Counter Intuitive probablity problems.

The following struck me as counterintuitive when I first encountered them.

First: Consider a randomly selected group of people. What is the probability that two will have the same birthday? Assume 365-day years, and that they need not be the same age. Before calculating, try guessing at the following.

Suppose there are 20 people in the group.
Suppose 25.
Suppose 35.
Suppose 180.

Next: Consider a circle with an inscribed equilateral triangle. What is the probability that a random chord will be longer than a side of the triangle?

Last: Given A > B > C, it is obvious than A > C if we are dealing with real numbers. Suppose that A, B, & C are probabilities and > stands for is more likely than? Is A more likely than C?

It would be nice if you labeled guesses as guesses and provided examples or analyses if you are not guessing.

I will watch the look count and read posts for a day or so before posting analyses of the above.

for the first problem.
I suppose you could interpret the qwestion in several ways, but here's mine. I've done it earlier another way but i think this way is more cool.

for exactly one group of exactly two persons to have the same birthday
n=amount of persons

some examples:

1:
0

2:
1/365 = 0.00273972602739726027397260273972603

3:
1/365 * 364/365 * 3 =0.0081966597860761869018577594295365

4:
1/365 * 364/365 * 363/365 * 6 = 0.0163034931635378402486266667009411

conclusion:
365^(-n-1) * 364!/(366-n)! * n nCr 2

for the last case, is the probabilities measured in numbers? And how do you define "is more likely than" then? well i don't quite understand the counterintuitive in it..

Kedaman: A total of 7 is more likely than a total of 12 when rolling standard casino dice.

With such dice 7 is more likely than 10 and 10 is more likely than 12. This seems to imply that 7 is more likely than 12, which happens to be true.

The mathematical property indicated by > is called transitive, because A > B > C implies A > C.

Is the probablility relationship is more likely than also transitive? If not, can you give an example? If so, can you prove it?

Give me a clue about the second one.

These are my thoughts so far:

1)If the circle has radius 1 then the equilateral triangle has sides of length sqrt(3).
2)The distribution of the length of the chords is not uniform.

Restated problem: Consider a circle with an inscribed equilateral triangle. What is the probability that a random chord is longer than a side of the triangel?

Clue: Think of some process which would generate a random chord. Then work out the probabilities relating to that process.

All the processes I have thought of result in a simple rational probablity like 1/4, 1/3, 1/2.

I'm not quite sure about casino dice, is it that you roll 2 dice and count the sum of the results?
In that case I don't think the sum should be called probability. The probability of getting 7 is
6/6^2=1/6, and so you could convert any result from 2 to 12 to a probability factor, and compare then using the aritmetic > and say that any sum A "is more likely than" sum B is equivalent with probability A > probability B. This thus implies sum A "is more likely than" sum B "is more likely than" sum C == probability A > probability B > probability C.

As only then length of the chord is relevant and the startpoint of the the chord doesn't matter as long as the end point is random, I place the startpoint at one corner of the triangle and the endpoint may be anywhere on the circle, so the probability is uniform all over the cirle. In the scope between the legs from the startpoint corner of the triangle, the length of the chord is greater. The probability of the endpoint to be between the legs is 1/3 due to symetry of the triangle.

Oh yer, well done keda. I shoulda thought of that (duh).

/David/

See the Great Rod Puzzle thread.

The birthday problem is a probability problem which is almost impossible to compute directly, but which can be handled by computing the probability of an event not happening. This is an often used technique. The logic goes as follows.

• Think of choosing people for the group in order, so we can refer to the first, the second, et cetera.
  
• 365/365 is the probability of the first person not matching another’s birthday.
  
• 364/365 is the probability of the second person not matching the first person’s birthday, since it can occur on any day other than the first person’s birthday.
  
• 363/365 is the probability of the third person not matching either of first two birthdays.

Thinking about the above leads to the following.

Probability(No match for a group of N) = 365! / (365 - N)! * 365^N
Probability(Match) = 1 - Probability(No match)

Example for a group of 5 follows.

P(No match in group of 5) = 365*364*363*362*361 / 365^5 or .972864
P(5) = 1 - .972865 or .027136

The following are some computed match probabilities.

