## A Periodic Look at the California LotteryAugust 30, 2011

Posted by Dan Ma in Commentary, Lottery, Number sense, Risk assessment.
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What are the odds of winning the California Lottery? I am talking about the winning of $1 million or more (the kind of winning that is a game changer in one’s personal life). How often are these million-dollar tickets won? Ten months ago I estimated that the odds of winning$1 million or more in the California Lottery were one in 36 million (see Taking another look at the California Lottery). The data were based on data from California Lottery that I obtained in November 2010 (see Shining a light on the California Lottery). Nothing happened in the last ten months indicates that the odds of winning has fundamentally changed.

## Governor Brown wants to take your cell phoneJanuary 12, 2011

Posted by Dan Ma in Commentary, Number sense.
Tags: , , , ,

No, this is not a government seizure of private cell phones. Jerry Brown wants your cell phone only if it was issued by the California state government. Even then, the chance of it having to turn it in is only 50%. The newly installed Governor Brown is only proposing to take away government paid cell phones from certain California state employees in an effort to save money. The potential saving is to the tune of $20 million. Trying to close a budget gap in the California state government that is expected to be about$28 billion, the governor needs to find money anywhere he can.

On the way to work this morning, I heard a report on the radio about the proposal from Governor Brown (here’s one link). The gist of the report is:

1. Governor Brown does not want all the cell phones back. He just wants half of them back. The total number of phones to be turned in by June 1: about 48,000.
2. The state government currently foots the cell phone bills for about 40% of its workforce. The total number of state employees with free cell phones: about 90,000.
3. The state government is currently paying on average $36 a month per cell phone. 4. Governor Brown’s proposed cell phone reduction order should save the state$20 million.
5. Even with the proposed saving, one fifth of the state workers will still use state-funded cell phones.

After I heard this report on the radio, I jotted down the nummbers and I found that these numbers are internally consistent and for the most part accurate. The numbers seem to hang together quite well. Definitely there is no glaring errors. In this one instance of budget cutting at least, Governor Brown and his staff got the numbers right.

Here’s how I looked at these numbers. I am going to discuss each of the points listed above.

