That wouldn't give much answer on the probability, because there is SOME probability that when I choose a random point on that line, I will at one time choose an integer.
But would 10/infinity be correct because in geometric sense, i'm not pointing out on the line graph with a pencil (that has a finite dimension)?
Confusion.
Last edited by nahya^^; Apr 2nd, 2003 at 11:23 AM.
Geometric probabilities are more subtle than one might expect.
As stated, the problem is ill defined. One must specify a process leading to the choice of a point.
No matter what process is chosen, the probability is obviously so close to zero that you might as well assign the value of zero.
To illustrate a problem with geometric probabilities, 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?
I can think of three processes leading to a random chord, each resulting in a different probability. There is no paradox involved, merely an ambiguous definition.
Processes are random. Sets or sequences of numbers are not random, although there are algorithms for generating pseudo random numbers suitable for use in simulating random processes. Terms like random chord or random point are ambiguous in the absence of a defining algorithm or process.
BTW: MathCad software has almost 20 pseudo random number generators, indicating that one size does not fit all when it comes to simulating random processes.
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the probability of a random real number being rational is 0. this is very much the same problem, and the probability is 0.
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The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!
Unless of course this is physically 'choosing' a point. In this case, there is a limit to how specific a choice can be. I.e. if we can only choose accurately to 0.05cm, and the thing in 10cm long, then the probability = 3/(10/0.05)= 3/200 = 1.5%
However, if choosing can be any number, then it is like we are choosing a real number. In that case it is zero.
BTW, what are the three 'random chord' things.
The only one i can think of is choosing a random point, choosing another random point, joining them to make a chord. In this case, it is 1/3 of it being longer than the side of the triangle.
One random process is to draw a tangent to the circle at any point on the circumference, then pick a random angle between 0 and 180 to make the chord. This gives 1/3 as the probability. 0-60 and 120-180 chords are shorter. 60-120 chords are longer.
Another process is to imagine a polar cordinate system centered on the circle. Pick a random angle (0-360) and a random distance between zero and the radius of the circle. Use the resulting random point as the center of the chord. This gives 1/4 as the probability. Points inside a circle of radius r/2 result in longer chords. Circle of radius r/2 has 1/4 the area of the full circle.
Pick a random point on a diameter and construct a chord perpendicular to the diameter at that point. Points within r/2 of the center result in a longer chord. This gives 1/2 as the probability.
I have been very sloppy in the above descriptions, for which I apologize. Think a little about the sloppy descriptions, and it should become clearer. A diagram which includes the inscribed triangle helps. For the last method, two triangles, making a star with 6 points helps.
Live long & prosper.
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I thought about that last one for a bit, and got a bit confused.
You are saying that, out of all chords perpendicular to a line (in this case the diameter chosen) that 1/2 of them are longer than the side of the triangle. I tried that star-of-david drawing, and yes, it seems to be right.
Then, looking at the same case, I reasoned that if you pick a point (as in case 1), then any of these perpendicular chords matches up with a chord you can draw from that point, so that case 3 is exactly the same as case 1, and the probability is 1/3.
However, i managed to spot the paradox.
Consider the semi-circle with the base as the chose diameter. Now, you are saying that if you choose any point in the 2 middle quarters of the diameter, a chord perpendicular to the diameter through this point will be longer than the sides. If you chose any point in the outer two quarters, then these chords will be less than the side length. => 50%-50%
However, although you are choosing the middle half of the diameter, are you choosing half of the points on the circumference?? NO!! That's the odd bit. If you draw the lines up from the 1/4 and 3/4 marks from the diameter to the circle, the lines cut the diameter in the ratio of 1:2:1, but cut the circumference in ratio 1:1:1. Really weird.....
Any suggestions?? Maybe the number of points on a circle is another infinity??
Sql_lall: You are correct, you are not choosing half the points on the circumference. Dividing the diameter into equal pieces does not result in the corresponding parts of the circumference being equal.
However, the process is random and results in a chord. The probability is 1/2.
I mentioned this process last because it does not seem as natural as the other two. When the question was proposed to me many years ago, I thought of the first two methods, but not the third.
The poser of the problem described the third method after getting answers from several of us. He said that it is a classic problem illustrating the ambiguities of geometric probabilities.
He said that most people come up with one of the first two methods and consider the problem a done deal, not realizing the ambiguity of the problem.
Probability in general is more subtle than most people realize. For example, consider making up the following set of three dice.
The red die has 8, 1, 6 (opposite faces have same number).
The yellow die has 3, 5, 7
The blue Die has 4, 9, 2
The numbers are chosen from the rows of a magic square.
Suppose we play a game using two of these dice. I bet on one die, you bet on the other. The dice are rolled. If my die has a higher number, you pay me $1.00 and vice versa.
When you calculate the probabilities, you discover the following.
The relationship is more probable than is not transitive like greater than, which (to me) is counterintuitive.