• P(10) = .116948
• P(20) = .411438
• P(22) = .475695
• P(23) = .507297
• P(30) = .706316
• P(40) = .891232

90% of the time a group of 40 will have at least two people with the same birthday. With a group of 23, the chance of matching birthdays is just over 50-50 or a little better than even money.

Most people (including myself) find it surprising that the chance of matched birthdays is so high for groups of 25 to 40.

The strange aspect of the random chord problem is that it seems possible to prove that the probability is 1/4, 1/3, or 1/2.

Random is a tricky concept with traps for the unwary.

I will not go into the proofs of the geometric statements made in the following discussion. I apologize for not being able to draw pictures, which make the analyses much easier to follow.

Kedaman came up with an analysis that results in a probability of 1/3. His analysis is equivalent to the following, which I find easier to visualize.

• Draw a tangent to the circle at a vertex of the inscribed triangle.
  
• Imagine a straight line centered at that vertex and rotating.
  
• Stop the rotating line at some random position.
  
• If the chord formed by the now stationary line is between the sides of the triangle, it is longer than a side.
  
• Note that the tangent line and the sides of the triangle form three 60 degree angles, giving 1/3 as the probability.

Another analysis considers a circle inscribed within the equilateral triangle, resulting in a large circle through the 3 triangle vertices, and a smaller triangle inside and tangent to the sides of the triangle.

• The radius of this smaller circle is 1/2 the radius of the circumscribing circle, and therefore has 1/4 the area.
  
• Pick a random point inside the circumscribing circle as the midpoint of the random chord.
  
• If the random midpoint is inside the smaller circle, the chord is longer than a side of the triangle.
  
• Since the area of the smaller circle is 1/4 the larger circle, this results in 1/4 as the probablilty.

For the final analysis, pick a diameter of the circle. Due to symmetry, any diameter will do.

• Pick a random point on the diameter, and construct a chord through that point perpendicular to the diameter.
  
• If the random point on the diameter is in the middle half of the diameter, the chord will be longer than the side of the triangle.
  
• This results in 1/2 being the probability.

Why three different probablilities?

The problem is that random chord is an ambiguous phrase. Each of the above processes are random, and each results in constructing a chord. If you use one process to construct the chord, you get the probability associated with that process. If you use another process, you get a different probability.

To repeat, the processes are random, but the term random chord is not defined in the absence of the process that results in its construction.

Kedaman: Sorry that I did a bad job of using casino dice to explain probabilities. I should have said something like the following.

Casino dice are cubes with a different number (1 through 6) on each face. The numbers are actually indicated by dots (like the numbers on dominos). In the game of Craps, a pair of dice are thrown, and the total of the face up numbers is noted.

Obviously the total must be 2-12 inclusive. The probability of a total of N is usually specified as P(N).

For casino dice rolls, it is easy to calculate that P(7) > P(10) > P(12), which implies that P(7) > P(12). As indicated by the probabilities, it is obvious that a total of 7 is more likely than a total of 10, which is more likely than a total of 12, and a total of 7 is also more likely than a total of 12.

The question to be considered is the following.

Quote:

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It is given that A, B, C are events and P(A), P(B), & P(C) are the associated probabilities of the events occurring.

If the probabilities indicate that event A is more likely than event B, and event B is more likely than event C, is it always true that event A is more likely than event C? If the events are rolling various totals with casino dice, the implication is valid.

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It just so happens that the implication is not true for all events which have associated probabilities. The following is an example using casino-like dice constructed as follows.

• The dice have different colors to aid in quickly distinguishing which is which.
  
• Each die has only three numbers. Opposites faces have the same number.
  
• The numbers 8, 1, 6 are used for the red die.
  
• The numbers 3, 5, 7 are used for the blue die.
  
• The numbers 4, 9, 2 are used for the yellow die.

Note: The above numbers are ordered to make it obvious that they are from the rows of a magic square. Imagine the numbers put into a 3X3 array and check the row, column, and diagonal totals.

Now imagine a game played with pairs of the above dice. One die is assigned to one player and another is assigned to the second player. The third die is not used. The game is simple, roll the pair of dice and note the face up number on each. The owner of the die with the higher total is the winner.

There are three possible pairs of three dice and three possible games. If you calculate the probabilities, you discover that the Red die is likely to beat the blue die, the blue die is likely to beat the yellow die, and the yellow die is likely to beat the red die. In each case the probability of winning is 5/9.