1. Currently there are about 96,000 state-funded cell phones in the hands of California state givernment employees. Half of these phones are to be turned in by June 1. That leaves 48,000 cell phones still being in the hands of state employees.
2. About 90,000 state employees have government cell phones. The count of 90,000 is about 40% of the total state workforce. This means there are currently about 225,000 state employees (see note 1 below).
3. The average monthly bill per cell phone is $36, making the average annual bill per cell phone$432.
4. The total annual expense for the 48,000 cell phones being cut would be $20,736,000 (see note 2). This amount is slightly over$20 million and is thus in line with the $20 million being reported. 5. Even with the proposed saving, one fifth of the state workers will still use state-funded cell phones. One fifth of 225,000 (from point 2) is 45,000. From point 1, about 48,000 government paid cells phones will still be used by state employees. Though the 45,000 is less than 48,000 by 3,000, the difference is not large. The only external number that I had to find is the total number of state employees in California. The number I obtained from the California State Controller’s Office is 237,576 (as of October 2010). The estimate obtained from the radio report is 225,000 (point 2 above). Overall, the numbers hang together well. The proposed cut makes sense. According to Governor Brown, some agency heads and managers need to be in touch with employees 24/7. But most state employees do not. However, not the entire amount of proposed saving will be realized this year since some of the cell phones may be under contract with cell carriers. I hope the savings will work out as planned. Note 1 To find 40% of 225,000, we multiply 225,000 by 0.40. $225000 \times 0.40=90000$ But in the radio report, the number 90,000 is given instead. To derive 225,000, we need to do the opposite, i.e. divide 90,000 by 0.40. $\displaystyle \frac{90000}{0.40}=225,000$ Note 2 $432 \times 48000=20,736,000$. ## Hope there will be no lottery winners this New Year’s EveDecember 30, 2010 Posted by Dan Ma in Lottery, Number sense, Risk assessment. Tags: , , , , , , , , add a comment According to a report in npr.org called Road Fatalities Dip Thanks To Safer Cars, Economy, an array of factors are making the road safer. According to a study by the Department of Transportation, the overall number of fatality on American roads has dropped dramatically, fallen by over 20% in the last few years. Two likely reasons for this dramatic drop are safer cars and a slower economy. However, even with the over 20% drop in fatality on the road, there is still one death every 15 minutes on the road. I always think of dying from a crash involving a drunk driver is a lottery. It is a negative lottery for sure since no one would want to win it. In a previous post (The lottery of drunk driving fatality), I discussed the statistic of one drunk driving fatality every 45 minutes. By comparison, the number of deaths on the roads due to all causes is three times higher than just deaths from drunk driving (in the lottery analogy it is three times more likely to win)! I hope in this holiday season, no one will win this negative lottery. Be safe on the road. Between drinking and driving, only do one of them! Now the quantitative stuff. As reported in Road Fatalities Dip Thanks To Safer Cars, Economy, there were almost 44,000 road-related deaths in 2005. In 2009, there were about 34,000 deaths. This is a 22% decrease. There are two ways to see this. One is to calculate the number of reduction in deaths, which is $44000-34000=10000$. Then divide $10000$ by $44000$. We have: $\displaystyle \frac{10000}{44000}=0.2273$, which is 22.73%. Another way to derive the 22.73% is to calculate the following ratio: $\displaystyle \frac{34000}{44000}=0.7727$ Then subtract one from this ratio and obtain $0.7727-1=-0.2273$, which indicates a 22.73% decrease in road-related deaths. The 2009 figure for the number of road-related deaths is 34,000. This comes out to be one death every 15 minutes. To derive this rate, 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 44,000 to obtain 15.46 minutes. Then round the answer to 15 minutes. 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 calculation at hand, there are 525,600 minutes in a year and there are 34,000 events. Thus on average there are 15 minutes allotted for each event. $\displaystyle \frac{365 \times 24 \times 60}{34000}=15.46=15$ The hope is that the denominator in the above ratio will keep getting smaller in the years to come. From 2005 to 2009, the denominator shrank from 44,000 to 34,000. I have a thought. Supose that in the next 5 years (2009 to 2013), there will be the same percent decrease in the road-related deaths as in the 5-year period from 2005 to 2009. What will be the value of the denominator? In other words, according to the same trend line, what will be the number of road-related deaths in 2013? The answer to the above question is obtained by reducing the 34,000 deaths in 2009 by 22.73%. Try the following: $\displaystyle 34000 \times (1-0.2273) = 34000 \times 0.7727=26271.8$ If the same trend that played out between 2005 and 2009 holds, the projection for 2013 would be about 26,000. Whether this is a realistic projection or not, I do not know. I will leave this to the experts who study traffic fatality. Let’s hope that the improvement will be as least no worse than this projection. ## Putting a value on higher educationDecember 19, 2010 Posted