Probability - General Questions

With the exception of rare positive expectation opportunities in blackjack and video poker, yes, that is what I'm saying.

Saying the odds of something happening are x to y means that the event in question will happen x times for every y times it doesn't happen. To make the conversion let p be the probability of some event. The odds could also be expressed as (1/p)-1 to 1. Lets look at an example. The probability of drawing a full house in five-card stud is 0.00144058. This could also be represented as 693.165 to 1.

Assuming the cells in the grid are chosen at random, then the odds of winning any one quarter would be 1/100. Assuming each quarter was an independent event, which they aren't, the odds of winning all four quarters would be (1/100)⁴ = 1 in 100 million.

I don't like using probabilities in that form but they are generally used in this kind of syntax, "The odds against drawing a royal flush are 649,739 to 1." That means there are 649,739 ways you can't draw a royal flush and 1 way you can. In your examples 12 to 1 is a probability of 1/13, or 7.69%, and 3 to 2 is 2/5, or 40.00%, so the 3 to 2 is the better chance of winning.

You are right, that article is incorrect. The probability of a 500-year flood in a period of x years is 1-e^-x/500. So the probability of at least one 500-year flood in 50 years is 9.52% and in 100 years is 18.13%.

Let p be the probability of the favorite winning. If -160 is a fair line then:

100*p - 160*(1-p) = 0
260p = 160
p = 160/260 = 8/13 = 61.54%.

So the expected return on a $145 bet at a -145 line would be (8/13)*100 + (5/13)*-145 = 75/13 = $5.77. So the player advantage would be $5.77/$145 = 3.98%.

Let’s define t as the true money line with no house edge and a as the actual money line. Following are the formulas for the player’s expected return:

A is negative, t is negative: (100*(t-a) / (a*(100-t))
A is positive, t is positive: (a-t)/(100+t)
A is positive, t is negative: (a*t + 10000)/((t-100)*100)

So in your case your expected return is 100*(-160 -(-145))/(-145*(100-(-160))) = 3.98%.

First, I am publishing this because author gives permission to do so at the bottom. This is a good example that correlation does not necessarily mean causation. It is easy to look back in time and find lots of coincidences. To make a case for anything a hypothesis should be stated before any evidence is gathered.

Follow-up (November 13, 2004): Another reader pointed out that this map started out as a joke but turned into an urban legend. As this link points out the hurricane paths in the graphic were simply not accurate and the actual hurricanes hit many Gore counties. It just goes to show you shouldn’t believe everything you read, especially on the Internet.

Thanks for the kind words. To be honest nobody cares much about click-through rates any longer. So don't feel obligated to click through the banners if it is just for show. To answer your question, in the United States the probabilities are very close to 50.5% boy and 49.5% girl. Assuming no other information is known by the betting community the player advantage on the boy bet would be .505*1.05 - .495 = 3.53%. It could be that somebody with inside knowledge is betting on a girl. Another theory is that some people incorrectly believe you can tell the gender by the shape of the mother’s belly, and these people are betting on a girl. Personally I’m going to leave this one alone.

No. Using this actuarial table from the CDC (Centers for Disease Control), the probability of a 72 year old white male making it to age 76 is 85.63%. That is about a 1 in 7 chance of death. The survival rate can be found by dividing the birth cohort at age 76 of 57,985, by birth cohort at age 72 of 67,719, from the white male table on page 14. The table used is a called a "period life table," which assumes 2003 rates of mortality will not change in the future, and is the most commonly used kind of actuarial table. A perfectionist might want to use a 1936 cohort life table, but I don’t think it would make much difference.

p.s. After posting this answer I have received several comments that my response did not take into consideration John McCain’s individual health situation. Working against him is being a cancer survivor. Working in his favor is access to the best medical care money can buy, he is obviously still in good shape mentally and physically for a 72-year old, and longevity, as evidenced by the fact that his mother is still alive. However, I never intended to factor in this information. It was Matt Damon who quoted actuarial tables, which is what I was referring to. All I am saying is that for the average 72-year old white male, the probability of surviving four more years is 86%. If forced, I would predict John McCain’s odds are even better than that.

Please see my companion site MathProblems.info, problem number 210, for the answer and solution.

