Probability - Puzzles - Wizard of Odds
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Probability - Puzzles
I am interested in finding out some specific information on the odds of rolling dice. If you have 6 dice and roll them all at once, the odds of rolling all ones are 1 in 46,656. My question is what are the odds of rolling one to five ones. I am really interested in finding out the formula that should be used to calculate this type of problem.
Ken
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%.
Number of Ones in Six Dice
Ones
Total
1.0000000
Eight golfers went to a new course. The caddy master put 8 bags on four carts at random. The golfers put 8 marked golf balls in a hat. The balls were thrown in the air. The 2 closes balls to each other were partners. In every case the partners’ golf bags were already on the same cart. What is the probability of that the golf bags were paired up correctly before the throw?
Anonymous
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.
A friend sent me this , I was wondering if there was a formula as to how this works.
Anonymous
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.
Let n bet any integer. Express n as d0*1 + d1*10 +  d2*100+  d3*1000+ ... + dn*10n, where dn is the first digit, dn-1 is the second, and so on.
n = [d0 + d1 + d2 + ... + dn ] + [d1*9 +  d2*99+  d3*999+  ...+ dn*999...9 ( a number with n nines)]
n = [d0 + d1 + d2 + ... + dn ] + 9*[d1*1 +  d2*11+  d3*111+  ...  dn*111...1 (a number with n ones)]
9*any integer is evenly divisible by 9. So if d0 + d2 + d2 + ... + dn , 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*[10x - 10y] (where x>=y and n is the digit) = +/-n *10y * (10x-y - 1) = 10y * (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.
Great site. I refer to it often as a gambler with an interest in probability and statistics, but this question actually pertains to my work. My HR Department insists that I rate my small staff (5 people) on a bell curve--one in the top 5% of all employees, one in the next 20%, one in the next 50%, one in the next 20%, and one in the bottom 5%. The company has approximately 5000 employees. What is the probability of such a small sample size fitting this distribution?
Anonymous
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%.
To the fellow who asked the question about order statistics ( column #100 ), I have two quibbles: one small and one large. Your method failed to make a finite population correction, which I grant is trivial with 5000 employees, but it certainly wouldn’t have been had there been 20 employees!
More importantly, however, you implicitly assume that managers have no effect on their employees. Suppose good managers, through judicious hiring and firing, or through above-average motivational skills, raise the average level of their employees. Without accounting for this effect we will either upward- or downward-bias the resulting probabilities. I’m sure you knew this, but I am sensitized to it because I do a lot of calculations like this in discrimination cases and failure to adjust for things we can adjust for (in this case a group-specific effect) can often lead people astray.
Anonymous
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.
Do there exist any famous unsolved problems facing gaming mathematicians? Like a Fermat's Last Theorem in the gambling world. If so, would you please share an example.
Anonymous
Good question. I can't think of any.
How does this work: www.1800gotjunk.com/genie/ ?
Anonymous
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.
I am about to take a professional licensure examination. The regulations provide that:
The examination shall consist of 7 subjects.
For each subject, 60 multiple choice questions shall be asked.
Each multiple-choice question shall have four possible answers, but only one correct answer.
In order to pass, an examinee must obtain a general average of at least 75% and must not have a grade lower than 65% in any subject.
My question is, if an examinee merely guesses all his answers, what is his chance of passing the exam? Stated differently, what is the probability of passing the exam by sheer luck?
Anonymous
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 a university’s football team has a 10% chance of winning game 1 and a 30% chance of winning game 2, and a 65% chance of losing both games, what are their chances of winning exactly once?
Anonymous
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%.
On the game show Let’s Make a Deal, there are three doors.  For the sake of example, let’s say that two doors reveal a goat, and one reveals a new car.  The host, Monty Hall, picks two contestants to pick a door.  Every time Monty opens a door first that reveals a goat.  Let’s say this time it belonged to the first contestant.  Although Monty never actually did this, what if Monty offered the other contestant a chance to switch doors at this point, to the other unopened door.  Should he switch?
Anonymous
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 .
I disagree with your answer to the Monty Hall question in the November 19, 2004 column. Assuming the car is behind door one there are actually four possibilities as follows, where the prize is behind door 1.
Player picks door 1 --> shown 2 --> switch to 3, lose
Player picks door 1 --> shown 3 --> switch to 2, lose
Player picks door 2 --> shown 3 --> switch to 1, win
Player picks door 3 --> shown 2 --> switch  to 1, win
As you can see the probability of winning is 50% whether you switch or not. Furthermore it just goes against common sense that switching would be better.
Anonymous
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.
