driving distance from Los Angeles to New York City is about 2800 miles (903 multiples of 3.1). Thus 2800 miles is approximately 903 x 5 = 4515 kilometers. For an even quicker estimate, let’s say the distance from LA to New York is 3000 miles. Then the distance in kilometers would be roughly 5000 kilometers. The same conversion may be done more formally using the following conversion factors. 1 Km = 0.621371192 miles 1 mile = 1.609344 km I would like to point out that for everyday purpose, the quick and dirty way of 5 K = 3.1 miles would work just fine. From 5 K = 3.1 miles, 1 K = 3.1/5 = 0.62 miles. Likewise, 1 mile = 5/3.1 = 1.62 K. These are quite close to the formal conversion factors indicated above. Unless you are an engineer building a highway, the approximation of 5 K = 3.1 miles would work just fine. In the United States, sometimes miles and kilometers are shown together. In some cases, the side-by-side information can actually clue you in on the conversion factors. The following is a warning sign in Hawaii. The sign indicates that 0.5 miles = 0.8 kilometers. Thus 1 mile = 1.6 kilometers. The following sign indicate that 100 kilometers = 62 miles. Thus, 1 kilometer = 0.62 mile. On rare occasions, road signs in the United States are kilometer only. See the following sign from Vermont on I-89. ## How big is the median household income? The U.S. Census Bureau announced on September 16, 2010 that real median household income in the United States in 2009 was$49,777. How big is this amount relative to, say, the income from earning the minimum wage? The current Federal minimum hourly wage is $7.25. But states are allowed to set their own minimum wage rates. The hourly minimum wage among the 50 states currently ranges from$5.15 to $8.55. Obviously, the median household income is more than the annual income from full time employment paying the minimum hourly wage. By how much? This post presents a back of the envelop calculation that does not require a calculator or math formula. For the interested reader, the precise calculation is also presented for comparison. The median household income of$49,777 as reported above is essentially $50,000. Suppose someone is making$1,000 a week. This will add up to $52,000 a year (since there are 52 weeks in a year). Assuming working 5 days a week and 8 hours each day, this means that the person makes about$200 each day ($1,000 divided by 5) and about$25 per hour ($200 divided by 8). So in this back of the envelop calculation, the annual median household income is equivalent to an hourly rate of about$25 per hour. Thus the median household income is about 3 to 5 times the income from full time employment paying the minimum wage.

The annual income of $52,000 is more than the median household income by about$2,000. So the estimate of $25 per hour is an overestimate of the hourly rate equivalent to the annual income of$49,777. The precise hourly wage is:

$\displaystyle \frac{49777}{52 \times 5 \times 8}=23.93$

Thus the estimate of $25 overestimates by about$1 per hour. But it still gives the essential information that the median household income is 3 to 5 times more than the minimum wage income (depending on the state). For a rough comparison, this quick and dirty calculation works quite well.