Not counting pushes, the probability of getting at least 88 wins out of 139 picks is 0.00107355, or 1 in 931. This is pretty underwhelming. I'm sure there are 930 other animals out there who did worse that nobody writes about. For more information on Princess, read the article N.J. camel predicts Giants over Patriots at ESPN.com.

I hope you're happy; I spent hours on this.

To answer the question it is important to quantify behavior under the Chelsea Handler red head hypothesis. Here are my assumptions.

1. A red head will never mate with another red head.
2. The female will always choose the male to mate with.
3. Everybody will mate and each mating will produce the same number of children.
4. The female redheads will get first dibs at a mate, choosing randomly among the non-red heads.
5. The female carriers (with one red-haired gene) will choose a mate randomly among the men left over by the red heads.
6. The negative females (neither red-haired gene) will chose randomly among the men left over by the red heads and the carriers.

First, let's review the chip stacks.

I hope you're happy. To answer this question, I bought a big roll of tickets at the Office Depot. Then I put 500 of them in a paper bag, half folded in half, at about a 90-degree angle, and the other half unfolded. Then I had six volunteers each draw 40 to 60 of them one at a time, with replacement, as I recorded the results. Here are the results.

Good question! Here is my answer and cursory solution. Also see my solution in PDF form.

This is known as the Saint Petersburg Paradox.

It is indeed true that the expected win of the game is ∞, while at the same time the probability is that the coin will eventually land on tails, leading to a finite amount of money. The calculation of the expected win is:

Expected win = pr(1 flip)×2 + pr(2 flips)×4 + pr(3 flips)×8 + pr(4 flips)×16 + pr(5 flips)×32 + pr(6 flips)×64 + ... =

(1/2)¹ × 2¹ + (1/2)² × 2² + (1/2)³ × 2³ + (1/2)⁴ × 2⁴ + (1/2)⁵ × 2⁵ + (1/2)⁶ × 2⁶ + ...

= ((1/2)*(2/1)) ¹ + ((1/2)*(2/1)) ² + ((1/2)*(2/1)) ³ + ((1/2)*(2/1)) ⁴ + ((1/2)*(2/1)) ⁵ + ((1/2)*(2/1)) ⁶ + ...

= 1¹ + 1² + 1³ + 1⁴ + 1⁵ + 1⁶ + ...

= 1 + 1 + 1 + 1 + 1 + 1 + ... = ∞

Where this is paradoxical is the player must win a finite amount of money, but the expected win is infinite. How can that be?

This is probably not a very satisfying answer, but there are lots of paradoxes when it comes to ∞. This may cause me to get some angry emails, but what lets me sleep at night, despite such infinity paradoxes, is that I believe that ∞ is a mathematical or philosophical concept that is unproven to exist in the real physical universe. This concept or theory of infinity carries with it built-in paradoxes.

For those who disagree with this, please tell me anything that is proven to have infinite quantity or measurement. Please don't say a black hole has infinite density unless you have evidence of its size.

To answer the initial question of how much should one pay to play this game, we should keep in mind that happiness is not proportional to the amount of money. Personally, I was taught in economics classes, and believe, that the utility, or happiness, from money is proportional to the logarithm of the amount of the money. Under this assumption, if you increase or decrease any two people's wealth by the same percentage, other than an initial wealth of zero, then they both experience the same change in happiness. For example, if Jim's wealth suddenly increases from $1,000 to $1,100 and John's wealth suddenly increases from $10,000,000 to $11,000,000 they both experience the same increase in happiness, because in both cases their wealth increased by 10%. Assuming that the happiness from money is indeed proportional to the log of the amount, then the following table shows the most somebody should be willing to pay according to his wealth before paying to play.

First, there is no positional advantage to acting last. Since the players are kept in a sound-proof booth while any previous players play, order doesn't matter.

Second, there must be a Nash Equilibrium to the game where a strategy to stand with a score of at least x points will be superior to any other strategy. The question is finding x.

What I did was ask myself what would be the strategy if instead of a card numbered 1 to 100, each player got a random number uniformly distributed between 0 and 1 and look for the point x where a perfect logician would be indifferent between standing and switching. With that answer, it is easy to apply the answer to a discrete distribution from 1 to 100.

I'll stop talking at this point and let my readers enjoy the problem. See the links below for the answer and solution.

Answer for a continuous distribution from 0 to 1.

Answer for a discrete distribution from 1 to 100.