The event of which the probability can be calculated for, needs the comparation of two dice, not distinguished for a single die. For a single dice you could calculate the probability of winning against any randomly choosen of the other dice and get 5/9 * 1/2 + 4/9 * 1/2 which is 1/2. Let's instead say > stands for "A's chance to win against B" is more probable", doesn't mean A and B are different events, but this comparation states a single event of which the probability is higher than 1/2. Putting them like A > B > C states two events, of which both probabilities are higher than 1/2. A > C now is a third event, more or less independent from both A > B and B > C, fails because of the characteristics of A and C. Using mathematical terms of probabilities A > B > C would be: P(A,B)>1/2 && P(B,C)>1/2 where && is boolean and operator and P(A,B) is a user defined function based on characteristis of the parameters that returns a probability factor. Now althought this works out, the illusion of our "user defined" > to be similar to the aritmetic > shouldn't be that consistent when you look at it. They aren't even close.

Quote:

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It is given that A, B, C are events and P(A), P(B), & P(C) are the associated probabilities of the events occurring.

If the probabilities indicate that event A is more likely than event B, and event B is more likely than event C, is it always true that event A is more likely than event C? If the events are rolling various totals with casino dice, the implication is valid.

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I think the implication has to be valid unless you try to undermine the logic behind it. Your sample didn't fit the case.

Kedaman: Your previous post was too complex for me to understand, although I can tell that you disagree with the 3-die game.

Do you disagree with the following?

Quote:

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The Red die is more likely to win that the Blue die. The Blue die is more likely to win than the Yellow die, but this does not imply that the Red die is more likely to win that the Yellow die.

In the long run for this game: Red beats Blue and Blue beats Yellow, but Yellow beats Red.

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Even if you do not agree with the way that I expressed the original problem and the later explanation, does the above not seem counterintuitive?

Does the above suggest that more likely than is not transitive like the =, >, & < operator?

BTW: If I thought that all of my posts were short, concise, and crystal clear, I would mention that I quite often find your posts long, confused, and difficult to understsand.

And I appreciate that they want me to explain further.

You have to excuse me, I'm often hardly understood when it comes to explaining my visions, most of the time I think it's because I think differently than others. For instance, I don't agree that the way you expressed the original problem seemed counterintuitive. However, many fans of strategy games (a computer game genre) would agree with me, since that is one of the main concepts that makes strategy games fun to play.

I'll try to explain a bit further, it's probably going to be lengthy, but I'll try to make it as clear as possible.

First I think English as well as any language are intended to be logically structured, this means they "could" be in fact converted into a mathematical language and further into a polymorphic programming language and compiled to machine code. I'm not going into how such program would function, but I'm picking an example of how a English phrase could be seen as piece of code in a polymophic programming language:

Quote:

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I can fly.

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First, This is an boolean expression, it can be validated if there are such capabilities. Second "can" is a binary operator, which means it takes two arguments, the person/thing that "can" and what it "can". Operators are actually similar to methods (functions), in their functionality, but are used so that it may be easier to read. "I can Fly" would look like: can(I,Fly). "I" and "Fly", are objects. "I" would refer to the owner of the statement and "Fly" would be a functionality object. For can(x,y) to limit the characteristics of the arguments you pass you implement strong typing, which means x has to be a person, or something that "can" do something and y has to be something that "can" be done. Finally, "can" returns a boolean value, and therefore the expression is boolean. This is possible to do in C++ and most other object oriented languages.

You might wonder what I'm aiming at. I'm aiming at the complexity of languages. Operators are used in programming languages, they are unnessesary but makes things easier to read, but also introduce some complexities. In math we have such things as a > b > c, which means something totally different in a programming language, since > is a binary operator, it would evaluate a > b to a boolean value and then try to compare it with c, doesn't make sense. a > b > c in math, is a single operator with a dynamic amount of arguments, and is equivalent with a > b && b > c. The transitivity is born due to the nature of dynamic operators, and does not exist in programming languages. In English, and other spoken languages there are even more complex operators, and such often cause unintended ambigueties. However I think we can oversee that if we suppose a decent language without ambigueties.

Anyway, as you saw how a operator used in math could be converted to a programming language, you could convert any logical language into a programming language as well.