by Dan Ma in Number sense. Tags: , , , , , , add a comment This post is a continuation of the previous post Putting a value on college education. Here we look at the earnings of various levels of higher education relative to “below high school” and relative to high school graduates. The data are obtained from a report in College Board that is called Education Pays 2010. This reports talks about the benefits of the value of higher education to both individuals and society. In Figure 1.1 of this report, I found information on the median earnings and tax payments of full-time year round workers ages 25 and older for various levels of education in 2008. The following table shows the median earnings. $\displaystyle \begin{pmatrix} \text{Education Level}&\text{Median Earnings in Dollars} \\{\text{ }}&\text{ }&\text{ } \\\text{Below High School}&\text{24,300} \\\text{High School Graduate}&\text{33,800} \\\text{Some college but no degree}&\text{39,700} \\\text{Associate Degree}&\text{42,000} \\\text{Bachelor's Degree}&\text{55,700} \\\text{Master's Degree}&\text{67,300} \\\text{Doctoral Degree}&\text{91,900} \\\text{Professional Degree}&\text{100,000}\end{pmatrix}$ To help us compare, I convert the median earnings into ratios using “below high school” as baseline. $\displaystyle \begin{pmatrix} \text{Education Level}&\text{Ratio}&\text{Percent Increase} \\{\text{ }}&\text{ }&\text{ } \\\text{Below High School}&\text{1.0000}&\text{0 percent} \\\text{High School Graduate}&\text{1.3909}&\text{39.09 percent} \\\text{Some college but no degree}&\text{1.6337}&\text{63.37 percent} \\\text{Associate Degree}&\text{1.7284}&\text{72.84 percent} \\\text{Bachelor's Degree}&\text{2.2922}&\text{129.22 percent} \\\text{Master's Degree}&\text{2.7695}&\text{176.95 percent} \\\text{Doctoral Degree}&\text{3.7819}&\text{278.19 percent} \\\text{Professional Degree}&\text{4.1152}&\text{311.52 percent}\end{pmatrix}$ For a full explanation on how the math is done, see Putting a value on college education. As an example, I show the calculation for the bachelor’s degree level. The following is the ratio of bachelor’s degree to the below high school level. $\displaystyle \frac{55700}{24300}=2.2922$ Thus, the median earning of college graduates is 2.2922 times the median earning of the below high school level, or equivalently 229.22% of the below high school level. Subtracting one from this ratio gives the percent increase going from below high school to bachelor’s degree. This means that the additional amount a college graduate can expect to earn is 129% of the median earning of the below high school level! In general, the ratio column in the above table gives information about the median earning at each level as a percentage of the baseline (below high school). The column for percent increase gives the percentage increase going from the baseline to each level, that is, the additional earning power at each level expressed as a percentage. It is clear that the higher the educational level, the higher the additional earning power. Now the same comparison but using the high school level as the baseline. $\displaystyle \begin{pmatrix} \text{Education Level}&\text{Ratio}&\text{Percent Increase} \\{\text{ }}&\text{ }&\text{ } \\\text{Below High School}&\text{0.7189}&\text{-28.11 percent} \\\text{High School Graduate}&\text{1.0000}&\text{0 percent} \\\text{Some college but no degree}&\text{1.1746}&\text{17.46 percent} \\\text{Associate Degree}&\text{1.2426}&\text{24.26 percent} \\\text{Bachelor's Degree}&\text{1.6479}&\text{64.79 percent} \\\text{Master's Degree}&\text{1.9911}&\text{99.11 percent} \\\text{Doctoral Degree}&\text{2.7189}&\text{171.89 percent} \\\text{Professional Degree}&\text{2.9586}&\text{195.86 percent}\end{pmatrix}$ With high school graduates as baseline, high school dropouts only earn about 71.89% of the median earning of the high school graduates. The additional earning power of the below high school level is -28.11%, which would actually be a decrease in earning power. Once again, a similar clear pattern here: learning has dollar value. The higher the educational level, the higher the additional earning power. On average, a college graduate can expect to earn 65% (64.79%) more than a high school graduate. The additional earning power for a professional degree is even greater, 195.86%, which is three times the addtional earning power for a college dregree. ## Putting a value on college educationDecember 17, 2010 Posted by Dan Ma in Number sense. Tags: , , , , , , add a comment Recently I read a piece in Time.com about high school dropout rates. According to this article, the economic outlook for high school dropouts is bleak. The statistics cited by the article are the average earnings and unemployment rates. In 2009, an average college degree holder earned$1,015 per week and an averaged high school dropout earned just $454 per week. The unemployment rate for the college degree holders is 5.2% and 14.6% for high school dropouts. The figures speak volume. They not only put a dollar value on college education. These numbers also give me an opportunity to talk about the concept of ratio, which is a handy way to compare two quantities. The numbers I want to focus on are the two weekly earnings of$1,015 and $454. How much larger is$1,015 over $454? In terms of absolute amount, the weekly earning of$1,015 is $561 more than$454 since $1015-454=561$. But the absolute amount does not tell the whole story. We can also look at the ratio of $1,015 over$454, which is obtained by $1,015 divided by$454.

