Probability - Puzzles

The probability of rolling x ones out of y dice is combin(y,x)*(1/6)^x*(5/6)^y-x. See my section on probabilities in poker for an explanation of the combin(x,y) function. For example the probability of rolling 4 ones is combin(6,4)*(1/6)⁴*(5/6)² = 0.803755%.

The formulaic answer for the number of combinations would be combin(8,2)*combin(6,2)*combin(4,2)/fact(4) = 25*15*6/24 = 105. Another way to solve the number of combinations would be to take one golfer at random. There are 7 possible people to pair him with. Then pick another golfer at random from the six left. There are 5 possible people to pair him with. Then pick another golfer at random from the four left. There are 3 possible people to pair him with. So the number of combination is 7*5*3 = 105. Thus the answer is 1 in 105.

Often these mind reading number puzzles work because an interesting mathematical oddity. If the sum of digits of a number is divisible by 9 then the number it self is divisible by 9. Let’s try it on the phone number of the Las Vegas Tropicana (702-739-2222). The sum of digit is 7+0+2+7+3+9+2+2+2+2 = 36. 36 divides evenly by 9, so 702739222 must also be divisible by 9. Here is a proof of this.

1. Let n bet any integer. Express n as d₀*1 + d₁*10 + d₂*100+ d₃*1000+ ... + d_n*10ⁿ, where d_n is the first digit, d_n-1 is the second, and so on.
2. n = [d₀ + d₁ + d₂ + ... + d_n ] + [d₁*9 + d₂*99+ d₃*999+ ...+ d_n*999...9 ( a number with n nines)]
3. n = [d₀ + d₁ + d₂ + ... + d_n ] + 9*[d₁*1 + d₂*11+ d₃*111+ ... d_n*111...1 (a number with n ones)]
4. 9*any integer is evenly divisible by 9. So if d₀ + d₂ + d₂ + ... + d_n , or the sum of digits, is divisible by 9, then the entire number must be divisible by 9.

Now that we have that proof out of the way we can look at this magic trick. The problem asks you to pick any number. Then rearrange the digits to make a second number. Then subtract the smaller number from the larger number.

The answer is always going to have a sum of digits divisible by 9. Why? For every digit in the original number it appears somewhere else in the other number. Going one set of digits at a time, changing all the other numbers to zero, we could boil down each set as +/- n*[10^x - 10^y] (where x>=y and n is the digit) = +/-n *10^y * (10^x-y - 1) = 10^y * (a number composed of only nines) = a number divisible by 9.

Let’s look at an example. Let the original number be 1965. Scramble it up to get 6951. 6951 - 1965 = 6*(1000-10) + 9*(100-100) + 5*(10-1) + 1*(1-1000) = 6*990 + 9*0 + 5*9 + 6*-999. Note that each part is divisible by 9, thus the number you get after subtracting must also be divisible by 9, and finally the sum of digits is also divisible by 9.

The trick then asks you to circle a number except 0 and enter the sum of all the other digits. The program then only needs to add a number to the number you entered so that the sum is divisible by 9. For example if you said the sum of your digits was 13 then you must have circled a 5, because 13+5 = a number divisible by 9.

The reason you can’t circle a zero is because if you did and then entered a number already divisible by 9 then the program wouldn’t know whether you circled a 0 or a 9.

Thanks for the compliment. This is a good problem. The probability that exactly one employee will be in the bottom 5% is 5*(.05)*(.95)⁴ = 0.203627. Given that one employee is in the lowest 5% the probability of exactly one in the next 20% is 4*(.2/.95)*(.75/.95)³ = 0.414361. Given these two underachievers the probability of exactly one in the next 50% out of the remaining 75% is 3*(.5/.75)*(.25/.75)² = 0.222222. The probability that one of the remaining two falls in the lower 20% of 25% is 2*(.2/.25)*(.05/.25) = 0.32. Taking the product of all these probabilities we get 0.006, or 3/5 of 1%.

Thank you for those good points. However the alternative to no control over the distribution of job performance ratings is rating inflation. The manager will be put in a position of giving out bloated ratings to keep his staff happy. As a government worker for ten years I speak with some experience on this. When I taught at UNLV there was no average class GPA standard but there were certain expectations about what a grading curve should look like at the end of the semester. At least in a college setting I thought that made for a reasonable policy. Perhaps in a business environment some sort of common sense medium would also be best.

Good question. I can't think of any.

Let’s express your number as 10t+u. You are asked to subtract each digit, leaving you with 10t+u-t-u = 9t, a number divisible by 9. Note how all the numbers divisible by 9 have the same item, which is the one the genie predicts.

To satisfy the 75% requirement the student must get at least 315 out of the 420 questions right. The expected number of correct answers from guessing is 420*0.25=105. The standard deviation is (420*0.25*0.75)^0.5 = 8.87412. So the candidate must exceed expectations by 210 questions, or 210/8.87412=23.66432 standard deviations. The probability of doing this is way off the charts. If every living thing on earth took this test, answering randomly, I doubt anyone or anything would pass. I won’t even get into the other requirement.

If we assumed the games were independent then the probability of losing both would be 90%*70%=63%. But since you say the probability of losing both is actually 65% (which is more than the 63%), that means the two events are correlated. If the probability of losing both is 65% and just losing game 2 is 70%, then the probability of winning game 1 and losing game 2 must be 5%. Using the same logic the probability of losing game 1 and winning game 2 must be 25%. That only leaves 5% for winning both games. So the probability of winning exactly once is 25%+5% = 30%.

Yes! The key to this problem is that the host is predestined to open a door with a goat. He knows which door has the car, so regardless of which doors the players pick, he always can reveal a goat first. The question is known as the "Monty Hall Paradox." Much of the confussion about it is because often when the question is framed, it is incorrectly not made clear the host knows where the car is, and always reveals a goat first. I think put some of the blame on Marilyn Vos Savant, who framed the question badly in her column. Let’s assume that the prize is behind door 1. Following are what would happen if the player (the second contestant) had a strategy of not switching.

• Player picks door 1 --> player wins
• Player picks door 2 --> player loses
• Player picks door 3 --> player loses

Following are what would happen if the player had a strategy of switching.

• Player picks door 1 --> Host reveals goat behind door 2 or 3 --> player switches to other door --> player loses
• Player picks door 2 --> Host reveals goat behind door 3 --> player switches to door 1 --> player wins
• Player picks door 3 --> Host reveals goat behind door 2 --> player switches to door 1 --> player wins

So by not switching the player has 1/3 chance of winning. By switching the player has a 2/3 chance of winning. So the player should definitely switch.

For further reading on the Monty Hall paradox, I recommend the article at Wikipedia.

Your mistake is to assume each of these events has a 25% possibility. Following is the correct probability of each event.

• Player picks door 1 (1/3) * shown 2 (1/2) = player loses (1/6)
• Player picks door 1 (1/3) * shown 3 (1/2) = player loses (1/6)
• Player picks door 2 (1/3) * shown 3 (1/1) = player wins (1/3)
• Player picks door 3 (1/3) * shown 2 (1/1) = player wins (1/3)

So losing events have a total probability of 2*(1/6) = 1/3 and winning events have a total probability of 2*(1/3)=2/3.