With five different toppings to choose from, how many different pizzas can you make, with any number of toppings?
Anonymous
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.
I saw in the paper last week that the latest earthquake that ravaged Indonesia hit on December 26th. It also showed that of the eight deadliest earthquakes to hit over the last 100 years, three of them have been on December 26th. I was wondering what the odds are of having three massive quakes hit on the same day knowing these facts: Earthquakes of this magnitude (8.0 or larger) happen only once per year. The last big quake was exactly one year ago, 12/26/03 in Iran (back to back probabilities?) I look forward to hearing from you.
Steve A. from Fort Collins, CO
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.
How many eggs do you start with if each day you sell 1/2 the eggs plus 1/2 an egg; after 3 days you have zero eggs? At the end of each day, the number of eggs is a whole number.
Anonymous
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 + ?
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.
Imagine a island that is inhabited by 10 people, and the politics is such that each day an islander is chosen at random to be chief for exactly one day; after the day has elapsed another islander is chosen at random (so the same islander who was just chief has a 1/10 chance of being chief again). The question to be solved: on average, how many days would have to elapse before each islander would have been chief at least once?
Anonymous
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.
The radius of the circle is 1.  The triangle is equilateral.  Find the area of each colored region.
Anonymous
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.
Say you won a contest where at halftime of an NBA game you got to shoot a free throw and if you make it you win $1 million. Further, you can keep shooting free throws, double or nothing, till you miss or choose to stop. If you’re a 75% free throw shooter, when would you stop? Can you ever? At some point the money starts to mean less and less. What would you do?
Pete from New York
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 remember that if 22 people are in a room the odds are even that two will celebrate the same birthday [month and day, not year]. I have forgotten how to do the math to prove this. Could you please provide it.
Dean from Bainbridge Island, WA
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%.
The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a standard deviation of $45. What is the probability that a randomly selected teacher earns more than $525 a week? I can’t remember how to calculate a probability from just the mean and SD without the population.
Sue from Queen Creek
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.
Two people are playing rock paper scissors. It is presumed the game doesn’t involve strategy. If you are playing ’best of 3’, and player A wins the first round, what are the odds that player B will win the game?
Andrew from Pewaukee,WI
Player B would need to win the next two (not counting ties) so the probability is (1/2)*(1/2) = 1/4.
Hi, I thought I would ask you this since I cannot find it anywhere on the web. I hope you answer this; What are the odds of existing? Whether it be on Earth or somewhere else in the universe?  It’s not a gambling question but an answer we should all know so we can appreciate what kind of odds we beat just being alive!
Andreas from Edmonton
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%.
Suppose a hotel has 10,000,000 rooms and electronic 10,000,000 keys.  Due to a computer mistake each key is programmed with a random code, having a 1 in 10,000,000 chance of being correct.  The hotel is sold out.  What is the probability at least one customer has a working key?
Danny from London, U.K.
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.
There are 75 multiple choice questions in an exam. Each question contains 4 possible answers only 1 is correct. The exam pass mark is 50%.  What are the chances of passing the exam by guessing each answer?
Wendy from London
1 in 635,241.
For in-running betting, if a tennis player has a chance "p" of winning a game, what chance does he have of winning a set?
Mike from Perth
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
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
You are in a boat with a rock, on a fresh water lake.  You throw the rock into the lake.  With respect to the land (shore), does the water level increase, decrease, or stay the same? My co-workers think that the water level will stay the same.
David
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.
How does this work?
20000x+250+2y-250 = 20000x+2y
(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.
There is a drawing for a $27,000 car, with tickets being sold at six for $500.00, or one for $100.00.  68 tickets have been sold, and tomorrow is the deadline for purchase.  I know that for a 50% probability of winning, I must spend $5666.44, and for a 66.66% of winning, I must spend $11,332.88 (Right?).  How much should I spend (or how many tickets must I buy) to virtually ensure that I "win" the car? (90%? 95%?)  Is this raffle worth playing, or must I spend the cost of the car?
Annette from Boise
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)
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.
I’m trying to compare the cost of replacing an old refrigerator now in order to save on electrical costs, vs. waiting until it dies to replace it.  I can calculate how much cheaper it is to run the new fridge vs. the old one: $37/yr., that’s easy.  But how do I factor in the cost of the new fridge?  Say the new fridge costs $425.  I can’t say that *all* of that $425 is a new expense, because I’ll have to replace the old fridge *someday*, if not now, so I’ll have that new-fridge expense at some point anyway.  Let’s say that a typical fridge lasts 14 years and my old fridge is 9 years old, so if I replaced it now I’d be replacing it in 5 years.  I tried to make a two-column table, comparing the cost of keeping the current fridge for 9 years and then replacing it, vs. replacing it now, but I didn’t know how to make an apples-to-apples comparison because I didn’t know for how far into the future to consider the costs, and because the fridges are replaced in different years.  How do I compare the economics of replacing now vs. replacing later?  By the way, this isn’t for my own situation, because my current fridge is probably 30 years old.  It’s for, uh, a friend.