For my solution, please click here (PDF).

This question was raised and discussed in my forum at Wizard of Vegas.

That is the same as asking what is the probability that 14 random cards contain all 10 black cards. There are combin(10,4)=210 ways to choose 4 red cards out of the 10 in the deck. There is of course only one way to choose all ten black cards. There are combin(20,14)=38,760 ways to choose 14 cards out of 20. So the answer is 210/38,760=0.005418, or 1 in 184.57.

Let's look at the gold standard of video poker, 9-6 Jacks or Better to answer your question.

The first step is to modify my calculator to include a line item for all 13 four of a kinds. Here is that modified return table:

For a question so easy to ask, the solution is rather involved. The way I did it, you will need to know this integral.

Here is the answer and my solution (PDF).

It was actually rather easy, especially for a course in combinatorial mathematics at MIT. Here is the wording of the problem:

"Draw all homeomorphically irreducible trees of size n=10."

Here is my attempt to put it in plain and simple English.

Using only straight lines, draw all figures where the sum of intersections and dead ends equals 10. You may not have any closed loops. You may also not have two equivalent figures. Any intersection must have at least three paths leading from it.

What do I mean by "equivalent," you might ask? It means you can move the pieces, while leaving the intersections alone, any way you wish and it won't create any new figures.

Here is an example:



I'll give you a hint. Unlike the answer in the movie, there are ten of them. Will got only eight of them. See if you can match or beat Will Hunting.

Show Spoiler

I show my logic to coming up with all ten at my MathProblems.info site, problem 220.

• MATHEMATICS IN GOOD WILL HUNTING II: PROBLEMS FROM THE STUDENTS PERSPECTIVE -- Academic paper on the problem.
• THE GOOD WILL HUNTING MATH PROBLEM -- Discussion about the problem in my forum.

The answer and solution can be found in my page of Math Problems, problem 225.

As a former actuary, you asked the right person. Hopefully the Society of Actuaries will not consider my answer an abuse of the profession. That said, to answer your question I consulted a 2014 Period Life Table from my former place of work, the Office of the Chief Actuary of the Social Security Administration.

A period life table shows, among other things, the probability of death for a person of any given age and gender in 2014. Using that information I created the following table, which shows both the probability of death and expected points for all ages from 0 to 100 and both genders.

The table shows that the maximum expected points is for a 90 year-old-man at 1.645220.

This question is raised and discussed in my non-gambling forum, Diversity Tomorrow.

Good question! I was just wondering this when I saw some skinny soda cans at a gaming show, which held the usual 355 milliliters that the standard size does. Surely both can't be right (and don't call me Shirley).

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

I agree that you hear the variance of a game bandied around more than it's standard deviation, which I have always found a little annoying. The reason I think gamblers should care about the volatility of a game is to associate a win or loss to a probability for a session of plays. For example, what would be a 1% bad loss after 200 hands of blackjack. To answer that, you would use the standard deviation of blackjack, which is about 1.15, depending on the rules.

The specific answer to that question is 1.15 × 200^0.5 × -2.32635 (which is the 1% point on the Gaussian curve) = -37.83 units south of expectations. Don't forget that due to the house edge you can expect to lose something. If we assume a house edge of 0.3%, then after 200 hands you could expect to lose 0.003*200 = 0.6 hands. So a 1% bad loss would be 0.6 + 37.83 = 38.43 hands.

If the population and age distribution were stable, then if the divorce probability was truly 50%, then we would expect to see a ratio of one divorce to two marriages in any given period of time, given a large sample size.

However, the population isn't stable. From this graph, it looks like the U.S. population is growing by 10.71% per decade. That comes to 1.02% per year. Let's just say 1% to keep it simple.

Map source: U.S. Census

According to fatherly.com, the average length of an unsuccessful marriage is 8 years.

If you were observing a 1 to 2 ratio of divorces to marriages in the present, what would be the average probability that any given marriage ends in divorce?

The divorces we're seeing now were from marriages 8 years ago, when the population was 92.35% what it is now. Simple math suggests the true probability of divorce is 54.14%.

Let's check that.

First, according to the CDC, there are 6.9 marriages to 1000 in population per year. This figure is not relevant to the question at hand, but I think helps see the numbers involved.