There is 1 way with 0 toppings, 5 ways with 1 topping, 10 ways with 2 toppings, 10 ways with 3 toppings, 5 ways with 4 toppings, and 1 way with 5 toppings. So the answer is 1+5+10+10+5+1 = 32. Another way to solve is either topping can be used or not. So the total is 2⁵ = 32.

After discovering the claim that the Florida hurricanes only hit Bush voting counties was a hoax (see the October 17, 2004 column) I am going to be more skeptical about such alleged coincidences. According to the National Earthquake Information Center of the top 11 earthquakes since 1990 only the recent one of 2004 hit on a December 26. The Iranian earthquake you mention was only 6.7 in magnitude, which is far from making the top eight.

Let’s let d (for day) be the number of eggs at the beginning of the day and n (for night) be the number at the end. The problem tells us that d/2 - ? = n. So, let’s solve d in terms of n.

d/2 = n + ?
d= 2n + 1
So on the third day n=0, so d=1.
On the second day n=1, so d=3.
On the third day n=3, so d=7.
So there you have it, you started with 7 eggs.

It will only take 1 day so that 1 person has served as chief. For the second day the probability of a new chief is 0.9. The expected number of days it will take to get a new chief, if the probability each day is 0.9 is 1/0.9 = 1.11. This is true for any probability: the expected number of trials until a success is 1/p. So after 2 people have served the probability of a new chief on the next day is 0.8. So the waiting period for a 3rd chief is 1/0.8 = 1.25 days. The answer is the sum of the waiting periods, which is 1/1 + 1/.9 + 1/.8 + ... + 1/.1 = 29.28968 days.

I don’t want to blow the answer for those who want to solve it for themselves. For the answer and solution visit my other web site mathproblems.info, problem 189.

At some point you should refuse a good bet because the stakes are too high. Personally I think a good measure of the enjoyment one gets from money is the log of the amount. The base of the log does not matter so let’s use 10. However we can’t take a log less than 10, so let’s say the enjoyment is 0 for any amount less than ten. So in your example let’s assume you have $0 before winning the $1,000,000 with your first throw. Now you have log(1,000,000) = 6 units of happiness. The expected value of your happiness taking another free throw is 0.75*log(2,000,000) + 0.25*0 = 4.975772. This is less than 6 so in this case you should take the million and walk. However it might be different if you already had some money. Let’s say you already have $200,000. Then your happiness by walking is log(1,200,000) = 6.07918. Your happiness by risking the million and taking another shot is 0.75*log(2,200,000) + 0.25*log(200,000) = 6.082075, so you marginally take the second shot. If you were to win that one your choice would be between log(2,200,000) = 6.34242 and 0.75*log(4,200,000)+0.25*log(200,000) = 6.29269. In this case you should not take a third shot and instead walk with the $2,000,000 win. The breakeven point for accepting the first double is an existing wealth of $191,487. To accept two doubles you should have $382,975 in other money.

I think I have answered this before but the 50/50 point is closer to 23. To make things simple let’s ignore leap years. The long answer is to order the 23 people somehow. The probability that person #2 has a different birthday from person #1 is 364/365. The probability person #3 has a different birthday from persons #1 and #2, assuming they are different from each other, is 363/365. Keep repeating until person 23. The probability is thus (364/365)*(363/365)*...*(343/365) = 49.2703%. So the probability of no match is 49.27% and of at least one match is 50.73%. Another solution is the number of permutations of 23 different birthdays divided by the total number of ways to pick 23 random numbers from 1 to 365, which is permut(365,23)/365²³ = 42,200,819,302,092,400,000,000,000,000,000,000,000,000,000,000,000,000,000,000 / 85,651,679,353,150,300,000,000,000,000,000,000,000,000,000,000,000,000,000,000 = 49.27%.

That would be $35 above average, or 7/9 standard deviations. The probability of being more than 7/9 standard deviations above expectations would be 1-Z(7/9) = 1- 0.78165 = 0.21835.

Player B would need to win the next two (not counting ties) so the probability is (1/2)*(1/2) = 1/4.

The probability that intelligent life exists anywhere in the galaxy I believe is very high. The Drake Equation seeks to estimate the number of incidents of intelligent life in the galaxy, which depending on the numbers you put into comes up with a figure of about a million. However there is also no good evidence that these civilizations have ever visited us or made contact. So the famous Fermi Question is "Where is everybody?" I do think the lack of evidence of other intelligent life casts some doubt on the Drake Equation but I would still put the number of intelligent civilizations in our galaxy in the ballpark of 1000. That is just our galaxy, there are billions of galaxies out there. However the distance between galaxies is so vast there is really not much point in discussing travel or communication between them. So to answer your question I would say roughly 99.9%.

The exact answer 1-(9,999,999/10,000,000)^10,000,000 = 0.632121. This is also the same as (e-1)/e to seven decimal places.

1 in 635,241.

As I understand the rules of tennis the winner of a set is the first to win six games, and by a margin of at least two games, except a 6-6 tie will result in a single tie-breaker game. The following table shows the probability of winning a set, given the probability of winning a game.

Probabilities in Tennis

Probability Game Win	Probability Set Win
0.05	0.000003
0.1	0.000189
0.15	0.001899
0.2	0.009117
0.25	0.028853
0.3	0.06958
0.35	0.138203
0.4	0.23687
0.45	0.361085
0.5	0.5
0.55	0.638915
0.6	0.76313
0.65	0.861797
0.7	0.93042
0.75	0.971147
0.8	0.990883
0.85	0.998101
0.9	0.999811
0.95	0.999997

The formula for any probability of winning a game p, and losing q, is 1*p⁶ + 6*p⁶*q + 21*p⁶*q² + 56*p⁶*q³ + 126*p⁶*q⁴ + 252*p⁷*q⁵ + 504*p⁷*q⁶

The water level relative to the shore will decrease. Inside the boat the rock is pressing down on the canoe and thus pushing up the water around it. The amount of water displaced is equal in weight to that of the rock. For example, a 10 pound rock will displace 10 pounds of water upward. When the rock is thrown overboard the weight will not matter but rather the volume of the rock. So the rock will push upward an amount of water equal in volume to the rock. The mass of a rock is greater than that of water so the rock displaces more water pushing down on it than in it. So the level of the lake will be higher with the rock in the canoe than at the bottom of the lake.

Let’s call the first three digits in your phone number x, and the last four y. Now let’s see what I have at each step.

1. Ready!
2. X
3. 80x
4. 80x+1
5. 250*(80x+1) = 20000x+250
6. 20000x+250+y
7. 20000x+250+2y
8. 20000x+250+2y-250 = 20000x+2y
9. (20000x+2y)/2 = 10000x+y

So that is of course going to equal your phone number. We need the 10000x to move the prefix four places to the left, and then we add on the last four digits.

You are right regarding the 1/2 and 2/3 probabilities. If you buy t tickets your probability of winning is t/(68+t). So for a 90% probability, solve for t as follows.