$\displaystyle (1) \ \ \ \ \ \ \ \ \ \ \frac{1015}{454}=2.236$

What to make of the ratio 2.236? First, it says that the average earning of a college graduate is 2.236 times of the average earning of a high school dropout. For each one dollar in income for a high school dropout, the average college graduate makes two dollars and twenty four cents. So on average a college degree holder makes over two times more than someone with less than a high school education. In this ratio, the baseline (or denominator) is the average weekly earning of a high school dropout.

We can also state the ratio 2.236 in terms of percent. Multiplying it by 100, we get 223.6. So the average earning of a college graduate is 223.6% of the average earning of a high school dropout. This is the same information as in the above paragraph, just that the scale is a little different. For each one hundred dollars in income for a high school dropout, the average college graduate makes two hundred and twenty three dollars and sixty cents.

The average college grad’s weekly paycheck is $561 more than the weekly paycheck of an average high school dropout. The additional amount of$561 as a percentage of the paycheck for the average high school dropout is:

$\displaystyle (2) \ \ \ \ \ \ \ \ \ \ \frac{561}{454}=1.236$

Note that the answer in (2) is just one less than the ratio in (1). Thus the ratio (1) above has another interpretation. Subtracting one from the ratio (1) gives the percent increase going from $454 to$1,015. The increase is 1.236 (or 123.6% after multiplying by 100). For example, if someone is currently making $454 a week and if his income is to increase to$1,015 a week, this represents a 123.6% raise! This is the average premium of a college education, i.e, the additional amount a college grad can expect to make.

We can also flip the ratio in (1) and obtain the following ratio.

$\displaystyle (3) \ \ \ \ \ \ \ \ \ \ \frac{454}{1015}=0.4473$

What to make of this ratio? Now the baseline is the college graduate. It tells us that the average earning of a high school dropout is only 0.4473 (or 44.73%) of the average earning of a college grad. For each one dollar earned by a college grad, the high school dropout only makes about 44.73 cents. For each one hundred dollars earned by a college grad, the high school dropout makes about fourty four dollars and seventy three cents.

If we subtract one from the ratio in (3), we get the percentage change going from $1,015 to$454 (percentage decrease in this case). Note that $0.4473-1=-0.5527$. Multiplying this by 100, we get negative 55.27%. This says that going from $1,015 to$454 represents a 55.27% decrease in income. So if someone is currently making $1,015 a week, and if his income is to decrease to$454 a week, this represents a 55.27% decrease in income.

To summarize, when we compare two quantities $P$ and $Q$ with $Q$ as a baseline, we can calculate the ratio $\displaystyle R=\frac{P}{Q}$. The ratio $R$ gives information about the quantity $P$ as a percentage of the baseline quantity $Q$.

The quantity $R-1$ is the percentage change going from the baseline quantity $Q$ to the quantity $P$. If $R-1$ is a positive number, then it is a percentage increase when going from $Q$ to $P$. Otherwise it is a percentage decrease. In our income example in this post, $R-1$ can be interpreted as the additional earning of college graduates expressed as a percentage.

## What is 32 plus 3?December 4, 2010

Posted by Dan Ma in Number sense.
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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. ## Taking another look at the California LotteryNovember 19, 2010 Posted by Dan Ma in Lottery, Number sense, Risk assessment. Tags: , , , , , , , add a comment 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 storeNovember 18, 2010 Posted by Dan Ma in Number sense. Tags: , , , , , add a comment 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:

$\displaystyle \begin{pmatrix} \text{Weight Reduction}&\text{Percentage Increase} \\{\text{ }}&\text{ }&\text{ } \\\text{1}&\text{4.35 percent} \\\text{2}&\text{9.09 percent} \\\text{3}&\text{14.29 percent} \\\text{4}&\text{20.00 percent} \\\text{5}&\text{26.32 percent} \\\text{6}&\text{33.33 percent} \\\text{7}&\text{41.18 percent} \\\text{8}&\text{50.00 percent}\end{pmatrix}$