Spanky McBluejay from Austin, TX
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.
If there are three people, then What is the probability that at least two persons have the birthday on the same date.
Manish from New Delhi
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.
Five persons are in a room. What is the probability that at least 2 of them were born in the same birth month?
Amy
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.
We’ve been given a challenge at work -- just for fun, and none of us can work it out. A farmer has 5 trailers full of sheep.  Four of the trailers contain sheep weighing 39kg and the 5th trailer contains sheep weighing 40kg. All of the sheep are identical.  He goes to the market. He wants to find out which of the trailers contains the sheep weighing 40kg, and he can only use the large weighing scales once!!! How does he do it? Please help, it is driving us all mad at my work place -- it’s a vet’s!!
Becca
The answer is at the end of the column.
Answer to sheep question
Anonymous
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.
On an airplane with 180 seats, what are the odds of me sitting next to the good looking girl I see who will be on the same flight?
Ted H. from Salt Lake City
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.
Three logicians are playing a game.  Each must secretly write down a positive integer.  The logician with the lowest unique integer will win $3.  If all three have the same number, each will win $1.  The logicians are selfish, and each wishes to maximize his own winnings.  Communication is not allowed.  What strategy will each logician follow?
Matthew from Fort Wayne, IN
The answer will appear in the next column.
I read that Warren Buffet (the world’s third richest man) complained that he only paid a 17.7% federal tax rate, while his secretary paid 30%.  This seems outrageous to me.  Can you comment?
Joe from Nashville
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)
Thank you for your entertaining collection of math puzzles.  My girlfriend and I came up with this variation on the pirate puzzle.  What if all the pirates are of equal rank, and in each round the proposer of the division is chosen by lot?  In this variation, assume that each pirate’s highest priority is to maximize his expected amount of coins received.  I have what I think is the solution, but perhaps you’d like to try your hand at it first.  Thanks again.
Jon S
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 need help with a puzzle called Eternity II.  The prize for solving the puzzle is a staggering $2,000,000, a considerable amount of money to me. Here's a link to an interview, including the game maker himself, Christopher Monckton (former adviser to Margaret Thatcher, among many things).  The game is obviously not really about gambling at all, but despite this fact, maybe you could add a word or two on your web page about it.
The game maker brags about the puzzle to be insolvable, in that link given above.  I'm starting to think that he's actually right, and that he himself is the only one who will eventually become rich from selling that (ridiculous but fascinating) game.  How would you, being a mathematician and all, go about solving this type of puzzle?
Robert
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.
Suppose the distance between two cities is 1000 miles.  In zero-wind, a plane can travel at 500 mph.  Will it take longer to make the round trip with no wind, or a direct 100 mph tailwind in one direction, and equal head wind the other way?
Kevin from Portland, OR
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.
I recently entered a raffle where there are 7,033 prizes and they say the odds of winning a prize are 1 in 13.  I bought 5 tickets.  What are my actual odds of winning something?  Also, there are 40 big prizes.  What are my odds of winning a big prize?
Anonymous from Mesa, AZ
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.
I have a puzzle that I’ve been trying to solve for a few months, with absolutely no progress. Time permitting, I’m hoping you can indulge me, as it’s been keeping me up at night :-).  Anyway, in the glossary of Beyond Counting -- Exhibit CAA , three sequences of numbers and letters are given as the glossary entry for "Magic Numbers." One of these numbers even graces the book’s cover, so I assume they’re of some importance.  Do you have any thoughts?
Anonymous
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.
I am curious to know what became of that Eternity II puzzle challenge. Was it solved? Are you still working on it?
Matt from Las Vegas
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.
I read with fascination the Wizard's blog about Arnold Schwarzenegger’s veto letter.  My question has to do with the governor’s ridiculous but predictable response. The governor stated that it was just a ’wild coincidence’. Notwithstanding the overwhelming circumstantial evidence (The bill’s sponsor and letter’s addressee was the person who had hurled insults at the governor a week earlier), do you have an estimate of what the odds are of an exactly seven-line letter spelling this phrase by chance? I think taking into account the letters used, it will be even more improbable than just assigning a 1 in 26 chance to each. It doesn't seem like U, Y, and especially K are common word-starting letters.
pocketaces
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