Suppose the population 8 years ago was 300,000,000. That would be 0.69% * 300 million = 2,070,000 marriages in that year.

If 54.14% of them end in divorce eight years later, then we would be seeing 2,070,000 * 54.14% = 1,120,698 divorces in the present.

1,120,698 / 2,070,000 = 50% observed ratio of divorces to marriages in the present.

Lest anyone say it, yes, I know that not all divorces end in exactly eight years. However, all things considered, I say that the bottom line won't be far from my 54.14% true divorce rate.

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

Consider everybody choosing one at a time. As each person picks, there will be two types of situations:

1. The name of the one picking has already been picked.
2. The name of the one picking is still in the bin of names.

For any given picker, let's say there are n people left to pick.

If the name of the one picking has already been picked, there is a 1/n chance that the picker will choose close a loop involving his name. For example, let's say Amy is picking. Amy's name is already held by Bob, Bob's name is already held by Charlie, and Charlie's name is still in the bin. With n names still in the bin, there is a 1/n chance that Amy picks Charlie's name, closing a loop.

If the name of the one picking has not already been picked, there is a 1/n chance Amy picks her own name, closing a loop.

Either way, if the picker doesn't close a loop, she is joining part of another chain, which will eventually be closed by somebody else. Each chain should be counted once only, when it gets closed.

Thus the answer is 1/100 + 1/99 + 1/98 + ... + 1/1 =~ 5.187377518.

An estimate for any sufficiently large number of players, n, is ln(n).

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

It is easy to choose these two, as probably the two most famous:

• 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ... = π/4
• 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ... = π^2/6

That is true, with 23 random people, the probability of at least one pair of people have a common birthday is 50.73%. This ignores leap day and assumes everybody has an equal chance of being born on each of the other 365 days (which is not actually the case, spring and fall birthdays are slightly more common).

The tables in answer to your question are quote long, so I'll put them in spoiler tags. Click on the buttons for the answers.

There are 42 ways the dealer could break down $10. Here they are:

Mathematicians call these partitions. Here are the number of partitions for starting quantities up to 405, which is as high as my computer can calculate (2^64).

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

Let's start by looking at a graph of n (x-axis) by f(n) (y-axis).

As you can see, the limit approaches ∞ from the left and -∞ from the right. Since it doesn't converse to the same place from both sides, there is no limit.

However, let's answer the question without graphing. L'Hôpital's rule says that if the limit of f(x)/g(x) = 0/0, then lim f(x)/g(x) = lim f'(x)/g'(x). So, let's solve for f'(x) and g'(x).

f'(n) = ((ln(1-n) - sin(n)) d/dn = -1/(1-n) - cos(n)
g'(n) = (1 - cos²(n)) d/dn = sin²(n) d/dn

Let's use the product rule to solve for sin²(n) d/dn

sin²(n) d/dn = sin(n) × sin(n) d/dn =
sin(n) × cos(n) + cos(n) × sin(n) =
2sin(n)cos(n).

Next, let's solve for f'(n) and g'(n) at n = 0.

f'(0) = -1/(1-0) - cos(0) = -2.
g'(0) = 2sin(0)cos(0) = 0

So, f'(0)/g'(0) = -2/0 = -∞. Thus, the limit of the original function doesn't exist.

I would like to compliment the writers of Mean Girls for getting the math in this movie perfect. Even serious mathematical movies, like Good Will Hunting, often completely blow the math.

First, I'll go over the number of permutations. This means that not only do the numbers matter but also their order on the card. There are permut(15,5) = 15!/(15-5)! = 15*14*13*12*11 = 360,360 possible permutations for the B, I, G, and O columns. For the N column, the number of permutations is permut(14,4) = 15!/(15-4)! = 15*14*13*12 = 32,760. Thus, the total number of permutations of bingo cards is 360,360⁴ × 32,760 = 552446474061128648601600000.

Second, I'll go over the number of combinations. This means that the numbers matter but not their order on the card. There are combin(15,5) = 15!/(5!*(15-5)!) = (15*14*13*12*11)/(1*2*3*4*5) = 3,003 possible combinations for the B, I, G, and O columns. For the N column, the number of permutations is combin(14,4) = 15!/(4!*(15-4)!) = (15*14*13*12)/(1*2*3*4) = 1,365. Thus, the total number of permutations of bingo cards is 3,003⁴ × 1,365 = 111007923832370565.