0.9 = t/(68+t)
0.9*(68+t) = t
61.2 = 0.1t
t = 612, or $51,000

For 95%...

0.95= t/(68+t)
0.95(68+t) = t
64.6 = 0.05t
t = 1292, or $107,666.67

Assuming the car is worth $27,000 to you, you should quit buying tickets as soon as the next ticket sold does not increase your probability of winning enough to warrant the price.

For a ticket to be worth the price it should increase your probability of winning by p, where...

27000*p=(500/6)
p=0.003086

Let's say t is the number of tickets your purchased where you are indifferent to buying one more ticket.

[(t+1)/(t+68+1)] − [t/(t+68)] = 0.003086
[(t+1)/(t+69)] − [t/(t+68)] = 0.003086
[((t+1)*(t+68))/((t+69)*(t+68))] − [(t*(t+69))/((t+68)*(t+69))] = 0.003086
[((t² +69t+68)/((t+69)*(t+68))] − [(t²+69t)/((t+68)*(t+69))] = 0.003086
68/((t+68)*(t+69)) = 0.003086
((t+68)*(t+69)) = 220.32
t²+137t+4692 = 22032
t² +137t - 17340=0
t=(-137+/-(137²-4*1*-17340)²)/2
t = 79.9326

Let's test this by plugging in some values for tickets purchased, assuming the player can always buy tickets at $500/6 = $83.33 each.

At 79 tickets your cost is 79*(500/6) = $6,583.33, your probability of winning is 79/(79+68) = 53.74%, your expected return is $27,000*0.5374 = $14,510.20, and your expected profit is $14,510.20 - $6,583.33 = $7,926.87.

At 80 tickets your cost is 80*(500/6) = $6,666.67, your probability of winning is 80/(80+68) = 54.04%, your expected return is $27,000*0.5405 = $14,594.59, and your expected profit is $14,594.59 - $6,666.67 = $7,927.92

At 81 tickets your cost is 81*(500/6) = $6,750.00, your probability of winning is 81/(81+68) = 54.36%, your expected return is $27,000*0.5436 = $14,677.85, and your expected profit is $14,594.59 - $6,750.00 = $7,927.85.

So we can see that the maximum expected win peaks at 80 tickets.

If you keep the current fridge then in five years you will have spent an extra $37*5 = $185 on electricity compared to a new one. If you replace it now you’ll be out $425 but assuming linear depreciation after five years it will still be worth $425*(9/14) = $273.21. So you will have lost $425*(5/14) = $151.79 due to depreciation. So the cost of depreciation of the new fridge is less than the additional electricity expense of keeping the old one, so I favor buying a new one now.

Ignoring leap day, the probability of all three different birthdays is (364/365)*(363/365) = 0.99179583. So the probability of at least one common birthday is 1 - 0.99179583 = 0.00820417.

To keep things simple let’s assume that each person has a 1/12 probability of being born in each month. The probability that all five people are born in different months is (11/12)*(10/12)*(9/12)*(8/12) = 0.381944. So the probability of a common month is 1 - 0.381944 = 0.618056.

The answer is at the end of the column.

Take one sheep from trailer 1, two from trailer 2, three from trailer 3, four from trailer 4, and zero from trailer 5. If all the sheep weighed 39 kg then the total weight would be 39 * 10 = 390 kg. However 0 to 4 sheep are one kg heavier. If the total weight is 391, then there is one heavy sheep on the scale; thus it must have come from trailer 1. Likewise, if the total weight is 392, then there are two heavy sheep on the scale, which must have come from trailer 2. In the same manner a weight of 393 means the heavy sheep are in trailer 3, a weight of 394 means the heavy sheep are in trailer 4, and a weight of 390 means the heavy sheep are in trailer 5.

It depends on the number of seats in a cluster. Most domestic flights have three seats on either side of the aisle. That would make 60 3-seat clusters. After the first one of you is seated, there will be two seats in the same cluster out of the remaining 179, so the chances of being in the same cluster are 2/179 = 1.12%. Then you can’t have somebody else in the middle seat. The chances of the third person being in the middle seat are 1/3. So the answer is (2/179)*(2/3) = 0.74%, or 1 in 134.25.

The answer will appear in the next column.

Normally I would say this is out of my area. However, as a former government actuary for eight years, I know a thing or two about taxes. From what I’ve read, most of Warren Buffet’s income is defined as capital gains, which is taxed at only a 15% rate. Like it or not, the tax laws allow it. What puzzled me is why his secretary was paying as much as 30%. According to this video, he was counting “payroll and income taxes.” By “payroll taxes” he obviously meant Social Security and Medicare taxes. Let’s see if 30% is a reasonable total federal tax rate for his secretary.

In 2007 the highest tax bracket was taxed at 35%, but that only applies to income above $349,700. The income up to that point is taxed much less. Let’s assume his secretary is single, with no dependent children, and her salary was $100,000. First, let’s subtract the minimum deductions. In 2007 the standard deduction for single filers was $5,350. The personal deduction was $3,400. So, we’re left with $100,000 - $5,350 - $3,400 = $91,250 in income subject to income taxes. For single filers in 2007, the tax rate was 10% on the first $7825 in income, then 15% up to $31,850, then 25% up to $77,100, and 28% up to $160,850. So, her income tax would have been =0.1×$7,825+0.15×($31,850-$7825)+0.25×($77,100-$31,850)+0.28×($91,250-$77,100) = $19,660.75. That is only 19.7% of her income. All my assumptions like her income, filing status, and not itemizing worked against her, or for a higher tax rate.

Now let’s do Social Security and Medicare. In 2007, the Social Security tax was 6.2%, up to incomes of $97,500, when it completely shuts off. The 2007 Medicare tax rate was 1.45%, with no cap. So, her combined Social Security and Medicare tax would have been 6.2%*97,500 + 1.45%*100000 = $7,495. Counting those taxes, her overall tax rate would have been ($19,660.75 + $7,495)/$100,000 = 27.2%. Still we’re 2.8% short of 30%.

My best guess is that she is also considering the fact that ultimately she is the one paying the employer’s matching Social Security and Medicare tax. For those who don’t know, Social Security and Medicare taxes are really double that deducted from your checks. The employer pays the other half. However, some, including me, would argue that ultimately it is the employee who pays both. If the employer didn’t have to pay that tax, he would have more money to pay his employees. It is easy to feel that way when you’re self-employed, like I am, and have to pay both shares. If you double the Social Security/Medicate tax, the rate is now ($19,660.75 + 2×$7,495)/$100,000 = 34.7%. I assume the 4.7% difference is because she makes less than $100,000, is married, has dependents, itemizes deductions, or some combination.

The Social Security and Medicare taxes would not apply much to Warren Buffet. First, the Social Security cap of $97,500 would be insignificant to him. Second, those taxes apply to wages, not capital gains, as he defines most of his income to be.

So, that is my best guess as to the math behind Mr. Buffet’s statement.

Update: Shortly after this column appeared I received the following response. In the interests of fairness, I present the following argument that Mr. Buffet is paying too much in taxes.