BTW: The 3X3 magic square is the subject of a wonderful example of an isomorphism between tic-tac-toe and a simple game played with nine playing cards. It is the simplest example of an isomorphism I have ever seen, yet it conveys the concept well.
Place the Ace through the nine of one suit face up. Two people play. The object of the game is to get a triplet of cards whose total is 15. Players take turns picking a card, which is not replaced. Very little practice is required to play the game perfectly, with the result being a tie.
The game is equivalent to playing tic-tac-toe on the 3X3 magic square. A preschool child can play tic-tac-toe, but might not be able to play the other game. He could play the other game by putting the cards in a tic-tac-toe arrangement.
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The relationship is more probable than is not transitive like greater than, which (to me) is counterintuitive.
I think its transitive. By definition the probability of the event X is n(X)/n(all events), a rational in other words, and rationals are transitive. I think we discussed this earlier too. Red, Yellow and Blue are not events, Red beating Blue, etc.. are. I can understand some people think its counterintuitive now even though I don't..
Agreed though is that "The probability of Red winning" is illdefined without knowing anything about the opponents.
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writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
Kedaman: It seems very obvious to me that a person betting on the red die would win over a person betting on the yellow die in a long term game. A casino would be most happy to back a game in which it won 5/9 or 55.55555% of the time. This is better odds than blackjack, craps, or roulette. A person betting on the red die versus the yellow die would be expected to bankrupt his opponent if the game was played long enough.
A person betting on the yellow die versus the blue die would also win 5/9 of the time, in the long run bankrupting his opponent.
A person betting on the blue die versus the red die would also win 5/9 of the time.
To me the above clearly indicates the following.
Red die is likely to beat yellow die
Yellow die is likely to beat blue die.
Blue die likely to beat red die.
In no way is the above similar to 7 > 5 > 3 implying 7 > 3, which indicates that greater than is a transitive operator or relationship, while is likely to beat is not transitive.
Perhaps your insight into probabilistic situations is far better than mine, but I was surprised when I analyzed the above game. What do you mean by the following?
"The probability of Red winning" is illdefined without knowing anything about the opponents.
In a game determined by probabilities, why must one know anything about the opponent, other than his ability to pay if he loses?
Is there something poorly defined about the above game? One person bets that the number rolled on the red die will be higher than the number rolled on the yellow die. Is this not a situation governed by probability, rather than the behavior of the players? Similarly for betting on the other possible pairs of dice.
Live long & prosper.
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Now, this only means that in a 1 on 1 battle, the stronger one will win.
If both armies have groups of both, then it will be even.
Similarly, if all three dice are thrown, there is a probability that one particular one will beat the other, but this is regardless of what the third one is.
Yep, the fun of strategy games. If it wasn't for this all armies would have just built the strongest types of weapons, but in the real world we have a variety of specialised weapons for specific targets.
A weapon is a tool, you can't evaluate its success without a target. Similarly a game between Red and Yellow is more likely to be won by red.
Sometimes its easier to visualise probabilities using a diagram like the attached one. This diagram shows all events as sets, {red beating yellow} here would be the union of {games won by red} and {games between red and yellow}.
The event of red winning is not defined here, but if we assumed that red only plays 9 games with yellow and 9 with blue then we can define red's winning
Use
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
Sql-Lall: In contests of intelligence or skill between humans, I would expect situations in which A beats B, B beats C, and C beats A. Humans are extremely complex.
The probability dice are a simple situation, which seems counter intuitive even after you analyze it.
Suppose you encountered some raw statistics indicating that the red die beats the yellow die 55% of the time and that the yellow die beats the blue die 55% of the time. Further suppose that there was no description of the numbers on the dice, and that the blue versus red statistics had been lost.
Would you bet on the red or the blue die?
Kedaman: Your diagrams seem to be lacking something. For example: Games-won-by-red has an area which relates to neither yellow nor blue, but red has no other opponents in this problem. What does that extra area represent? Another example: Games-between-red-and-yellow indicates 5 wins for red and 4 for yellow. There can be no ties, what does the rest of that circle represent?
Live long & prosper.
The Dinosaur from prehistoric era prior to computers.
Eschew obfuscation!
If a billion people believe a foolish idea, it is still a foolish idea!
VB.net 2010 Express
64Bit & 32Bit Windows 7 & Windows XP. I run 4 operating systems on a single PC.
Your diagrams seem to be lacking something. For example: Games-won-by-red has an area which relates to neither yellow nor blue, but red has no other opponents in this problem. What does that extra area represent?
It represents games won by red against neither blue or yellow, which if you now define none, is an empty set.
But its not really relevant to the problem is it?
There can be no ties, what does the rest of that circle represent?
This is definitely an empty set - its still an event but as you define no ties, this event can't happen. Maybe it would be intuitive to draw figures that have area proportional to their probability, but with more than two circles you get problems, the circles have only symbolic meaning, but it illustrates boolean calculus and set theory well in my opinion. I put numbers in the events that are defined and relevant to the problem, and as you can see all of them are intersections of a game and a winner, the winner of the game.
Use
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.