On the show, Sheldon asks himself how may UNIQUE bingo cards exist. Based on the later incorrect formulas, I assume he means permutations. In other words, two cards with the same numbers but in different positions would both be unique.

The image above shows Sheldon's formula for the B, I, G, and O columns. He initially gets the formula right at 5! × combin(15,5). However, he incorrectly simplifies it to 15!/(15!-5)!. The second exclamation point shouldn't be there. It should read 15!/(15-10)!. However, he then gets back to the correct answer at 360,360.

We have exactly the same problem with the N column. The formula should be 15!/(15-4)!, not 15!/(15!-4)!. The second exclamation point ruins it.

The ironic thing is later in the Episode, Sheldon become obsesses with errors in the chronology of the Lord of the Rings, much as I'm obsessing with this.

Let's start by defining a couple variables:

• s = kg of salt in the tank
• t = minutes since the salt was dumped in the tank

We're given that 10% of the salt is drained away per minute. To put that in mathematical language:

ds/dt = (-10/100) × s

Let rearrange that to:

ds = (-10/100) × s dt

-10/s ds = dt

Integrating both sides:

(1) -10×ln(s) = t + c

Next, let's find the dreaded constant of integration. To do that, we're given that s = 10 when t = 0. Plugging that into formula (1) above gives us:

-10 × ln(10) = 0 + c

So c = -10×ln(10)

Putting that into equation (1) gives us:

(2) -10×ln(s) = t -10×ln(10)

The question at hand is how much salt will be in the tank at t=30. Solving for s when t=30:

-10×ln(s) = 30 -10×ln(10). Next divide both sides by -10...

ln(s) = -3 + ln(10)

s = exp(-3 + ln(10))

s = exp(-3) × exp(ln(10))

s = exp(-3) × 10

s =~ 0.4979 kg of salt.

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

The key to problems like this is in setting them up. I recommend trying to boil the problem down to as few unknowns as possible. In this case, we can express the unknown distances on the square to just three, as follows:

It's easier to deal with rectangles than triangles. Given we know the area of three triangles, we can double the size and double the areas. That gives us:

• ab=10
• ac=16
• (a-b)(a-c)=14

Let's factor (a-b)(a-c):

a² - ab- ac + bc = 14

a² - 10 - 16 + bc = 14

(1) a² + bc = 40

Let's express b and c in terms of a, to get this down to a single variable:

b = 10/a

c = 16/a

Substituting those values for b and c in equation (1):

a² + (10/a)*(16/a) = 40

a² + 160/a² = 18

Next, let's get rid of that a² in the denominator by multiplying everything by a².

a⁴ + 160 = 40*a²

a⁴ - 40*a² + 160 = 0

Let's define a new variable y = a²

y² - 18y + 32 = 0

Next, let's solve for y using the quadratic formula:

y = (40 +/- sqrt(1600-640))/2

y = (40 +/- sqrt(960))/2

y = (40 +/- 8*sqrt(15))/2

y = 20 +/- 4*sqrt(15)

The area of the entire square is a², which conveniently is equal to y. From the equation above, if the +/- is negative, then y = apx 4.5081, which is obviously wrong, since we know the area is at least 20, not even including x. So that area of the square must be 20 + 4*sqrt(15).

The three triangles we're given have an area of 5+7+8=20. Subtracting that from the total area of the square leaves us the area of x: 20 + 4*sqrt(15) - 20 = 4*sqrt(15) = apx 15.4919.

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

Note my t-shirt in this picture. The movie theater cashier complimented it when I went to see Uncut Gems. I thanked her by torturing her with this problem, only with triangles of areas 2, 3 and 4. After the movie, I checked on her and she still hadn't solved it, but seemed to be trying. So I wrote out the following solution at the Suncoast bar for her. She actually seemed to appreciate it. I think that young woman will go far in life.

It will be very useful to know the infinite series in the following hint.

For the answer only, click the following button.

Click the button below for the solution.

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

Click the button below for the answer.

Click the button below for the solution.

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

Acknowledgement: I got a variation of this problem from Presh Talwalkar of Mind Your Decisions.

This problem is adapted from a similar problem by Presh Talwalkar of Mind Your Decisions.

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

Click the button below for the answer.

Click the button below for the solution.

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