I read with interest your answer to the ’outraged’ person who thinks it is so unfair that Warren Buffet pays less percentage in taxes than his secretary. I was disappointed in your answer, which does not correct the misinformation that implies that Mr Buffet pays less tax than his secretary.

First, as you noted, investment income is indeed taxed at 15%. This is in effect double taxation as the earned income that Mr. Buffet invested was taxed at his marginal rate of 36%. Comparing apples to oranges (work income vs investment income).

Second, one should not look at the percentage. In gambling terms, one should look at the ’payout’ instead. I am very certain that Mr. Buffet paid millions of dollars in taxes in the same year that his secretary paid thousands of dollars. Shouldn’t your reader be more outraged that one citizen of the country is paying 1000’s of times more than other citizens for the same government services? Once could just as easily say "I heard that Warren Buffet paid 1,000,000 times more taxes than his secretary, that is outrageous!"

Just thought I’d point out that only looking at "percentage" and not "actual payout" is a fallacy. Similar to many of your gambling fallacies.

Best Regards,

Kevin A. (Dallas)

You’re welcome. If there are only two pirates left, then the one chosen to make a suggestion has no hope, because the other pirate will vote no. The one drawn will get zero, and the other all 1000. So, before the draw, the expected value with two pirates left is 500 coins.

At the three pirate stage, the drawn pirate should suggest giving one of the other pirates 501, and 499 to himself. The one getting 501 will vote yes, because it is more than the expected value of 500 by voting no. Before the draw, with three pirates left, you have a 1/3 chance each of getting 0, 499, or 501 coins, for an average of 333.33.

At the four pirate stage the drawn pirate should choose to give 334 to any two of the other pirates, and 332 to himself. That will get him two ’yes’ votes from the pirates getting 334 coins, because they would rather have 334 than 333.33. Including your own vote, you will have 3 out of 4 votes. Before the draw, the expected value for each pirate is the average of 0, 334, 334, and 332, or 1000/4=250.

By the same logic, at the five pirate stage, the drawn pirate should choose to give 251 to any two pirates, and 498 to himself. Unlike the original problem, it isn’t necessary to work backwards. Just divide the number of coins by the number of pirates, not including yourself. Then give half of them (rounding down) that average, plus one more coin.

I hope you’re happy; I’ve been obsessed by this puzzle for the last month or so. I was lucky (or perhaps unlucky) to find the 256-piece puzzle at the local Borders book store, but I had to buy the four clue puzzles on eBay, from a guy in Australia.

I wrote a program that can easily solve the four clue puzzles. It solved the 72-piece clue puzzle #4 in less than a second. The way I did it was with a simple brute-force recursive program. I mapped out a path on the board, starting with the border. At each position, the program looped through all the unused pieces, looking for one that fit. If it found one, it moved to the next square, if it didn’t, it moved back a square.

I have had two computers crank away at the 256-piece $2 million puzzle for weeks, and neither have come anywhere close. I tend to agree with what the creator said in that video, that if you hooked up ten million of the world’s fastest computers, they still might not find the solution by the death of the universe. You would think I would have heeded his warning before starting, but in the face of a good puzzle, all consideration for practical use of my time goes out the window.

I have lots of ideas for shortcuts, but even if they sped up my program by a factor of a billion, it still probably wouldn’t help. I’m going to be extremely impressed if anybody solves this thing. What really nags at me is I feel there is some undiscovered branch of mathematics that could solve puzzles like this easily. Until then, I think glorified trial and error is the best we can do to solve it. Today’s computers are simply too slow, and the number of combinations too vast, for that to have much of a chance of success.

In zero-wind it will take 2 hours each way, for a total of 4 hours. With the tailwind, the plane will travel at 600 mph, making the trip in 1000/600 = 1.667 hours. With the headwind, the plane will travel at 400 mph, taking 1000/400 = 2.5 hours. So, in the wind, the total time is 4.167 hours, or 10 minutes longer.

This just goes to show that it is dangerous to average averages. You can’t say the average rate of a trip is 500 mph, if it is 400 mph one way and 600 mph the other, because the 400 mph leg is over a longer period of time.

If this isn’t intuitive, consider a 500 mph wind. The plane would take 1 hour only with the wind, but it would stay in place the other way, taking forever.

For the sake of simplicity, let’s ignore the fact that the more tickets you buy the lower the value of each ticket becomes because you compete with yourself. That said, the probability of losing all five tickets is (12/13)⁵ = 67.02%. So the probability of winning at least one prize is 32.98%. There are 7033×13=91,429 total tickets in the drum before you buy any. 91,429-40=91,389 are not big prizes. The probability of not winning any big prizes with five tickets is (91,389/91429)⁵ = 99.78%. So the probability of winning at least one big prize is 0.22%, or 1 in 458.

It isn’t often I say this, but I have no idea. As you noted in another e-mail, they take the format of serial number on US currency, two letters, with a ten-digit number in between. Out of respect for copyright, I won’t indicate what the numbers are here.

Thanks for asking. No, I haven’t touched that thing since I wrote about in the November 17, 2008 Ask the Wizard column. According to their web site, they will have "scrutiny dates" on December 31, 2009, and 2010 if necessary. In my opinion, it will never be solved.

Update: The Eternity II web size appears to no longer exist.

If you are easily offended, please skip to the next question.

For the benefit of my readers who didn’t read that blog, look at the first letter of each line in this memo by California governor Arnold Schwarzenegger (PDF), starting with the line beginning with the letter F.

This was discussed in my companion site Wizard of Vegas. To find an answer, I found a frequency of each letter of the first word in the English Language at Wikipedia.

Word Frequency by First Letter

Letter	Frequency
A	11.60%
B	4.70%
C	3.51%
D	2.67%
E	2.00%
F	3.78%
G	1.95%
H	7.23%
I	6.29%
J	0.63%
K	0.69%
L	2.71%
M	4.37%
N	2.37%
O	6.26%
P	2.55%
Q	0.17%
R	1.65%
S	7.76%
T	16.67%
U	1.49%
V	0.62%
W	6.66%
X	0.01%
Y	1.62%
Z	0.05%

To estimate the probability that Arnold’s message was indeed just a coincidence would be Prob(F) × Prob(U) × ... × prob(U) = 0.0378 × 0.0149 × 0.0351 × 0.0069 × 0.0162 × 0.0626 × 0.0149 = 1 in 486,804,391,348. This doesn’t even factor in the fact that a line break conveniently was in the place of the space between the two words.

I’d like to thank Eliot J. and Jonathan F. for their input into this solution.

Let’s define some variables first, as follows:

n = number of your bags
b = total number of bags

As the number of total bags gets larger the answer will get closer to b×n/(n+1). For a large plane, that will give you a good estimate. However, if you want to be exact, the answer is

[b×combin(b,n)-(sum for i=n to b-1 of combin(i,n))]/combin(b,n)

For example, if there are 10 total bags, and four of them are yours, then the expected wait time =

[10×combin(10,4)-combin(4,4)-combin(5,4)-combin(6,4)-combin(7,4)-combin(8,4)-combin(9,4)]/combin(10,4) = 8.8 bags.

