threesome

Originally posted by luskin
Well for one thing it's far more plausible, coherent and logically consistent than anything you or forkedknight have put up so far.

Oh and by the way the post of yours that I was referring to was not the one with the smiley, and you surely know that. You are becoming increasingly desperate in your attempt to worm out.

OK, this is getting ridiculous. Thank you for your puzzler.

Originally posted by luskin
Another artificially contrived 'solution'.

If A finishes ahead of B 2 out of 3 and B finishes ahead of C 2 out of 3, is it really likely that A will only be ahead of C 2 out of 3???

NO. Only IF you force it!

Let me remind you of your opening post

" any type of three way contest you care to imagine"

Originally posted by forkedknight

No less contrived than your model

This is exactly the point you can't or won't understand.

A model is supposed to represent the problem. My model was not contrived to produce a particular desired result. It was designed to represent the situation where you have one player who is twice as good as another player, who in turn is twice as good as a third player. That's the reason for the 4:2:1 ratio that is built into it. It is, if I may say so myself, a simple and elegant way of representing the relative abilities of the three players.
On the other hand your model is an ugly monstrosity that bears absolutely no resemblance whatever to the puzzle I posed. Of course it was not intended to resemble or represent anything at all. It's sole purpose is to churn out the particular result you chose.
Wolfgang's model at least has the decency of being somewhat simpler and prettier, but it doesn't accurately represent the relative abilities because it operates on a 2:1:1 ratio instead of the necessary 4:2:1

Originally posted by wolfgang59
Let me remind you of your opening post

" any type of three way contest you care to imagine"

Yes, it was probably a mistake to leave the door so wide open. Still, I think you knew that I had in mind a game such as golf.

Originally posted by luskin
This is exactly the point you can't or won't understand.

A model is supposed to represent the problem. My model was not contrived to produce a particular desired result. It was designed to represent the situation where you have one player who is twice as good as another player, who in turn is twice as good as a third player. That's the reason for the 4:2:1 ...[text shortened]... the relative abilities because it operates on a 2:1:1 ratio instead of the necessary 4:2:1

Perhaps you will like this model better, it fulfils all the requirements and I hope the ball ratio is more pleasing to you ....

randomly drawing balls from a barrel as per your model .

4 balls labeled C; 2 labeled B; and 1 A

The result of each 'game' is then determined by the following process.

1) Randomly draw one ball - this is the losing ball - 3rd place.
2) Remove from the barrel any others with the same label as the one just drawn
3) Randomly draw a second ball - this is the runner up - 2nd place
4) Remove from the barrel any others with the same label as the one just drawn
5) The remaining ball in the bag is the winner - 1st place

This model does fulfil the requirements that
A wins over B 2/3 and
B wins over C 2/3

Probability of BAC = 4/7*1/3=4/21

And probability of ACB = 2/7*4/5=8/35

Again it proves NOTHING (except that proofs from models are invalid)

Originally posted by luskin
Yes, it was probably a mistake to leave the door so wide open. Still, I think you knew that I had in mind a game such as golf.

I'm a bit confused why anyone would have reasonably thought you had in mind a game such as golf, given that your opening post implied we should be considering only games where there is "no possibility of ties or draws." Ties or draws are possible in golf.

Originally posted by wolfgang59
Perhaps you will like this model better, it fulfils all the requirements and I hope the ball ratio is more pleasing to you ....

randomly drawing balls from a barrel as per your model .

4 balls labeled C; 2 labeled B; and 1 A

The result of each 'game' is then determined by the following process.

1) Randomly draw one ball - this is the losing b ...[text shortened]... of ACB = 2/7*4/5=8/35

Again it proves NOTHING (except that proofs from models are invalid)

Sorry about the delay; I'm waiting to see if forkedknight might want to follow you down this new hole you've dug for yourself. Somehow, I think he'll be wise enough to see the trap and not go anywhere near it.
Take a few days or even a week to ponder over just what it is you've actually done there. I'll give you a hint in 3 or 4 days, if you haven't figured it out by then, or if someone else hasn't pointed it out to you.

Originally posted by LemonJello
I'm a bit confused why anyone would have reasonably thought you had in mind a game such as golf, given that your opening post implied we should be considering only games where there is "no possibility of ties or draws." Ties or draws are possible in golf.

Well golf was mentioned earlier in the thread that's why. You haven't looked far enough back. As for ties, they can always be resolved in just about any game as far as I know, and in golf they are resolved usually by playoff if needed. And for the purpose of this puzzle it is needed. It just makes things simpler if you assume it can't happen in the first place.

Originally posted by wolfgang59
Perhaps you will like this model better, it fulfils all the requirements and I hope the ball ratio is more pleasing to you ....

randomly drawing balls from a barrel as per your model .

4 balls labeled C; 2 labeled B; and 1 A

The result of each 'game' is then determined by the following process.

1) Randomly draw one ball - this is the losing b ...[text shortened]... of ACB = 2/7*4/5=8/35

Again it proves NOTHING (except that proofs from models are invalid)

That "last becomes first" type of logic could be useful for many things. Want to change the result of an election? Just simply decree that the candidate who gains the least number of votes will be defined as the winner! That's the equivalent of what you've done in this masterpiece of ill logic.

Now here is another way to solve the problem, which hopefully will be more suited to your train of thought.