"more likely than" doesn't fit the case, basically because it can be converted to

Quote:

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(probability)(EVENT)arg1 > (probability)(EVENT)arg2

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the parentesis around "probability" indicates type casting, which means the events you pass as arguments are converted into probabilities. Note that the arguments has to be passed as events, not probabilities, otherways they are converted to events (if that is possible) or the the expression is invalid. Note also that you can't pass a probability; "1/3 is more likely than 1/4" doesn't sound right to you does it?

I suggested

Quote:

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A's chance to win against B is more probable"

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would fit the 3-die game situation better, maybe "A is more likely to win than B" would sound better, but this is somewhat invalid English, it doesn't directly imply you A is playing against B. It should be "A is more like to win AGAINT B". Assigning ">" dynamic operator this meaning, A > B > C would be converted into:
and further into:

Quote:

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((Probability)WinAgainst(A,B) > 1/2) && ((Probability)WinAgainst(B,C) > 1/2)

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where WinAgainst is a function that returns the EVENT of A winning B.

However the complexity doesn't end here, this is just optimized version from the following:

Quote:

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#define MoreLikely(X,Y) ((probability)X > (probability)Y)

MoreLikely(WinAgainst(A,B),WinAgainst(B,A)) && MoreLikely(WinAgainst(A,B),WinAgainst(B,A))

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This might also just be why I don't see it counterintuitive :)

• I agree it is not transitive, but
• I don't agree it should be expressed as "more likely than" simply because there's too much omited.
• I don't agree it's counterintuitive whether everything is expressed explicitely or not, but that's another story.
• I think it shouldn't be thought as counterintuitive when it is expressed explicitely.

In case there's something that passed your eyes without being understood (hopefully nothing got missunderstood) just ask.

last quote should of mine should be

Quote:

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#define MoreLikely(X,Y) ((probability)(EVENT)X > (probability)(EVENT)Y)

MoreLikely(WinAgainst(A,B),WinAgainst(B,A)) && MoreLikely(WinAgainst(B,C),WinAgainst(C,B))

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and this quote of yours got unquoted

Quote:

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It is given that A, B, C are events and P(A), P(B), & P(C) are the associated probabilities of the events occurring.

If the probabilities indicate that event A is more likely than event B, and event B is more likely than event C, is it always true that event A is more likely than event C? If the events are rolling various totals with casino dice, the implication is valid.

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the 3-die game example differs from the casino dice example, in that the A B and C are not events but characteristics of the dice, and that the events of winning are functions of two of these, resulting in only two, instead of three of them. So there's also only two probabilities involved.

Kedaman: You are unique! Everyone, but you, who has ever understood the 3-Die game as described has viewed it as counterintuitive.

I think strategy gamers would think like me.

In Dune 2000, a strategy game by westwood (based on the movie) you use infantry and armored vehicles to try to get to the opponents base and destroy it. Vehicles are more expensive than infantry but also much faster and stronger, you would think you could just use them and train no infantry, but that wouldn't be fun would it? Now there are two types of weapons, one of which one is most effective against infantry but no use against vehicles, machine guns. And the other is rockets, most effective against vehicles, but not much effective against infantry. Since each vehicle or infantry, commonly called units, could only carry one type of weaponry, two types of infantry and two types of vehicles were used:

Name: Weaponry: Unit Type:
Infantry Machine gun infantry
Trooper: Rockets infantry
Trike: Machine gun vehicle
Quad: Rockets vehicle

In the game, things were a bit more complicated, but basically this scheme was used.

When you saw a Quad approaching your base, you would send a trooper to destroy it, Infantry was sent against troopers, Trikes against Infantry and Quads against Trike's. So the richness of diversity and strategic use of it lead to fast victory.

In most Strategy games, things are more complicated than this. Stealth, speed, weapon range and accuracy, armor, manouverability, size, crossable terrain and expense are often involved. Unit types could be infantry, light vehicle, tank, airborne and seaborne units. But basically there's a specific type of units which is mostly wounerable against a weapon particulary designed to fight against them.

The 3-die game is a similar case, the characterity of the die is not single dimensioned, as a number always loose against a higher number, the win-rate depends on each numbers closeness to the opponents number, and in which order. A exreeme high number winning against a extreeme low number is not particulary effective if you compare to a number one step smaller. And so the dice have different win-rate depending their opponents closeness of numbers.