Solution:

The number of ways to pick n out of b bags is combin(b,n). So, the probability that all your bags come out within the first x bags is combin(x,n)/combin(b,n). The probability that your last bag is the x^th bag to come out is (combin(x,n)-combin(x-1,n))/combin(b,n), for x>=n+1. For x=n it is 1/combin(b,n).

So, the ratio of the expected wait time to the total wait time is:

n×combin(n,n)/combin(b,n) +
(n+1)×(combin(n+1,n)-combin(n,n))/combin(b,n) +
(n+2)×(combin(n+2,n)-combin(n+1,n))/combin(b,n) +
.
.
.
+
(b-1)×(combin(b-1,n)-combin(b-2,n))/combin(b,n) +
b×(combin(b,n)-combin(b-1,n))/combin(b,n)

Taking a telescoping sum, this can be simplified to:

[b×combin(b,n)-combin(b-1,n)-combin(b-2,n)-...-combin(n,n)]/combin(b,n)

A reader later wrote in saying that the answer can be simplified to n×(b+1)/(n+1). This can be shown by induction, a legitimate method, but always leaves me emotionally unsatisfied.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

This is a good confidence interval kind of problem. In 100 sales the expected turtles sold will be 14.29. The standard deviation is sqrt(100×(1/7)×(6/7)) = 3.50.

Let t be the number of turtles made, and x the number sold.

pr(x<=t)=0.9
pr(x-14.29<=t-14.29)=0.9
pr((x-14.29)/3.5)<=(t-14.29)/3.5))=0.9

The left side of the inequality follows a standard normal distribution (mean of 0, standard deviation of 1). This next step takes an introductory statistics course, or some faith, to accept.

(t-14.29)/3.5 = normsinv(0.9) This is the Excel function.
(t-14.29)/3.5 = 1.282
t-14.29 = 4.4870
t = 18.77

Nobody is likely to buy 0.77 of a turtle statue, so I would round up to 19. According to the binomial distribution, the probability of selling 18 or less is 88.35%, and 19 or less is 92.74%. This question was raised and discussed in the forum of my companion site Wizard of Vegas.

"Scroll down 100 lines for the answer.

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There were 15,621 coconuts in the original pile. Scroll down another 100 lines for my solution.

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Let c be the number of coconuts in the original pile and f be the final share for each sailor after the last division.

After sailor 1 takes his share and gives the monkey his coconut there will be (4/5)×(c-1) = (4c-1)/5 left.

After sailor 2 takes his share and gives the monkey his coconut there will be (4/5)×(((4c-1)/5)-1) = (16c-36)/25 left.

After sailor 3 takes his share and gives the monkey his coconut there will be (4/5)&times(((16c-36)/25)-1) = (64c-244)/125 left.

After sailor 4 takes his share and gives the monkey his coconut there will be (4/5)×(((64c-244)/125)-1) = (256c-1476)/625 left.

After sailor 5 takes his share and gives the monkey his coconut there will be (4/5)×(((256c-1476)/625)-1) = (1024c-8404)/3125 left.

In the morning each sailor’s share of the remaining pile will be f = (1/5)×(((1024c-8404)/3125)-1) = (1024c-11529)/15625 left.

So, the question is what is the smallest value of c such that f=(1024×c-11529)/15625 is an integer. Let’s express c in terms of f.

(1024×c-11529)/15625 = f
1024c - 11529 = 15625×f
1024c = 15625f+11529
c = (15625f+11529)/1024
c = 11+((15625×f+265)/1024)
c = 11+15×f+(265×(f+1))/1024

So, what is the smallest f such that 265×(f+1)/1024 is an integer? 265 and 1024 do not have any common factors, so f+1 by itself is going to have to be divisible by 1024. The smallest possible value for f+1 is 1024, so f=1023.

Thus, c = (15625×1023+11529)/1024 = 15,621.

Here is how many coconuts each person, and monkey, received:

Coconut Problem

Sailor	Coconuts
1	4147
2	3522
3	3022
4	2622
5	2302
Monkey	6
Total	15621

David Filmer, the one who challenged me the with question, already knew the answer. Actually, he asked me the formula for the general case of s sailors, but I had enough trouble with the specific case of 5 sailors. David notes the answer for the general case is c = s^s+1 - s + 1.

I’ll leave that proof to the reader.

Here are some links to alternate solutions to the problem:

• mathworld.wolfram.com/MonkeyandCoconutProblem.html
• www.ugcs.caltech.edu/~brothste/coconuts.pdf
• www.puzzle.dse.nl/teasers/coconut_chaos_us.html

This appears to be a mathematical paradox, but is really just an abuse of the expected value formula. As you noted in the question, it seems like the other envelope should have 25% more than the one you chose. However, if you buy that, then you may as well pick the other envelope to begin with. Furthermore, you could use that argument to switch back and forth forever if you don’t get to open the envelopes before deciding to switch. Clearly there must be some flaw in the expected value argument. The question is, where is the flaw?

I have spent a lot of time reading about this problem and discussing it over the years. I’ve heard and read many explanations about why the y=.5x + .5*2x = 1.25x argument is wrong. Many have used many pages of advanced mathematics in the explanation, which I don’t think is necessary. It is a simple question that calls out for a simple answer. So, this is my crack at it.

You must be very careful with what you do with the stated fact that one envelope has twice the money as the other one. Let’s call the amount in the smaller envelope S, and the larger one L. So we have:

L=2×S
S=0.5×L

Notice how the 2 and 0.5 factors are applied to different envelopes. You can’t take both factors and apply them to the same amount. If the first envelope has $100 then if it was the smaller envelope, the other one will have $200. If the $100 was the larger envelope, then the other one will have $50. So the other envelope will have $50 or $200. However, you can’t jump from there to say there is a 50/50 chance of each. This is because that would be applying the 0.5 and 2 factors to the same amount, which you can’t do. Without knowing the prize distribution to begin with, you can’t assign possible amounts to the second envelope.

If the 0.5x/2x argument is wrong, then what would be the correct way to set up the expected value of the other envelope? The way I would do it is to say that the difference between the two envelopes is L-S = 2S-S = S. By switching you’ll either gain or lose S, whatever it is. If the two envelopes have $50 and $100, then switching will gain or lose $50. If the two envelopes have $100 and $200, then switching will gain or lose $100. Either way, the expected gain by switching is 0. I think I could say that if the first envelope has $100, then there is a 50% chance the difference in the other envelope is $50, and a 50% chance it is $100. So the expected difference is $75. Thus, the expected value of the other envelope is 0.5×($100+$75) + 0.5×($100-$75) = 0.5×($175+$25) = $100.

I hope that makes some sense. This problem always induces lots of comments. If you have one, please don’t write to me directly, but post it in my Wizard of Vegas forum. The link is below.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Links

• Wikipedia
• David Chalmers
• www.cut-the-knot.org
• physorg.com

This puzzle is discussed on the BBC show Quite Interesting. Scroll down 100 lines for the answer and solution.

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Here are my probabilities of A winning according to each initial target. As you can see, A’s probability of winning is maximized by deliberately firing in the air.