In saying that the probability of A>B=2/3 ;(and B>C=2/3) is the same as saying that :
1) A does better than B in 2 games to 1 or;
2) A is superior to B twice as often as B is superior to A or;
3) A is twice as good a player as B
Word it however you like, but in the language of math it amounts to

A/B= 2/1 ; and B/C= 2/1

And then (A/B)(B/C)=A/C= 4/1
And there you have the missing piece of the puzzle
i.e. the probability of A>C=4/5

So A/B=2 and A/C=4
or A=2B=4C
This can now be expressed in terms of each players probability of 'winning' any given game.

p(A)=4/7
p(B)=2/7
p(C)=1/7

I'm sure you can take it from there, and finish off the necessary calculations to determine the probability of a game resulting in B1st ; A2nd ; C3rd
Good luck

Originally posted by luskin
That "last becomes first" type of logic could be useful for many things. Want to change the result of an election? Just simply [b]decree that the candidate who gains the least number of votes will be defined as the winner! That's the equivalent of what you've done in this masterpiece of ill logic.

Now here is another way to solve the problem, wh ...[text shortened]... lations to determine the probability of a game resulting in B1st ; A2nd ; C3rd
Good luck[/b]

You know nothing about logic.
You know nothing about Math.
I cannot be bothered trying to educate you.

Oh and btw ... there are electoral systems where candidates are eliminated through getting the most votes.

Originally posted by luskin
As a model for simulating the problem, I used the old idea of randomly drawing balls from a barrel.

4 balls labeled A; 2 labeled B; and 1 C

The result of each 'game' is then determined by the following process.

1) Randomly draw one ball
2) Remove from the barrel any others with the same label as the one just drawn
3) Randomly draw a second ball
...[text shortened]... 3

By that method we get the probability of BAC = 2/7*4/5 =8/35

And p of ACB=4/7*1/3=4/21

p(A>B) = 2/3
p(B>C) = 2/3

BAC --> A<B and B>C
p(A<B,B>C) = 1/3 * 2/3 = 2/9
A is smaller than B, and B is bigger than C, so there are two options,
BAC and BCA.
p(BAC) + p(BCA) = 2/9

Edit:
p(ABC) = 2/3 * 2/3 = 4/9
p(CBA) = 1/3 * 1/3 = 1/9
p(BAC) + p(BCA) = 1/3 * 2/3 = 2/9
p(ACB) + p(CAB) = 2/3 * 1/3 = 2/9

Originally posted by Thomaster
p(A>B) = 2/3
p(B>C) = 2/3

BAC --> A<B and B>C
p(A<B,B>C) = 1/3 * 2/3 = 2/9
A is smaller than B, and B is bigger than C, so there are two options,
BAC and BCA.
p(BAC) + p(BCA) = 2/9

Edit:
p(ABC) = 2/3 * 2/3 = 4/9
p(CBA) = 1/3 * 1/3 = 1/9
p(BAC) + p(BCA) = 1/3 * 2/3 = 2/9
p(ACB) + p(CAB) = 2/3 * 1/3 = 2/9

Ah, that won't do. Next try:

There are six solutions:
ABC
ACB
BAC
BCA
CAB
CBA

A > B in 2/3 of the games, so p(ABC)+p(ACB)+p(CAB)= 2/3
B > C in 2/3 of the games, so p(ABC)+p(BAC)+p(BCA)= 2/3
Conclusion:
p(ACB)+p(CAB) = p(BAC)+p(BCA)

A > B in 2/3 of the games, so p(ABC)+p(ACB)+p(CAB)= 2/3
B < C in 1/3 of the games, so p(CBA)+p(ACB)+p(CAB)= 1/3
Conclusion:
p(ABC)=p(CBA)+ 1/3

Originally posted by wolfgang59
Whoops
amended:-

CBA 1/9 0/9
ABC 4/9 3/9
ACB 1/9 1/6
BAC 1/9 1/6
BCA 1/9 1/6
CAB 1/9 1/6

So I reckon (using my assumptions) BAC lies beyween 1/6 and 1/9

I think your draw-and-tossgame proves these numbers are wrong.

Then what are the bounds. Since it is most likely that A > B and B > C, nothing can be more likely than ABC. Same reasoning for CBA, nothing can be less likely.

Let's say p(CBA) = z
and p(ACB)+p(CAB)=2a
This means:
(z + 1/3) + 2a + 2a + z = 1
2z + 4a = 2/3

z, p(CBA), can at lowest be 0, so:
0 < p(CBA)
1/3 < p(ABC)
Now what is the upper bound.
z < a < z + 1/3, and 4a + 2z = 2/3,
Upper bound for z when a is as lowest as possible, so a=z
6z = 2/3
z = 1/9

0 < p(CBA) < 1/9
1/3 < p(ABC) < 4/9

Wolfgangs model also proved, that one of the other can be 0.
The upper bound is (with p(ABC) at its minimum):
2/3/2= 1/3

0 < p(ACB) < 1/3
0 < p(BAC) < 1/3
0 < p(BCA) < 1/3
0 < p(CAB) < 1/3

Originally posted by Thomaster
Same reasoning for CBA, nothing can be less likely.

Playing monopoly, player A drew a card. The player with most money, player C and he play a game. Player B throws a dice, if it's 3 or higher, player C gives player A €50,- and if it's 1 or 2, player A gives player C €50,-

Now there is 4/6 = 2/3 chance for ABC and 2/6 = 1/3 for CBA.

0 < p(CBA) < 1/3
1/3 < p(ABC) < 2/3
0 < p(ACB) < 1/3
0 < p(BAC) < 1/3
0 < p(BCA) < 1/3
0 < p(CAB) < 1/3

and

p(ACB)+p(CAB) = p(BAC)+p(BCA)
p(ABC)=p(CBA)+ 1/3

Edit:
Apparantly, p(ACB), p(BAC), p(BCA), and p(CAB) can be lower than p(CBA), but they can't be higher than 1/3. With 1/3 being the minimum of p(ABC) only 2/3's left. Since p(ACB)+p(CAB) = p(BAC)+p(BCA), both sides are at most (2/3)/2 = 1/3.