Truel Odds

Strategy	Prob. Win
Air	13.887%
A	0.000%
B	12.560%
C	13.094%

For the solution, let’s use the notation Pr(X) to denote the probability of group X, and only group X, remains after a round. Let’s use the terminololy Pr(X*) to denote the probability of group X eventually winning the round, after repeating until the game state changes by somebody getting hit. Let Pr(X**) be the probability that player X is the sole survivor. To find the final probabilities, let’s look at the two-player states first. It is obvious that each will shoot at the other.

A vs. B

• Pr(A) = 0.1
  
• Pr(B) = 0.9×0.6 = 0.54
  
• Pr(AB) = 0.9×0.4 = 0.36

If both survive then they will repeat until there is one survivor only. So the probabilities of being the final survivor are:

• Pr(A*) = Pr(A)/(1-Pr(AB)) = 0.1/0.64 = 0.15625
  
• Pr(B*) = Pr(B)/(1-Pr(AB)) = 0.54/0.64 = 0.84375

A vs. C

• Pr(A) = 0.1
  
• Pr(C) = 0.9×0.9 = 0.81
  
• Pr(AC) = 0.9×0.1 = 0.09

If both survive then they will repeat until there is one survivor only. So the probabilities of being the final survivor are:

• Pr(A*) = Pr(A)/(1-Pr(AC)) = 0.1/0.91 = 0.10989011
  
• Pr(C*) = Pr(B)/(1-Pr(AC)) = 0.81/0.91= 0.89010989

B vs. C

• Pr(B) = 0.6
  
• Pr(C) = 0.4×0.9 = 0.36
  
• Pr(BC) = 0.$×0.1 = 0.04

If both survive then they will repeat until there is one survivor only. So the probabilities of being the final survivor are:

• Pr(B*) = Pr(A)/(1-Pr(BC)) = 0.6/.96 = 0.625
  
• Pr(C*) = Pr(B)/(1-Pr(BC)) = 0.36/.96= 0.375

Now we’re ready to analyze the three-player case. Let’s consider the situation where A aims at B.

Three Player — A Aims at B

If A hits B then C will definitely survive, and may or may not hit A. So two possible outcomes of hitting B are AC and C. If A misses B then B will aim at the greater threat C. If B hits C then A and B will survive. If B misses C then C will aim at the greater threat B. If C misses B then all three will survive. If C hits B then A and C will survive. So the possible outcomes are C, AB, AC, and ABC.

• Pr(A) = 0.
• Pr(B) = 0.
• Pr(C) = 0.1 × 0.9 = 0.09. This is achieved by A hitting B, and then C hitting A.
• Pr(AB) = 0.9 × 0.6 = 0.54. This is achieved by A missing B, and then B hitting C.
• Pr(AC) = 0.1 × 0.1 + 0.9 × 0.4 × 0.9 = 0.334. This can be achieved two ways. The first is A hitting B, and then C missing A. The second is A missing B, B missing C, and then C hitting B.
• Pr(BC) = 0.
• Pr(ABC) = 0.9 × 0.4 × 0.1 = 0.036. This is achieved by all three missing.

By the same logic as the two-player cases, we can divide each outcome by (1-Pr(ABC))=0.964 to find the probabilities of each state, assuming that the state of the game did change after the round.

• Pr(C*) = 0.09/0.964 = 0.093361.
  
• Pr(AB*) = 0.54/0.964 = 0.560166.
  
• Pr(AC*) = 0.334/0.964 = 0.346473.

From the two-player cases, we know if it comes down to A and B then A will win with probability 0.15625, and B 0.84375. If it comes down to A and C then A will win with probability 0.109890, and C 0.890110.

• Pr(A**) = (0.560165975 × 0.15625) + (0.346473029 × 0.10989011) = 0.125600. A can be the winner two ways: (1) getting to the AB state, and then winning, or (2) getting to the AC state and then winning.
• Pr(B**) = 0.560166 × 0.84375 = 0.472640. B will be the winner if it gets to the AB state, and then B wins.
• Pr(C**) = 0.093361 + (0.346473 × 0.890110) = 0.401760. C can win by A killing B, and then C killing A in the first round, or by it getting to state AC, and then C winning.

So, if A’s strategy is to aim at B at first, then his probability of being the sole survivor is 12.56%.

Three Player — A Aims at C

If A hits C then B will definitely survive, and may or may not hit A. So two possible outcomes of hitting C are AB and B. If A misses C then B will aim at the greater threat C. If B hits C then A and B will survive. If B misses C then C will aim at the greater threat B. If C misses B then all three will survive. If C hits B then A and C will survive. So the possible outcomes are B, AB, AC, and ABC.

• Pr(A) = 0.
• Pr(B) = 0.1 × 0.6 = 0.06.
• Pr(C) = 0.
• Pr(AB) = (0.1 × 0.4) + (0.9 × 0.6) = 0.04+0.54 = 0.58. This can be achieved two ways. The first is A hitting C, and then B missing A. The second is A missing B, and then B hitting C.
• Pr(AC) = 0.9 × 0.4 × 0.9 = 0.324. This is achieved by A missing C, B missing C, and C hitting B.
• Pr(BC) = 0.
• Pr(ABC) = 0.9 × 0.4 × 0.1 = 0.036. This is achieved by all three missing.

By the same logic as the two-player cases, we can divide each outcome by (1-Pr(ABC))=0.964 to find the probabilities of each state, assuming that the state of the game did change after the round.

• Pr(B*) = 0.06/0.964 = 0.062241.
  
• Pr(AB*) = 0.58/0.964 = 0.601660.
  
• Pr(AC*) = 0.324/0.964 = 0.336100.

By the same logic as the solution for the A aims at B case:

• Pr(A**) = (0.601660 × 0.15625) + (0.336100 × 0.10989011) = 0.130943.
• Pr(B**) = 0.062241 + 0.601660 × 0.84375 = 0.569891.
• Pr(C**) = 0.336100 × 0.890110 = 0.299166.

So, if A’s strategy is to aim at C at first, then his probability of being the sole survivor is 13.09%.

Three Player — A Misses Deliberately

After A deliberately misses then B will aim at the greater threat C. If B hits C then A and B will survive. If B misses C then C will aim at the greater threat B. If C misses B then all three will survive. If C hits B then A and C will survive. So the possible outcomes are AB, AC, and ABC.

• Pr(A) = 0.
• Pr(B) = 0.
• Pr(C) = 0.
• Pr(AB) = 0.6. This is achieved by B hitting C.
• Pr(AC) = 0.4 × 0.9 = 0.36. This is achieved by B missing C, and then C hitting B.
• Pr(BC) = 0.
• Pr(ABC) = 0.4 × 0.1 = 0.04. This is achieved by all three missing.

By the same logic as the two-player cases, we can divide each outcome by (1-Pr(ABC))=0.96 to find the probabilities of each state, assuming that the state of the game did change after the round.

• Pr(AB*) = 0.6/0.96 = 0.625.
  
• Pr(AC*) = 0.36/0.96 = 0.375.

By the same logic as the solution for the A aims at B case:

• Pr(A**) = (0.625 × 0.15625) + (0.375 × 0.109890) = 0.138865.
• Pr(B**) = 0.625 × 0.84375 = 0.527344.
• Pr(C**) = 0.375 × 0.890110 = 0.333791.

So, if A’s strategy is to aim at C at first, then his probability of being the sole survivor is 13.89%.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

There are several ways that are about equally as good. However, the one I find the easiest to understand is called the merge sort. Here is how it works:

1. Divide the list in two. Keep dividing each subset in two, until every subset is size 1 or 2.
2. Sort each subset of 2 by putting the smaller member first.
3. Merge pairs of subsets together. Keep repeating until there is just one sorted list.

The way to merge two lists is to compare the first member of each list, and put the smaller one in a new list. Then repeat, and put the smaller one after the smaller member from the previous comparison. Keep repeating until the two groups have been merged into one sorted group. If one of the original two lists is empty, then you can append the other list to the end of the merged list.

The following table shows the maximum number of comparisons necessary according to the number of elements in the list.

Merge Sort

Elements	Maximum Comparisons
1	0
2	1
4	5
8	17
16	49
32	129
64	321
128	769
256	1,793
512	4,097
1,024	9,217
2,048	20,481
4,096	45,057
8,192	98,305
16,384	212,993
32,768	458,753
65,536	983,041
131,072	2,097,153
262,144	4,456,449
524,288	9,437,185
1,048,576	19,922,945
2,097,152	41,943,041

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

My apologies for the dated reply on this. I wrote the following before the election.

First, I’m going to get rid of that other 6%, who are either undecided or will waste their votes on a third party candidate or "none of the above," which is an option in Nevada. Some may disagree with this assumption. To be honest, another reason for ignoring them is the math gets more complicated with more than two candidates. So, after rounding, that would leave us with 306 votes for Angle, and 281 votes for Reid, for a total of 587 in the sample.

I’m going to use the standard normal approximation to answer this question. If I were going to be a perfectionist, then I would use the T distribution, because the actual mean and variance are not known. However, in my opinion, a sample size of 587 is perfectly fine for the normal distribution.

Sample size = 306+281 = 587.

Angle sample mean is 306/587 = 0.521295.

The estimated standard deviation of the mean is (0.521295 × 0.478705 / (587-1))^0.5 = 0.0206361.

Angle’s share above 50% is (0.521295-0.5)/0.0206361 = 1.031917 standard deviations.

According to the normal distribution, the probability of Reid finishing 1.031917 standard deviations above expectations is 0.151055. This can be found in Excel with the function NORMSDIST(-1.031917). So Angle’s probability of winning is 1-0.151268 = 84.89%.

To create a 95% confidence interval, note that the 2.5% point on either side of the Gaussian curve is at 1.959964 standard deviations from the mean. This can be found in Excel with the function NORMSINV(0.975). As already noted, the estimated standard deviation of the sample mean is 0.0206361. So there is a 95% chance that either candidate will get within 0.0206361×1.959964 = 0.040446 standard deviations of the poll results. So Angle’s 95% confidence interval is 0.521295 +/- 0.040446 = 48.08% to 56.17%.

I’m told it would be mathematically incorrect to phrase that as "Angle’s share of all Angle/Reid votes has a 95% chance of falling between 48.08% and 56.17%." That was how I originally phrased my answer, but two statisticians recoiled in horror at my wording. To paraphrase their response, they said I had to use the passive voice, and say that "48.08% and 56.17% will surround Angle’s share with 95% probability." To be honest with you, it sounds the same to me. However, they stressed that the confidence interval is random and Angle’s share is immutable, and that my original wording implied the opposite. Anyway, I hope the frequentist statisticians out there will be satisfied with the second wording.

The "margin of error" is half the difference between the two ends of the 95% confidence interval. In this case (56.17% - 48.08%)/2 = 4.04%.

As a follow-up, here are the actual results:

Reid: 361,655
Angle: 320,996
Other: 21,979

So, not counting the "other" votes, Reid got 53.0% and Angle 47.0%. That is a comfortable 6% win for Reid. It begs the question of why the poll was so far off. Was it chance? Did voters change their minds? Or was it a bad poll to begin with? I leave those questions to the reader (I hate it when textbooks say that).

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

You’re confusing the mean and the median. Let’s look at my situation as an example. I’m a 45-year-old male. My life expectancy is 78.11 years, yet I have a 50.04% chance of making it to age 80.

My age at death will be like throwing a dart at this graph. Notice how the left tail is a lot fatter than the right. That means my probability of death right now is quite low. However, as I get older, the probability of death in the next year will keep getting higher. For example, for a 45-year-old male the probability of surviving to 46 is quite high at 99.64%. However, at age 85 the probability of making it to 86 is 89.21% only. It is like nature slowly pushing a knife in your back. At first it probably won’t kill you, but with each passing year, the odds slowly increase that it will. However, once you get to the late seventies nature says enough with the games and really starts shoving it in.

So if a lot of 45-year-old men throw darts at this graph, 49.96% will hit between 45 and 79, and 50.04% will hit between 80 and 111. However, the lucky half who make it on the right side of the graph will probably not live much past 80. Once a male reaches 80 he can expect to live 7.78 more years only. Meanwhile, many in the unlucky half who don’t make it to 80 will die much younger than that. So it is the many young deaths that pull down average life expectancy.

For a similar situation, consider a die numbered 10, 20, 30, 31, 32, 33. The average is 26, yet there is a 2/3 chance of rolling more than that.

As an example of how the mean and median are different, suppose we add two more deaths to the sample. One death at 46 and one at 81. The probability of making it to 80 doesn’t change, but the average life expectancy at age 45 would go down.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Yes, it will, after e^100,000 -1 seconds. See my mathproblems.info site, problem 206, for two solutions.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Good question. To answer it I considered a Toyota Highlander, a vehicle I am thinking of purchasing. The retail cost of the standard hybrid model is $37,490. For the same four-wheel-drive vehicle non-hybrid the cost is $29,995. So the hybrid engine adds $7,495 to the cost.

The gas mileage of the hybrid is 28 mph, both city and highway. The mileage of the non-hybrid is 17 city and 22 highway. Let’s take the average at 19.5.

The general formula for the number of miles to break even is h×m_h×m_r/(g×(m_h-m_r)), where

h = Additional cost of the hybrid.
g = Cost for a gallon of gas.
m_r = Mileage for non-hybrid (the "r" is for a regular car).
m_h = Mileage for a hybrid.

The following table uses this formula to find the break even point for various prices of gas from $2 to $5 per gallon.

Hybrid Break Even Point

Cost of Gas	Number of Miles
$2.00	240,722
$2.25	213,975
$2.50	192,577
$2.75	175,070
$3.00	160,481
$3.25	148,136
$3.50	137,555
$3.75	128,385
$4.00	120,361
$4.25	113,281
$4.50	106,987
$4.75	101,357
$5.00	96,289

So, at the current price of $3.00 per gallon here in Vegas, you would need to put more than 160,481 miles on the vehicle to come out ahead. This does not consider other expenses that may be associated with a hybrid, such as the expensive replacement cost of batteries, nor any perceived green points for consuming less fossil fuel.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

1/2.

If we used keno as a comparison, everybody would have 40 genes, each represented by a keno ball. However, each ball would have unique number. When two people, who are not related, mate it is like combining 80 balls between the two of them into a hopper, and randomly choosing 40 genes for the offspring of the mating.

So when you were conceived, you got half the balls in the hopper, and the other half were wasted. When your brother or sister was conceived he/she got half from the balls drawn when you were born, and half that were not drawn. So you are 50% genetically identical. Much for the same reason that if 40 numbers were drawn in keno, two consecutive draws would average 20 balls in common.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Let’s let c stand for the number of chairs made per hour, and t the number of tables. Revenue per hour will be 10×c + 20×t.

The 10 saws result in 600 minutes of sawing per hour. We were given that it takes a chair 10 minutes of saw time, and a table 5. So that limits hour hourly production to:

(1) 10c + 5t <= 600

The 6 lathes result in 360 minutes of lathing per hour. We were given that it takes a chair 5 minutes of saw time, and a table 5. So that limits hour hourly production to:

(2) 5c + 5t <= 360

The 18 sanding machines result in 1080 minutes of sanding per hour. We were given that it takes a chair 5 minutes of saw time, and a table 20. So that limits hour hourly production to:

(3) 5c + 20t <= 1080

The following graph shows the three constraints placed by the three sets of machinery. The factory may produce any combination of chairs and tables that is under all three lines. The question is where under the three lines results in the greatest revenue.

It stands to reason that the answer would be the intersection of two lines, make all chairs, or make all tables. So let’s find where the lines intersect. First, let’s find where equation (1) and (2) intersect. We can change the <= expression to just =, to use the machines to their maximum potential.

(1) 10c + 5t = 600
(2) 5c + 5t = 360

Subtract (2) from (1):

5c = 240
c = 48

Plugging 48 for c into equation (1):

10×48 + 5t = 600
5t = 120
t = 24

So, equations (1) and (2) meet at 48 chairs and 24 tables.

Next, let’s find where equations (2) and (3) meet:

(2) 5c + 5t = 360
(3) 5c + 20t = 1080

Subtracting (2) from (3):

15t = 720
t = 48

Putting that into (2) or (3) we can solve for c, which is 24.

So, equations (2) and (3) meet at 24 chairs and 48 tables.

We don’t need to bother finding where equations (1) and (3) meet, because we can see from the graph that where the saw and sanders lines meet is outside of the lathe constraint.

It is also possible that making chairs only is the right answer. The graph shows that the saws is the biggest limitation to making only chairs. From equation (1) if we put in 0 for the number of tables we get c=60.

Another possibility is making tables only. The graph shows that the sanders will be the biggest limitation. Putting in 0 chairs into equation (3) we find we can make no more than 54 tables.

The following graph shows the total revenue for each viable answer. Remember, the revenue is $10 per chair and $20 per table.

Total Hourly Revenue

Chairs	Tables	Revenue
0	54	$1,080
24	48	$1,200
48	24	$960
60	0	$600

So, the greatest revenue possible is $1,200, from making 24 chairs and 48 tables. Note that this would leave some saws unused part of the time. This question is also raised and discussed in the forum of my companion site MathProblems.info, problem #28.

Here is the answer and solution (PDF).

For discussion about this problem, please visit my forum at Wizard of Vegas.

For more discussion on this question, please visit my forum at Wizard of Vegas.

It would be easily to estimate the answer as 1-(51/52)⁵² = 0.63568648. However, estimates are so intellectually unsatisfying. So, let's find an exact solution!

Step 1: To start, consider the number of ways the second deck can be ordered where the first card is number 1. The answer is the number of ways to order the other 51 cards, which is 51! = 1551118753287382280224243016469303211063259720016986112000000000000.

Any card can match the first deck, so we must do this for all 52 cards. This gives us 52*51! = 52! combinations where at least one card matches.

Step 2: However, step 1 double counts every situation where two cards match. For example, if the first two cards were 1 and 2, we would have counted the 50! ways to arrange the other cards twice, once for 1 as the first card and a second time for 2 as the second card. The number of ways to choose 2 cards out of 52 is combin(52,2) = 1326. For each combination of two cards, there are 50! = 30414093201713378043612608166064768844377641568960512000000000000 ways to order the other cards. Thus, for step 2 we need to subtract out combin(52,2)*50! = (52*51/2!)*50! = 52!/2! combinations.

Step 3: Next, consider the situation where the first three cards in the random deck are 1, 2, and 3 in order. There are 49! ways to order the other 49 cards. We would have counted them three times in the initial step of counting for at least one matching card. Then we would have subtracted out all combin(3,2)=3 ways to choose 2 out of these three cards in the second step. So this situation would have been counted 3-3=0 times, so we need to add them back in. There are combin(52,3) such situations of choosing at least 3 cards that match. So we need to add back in combin(52,3)*49! = 52*51*50*49!/3! = 52!/3! combinations back in.

Step 4: Next, consider the situation where the first four cards in the random deck are 1, 2, 3, and 4 in order. There are 48! ways to order the other 48 cards. We would have counted them four times in the initial step of counting for at least one matching card. Then we would have subtracted out all combin(4,2)=6 ways to choose two out of these four cards in step 2. Then we would have added all combin(4,3)=4 ways to choose 3 out of these four cards. So we are at 4-6+4=2 ways each such situation would have been counted. So we need to subtract one of those ways out, so that each situation will be counted once. There are combin(52,4)*48! = 52*51*50*49*48!/4! = 52!/4! such situations that need to be added back in.

We will keep doing this, alternating adding and subtracting to correct for double counting.

In the end, the number of situations where at least one card in the random deck matches the ordered deck = combin(52,1)*51! - combin(52,2)*50! + combin(52,3)*49! - combin(52,4)*48! ... - combin(52,52)*1! = 52!/1! - 52!/2! + 52!/3! - 52!/4! ... - 52!/52! = x = 333239808909468890675694068318655265019682314241643033726180828783.

There are 52! = y = 527177615496365219422618541545122659969212453861982208000000000000 total ways to order 52 cards.

Thus, the answer is x/y = 0.6321205588285576784044762298

The probability that there are no matches is 1-(x/y) = 0.3678794411714423215955237702.

If this number looks familiar, it should. 1/e = 0.3678794411714423215955237702.

So, the answer can be VERY closely estimated as 1-(1/e).

Acknowledgements

Mathematical calculations were done in Pari/GP

This problems was asked and discussed in my forum at Wizard of Vegas.

Click the button below for the answer.

Here is my solution. (PDF)

This question is asked and discussed in my forum at Wizard of Vegas.

I got this problem from Presh Talwalker of Mind Your Decisions.

This question is asked and discussed in my forum at Wizard of Vegas.

Good question. Here are my answers:

Here is my solution to both (PDF).

This question is asked and discussed in my forum at Wizard of Vegas.

Here is my solution (PDF).

This question is asked and discussed in my forum at Wizard of Vegas.

 

Here is my solution (PDF).