threesome

Originally posted by luskin
Golf is a good example. In fact it [b]is the game I had in mind.

If you had 3 golfing friends,A,B and C, who had been playing as a threesome once or twice a week for the last 50 years, and they tell you that A beats B 2 out of 3, and B beats C 2 out of 3, and then they ask you to estimate how often they have finished in the BAC order, would you [i]real ...[text shortened]... e time and CBA 1/3??? I don't think so, unless you know that B just hates being first or last.[/b]

We give up.

Give us any solution for all 6 outcomes where they play a total of 900 games.

Originally posted by wolfgang59
We give up.

Give us [b]any solution for all 6 outcomes where they play a total of 900 games.[/b]

A solution for 900 games would involve fractions, and so a bit messy. I can give you a neat and tidy solution for 945 games, or 840, or any multiple of 105. Does that give any clue?

Originally posted by luskin
A solution for 900 games would involve fractions, and so a bit messy. I can give you a neat and tidy solution for 945 games, or 840, or any multiple of 105. Does that give any clue?

I can give a solution for 900 games ... but choose whatever total you want ... I just want to see a solution of yours!

Originally posted by wolfgang59
I can give a solution for 900 games ... but choose whatever total you want ... I just want to see a solution of yours!

OK here is the solution for 945 games

ABC 360
ACB 180
BAC 216
BCA 54
CAB 90
CBA 45

Originally posted by wolfgang59
I can give a solution for 900 games ... but choose whatever total you want ... I just want to see a solution of yours!

I can give a solution for 900 too, but as I said earlier it will contain fractions, not all integers as is possible with multiples of 105.

Originally posted by luskin
OK here is the solution for 945 games

ABC 360
ACB 180
BAC 216
BCA 54
CAB 90
CBA 45

Well Done!

Can you tell us why BAC is higher than ACB?

Originally posted by wolfgang59
Well Done!

Can you tell us why BAC is higher than ACB?

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

This model does fulfil the requirements that A will come out before B 2/3 and B comes out before C 2/3

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

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

But your model is not the only one that will fulfil the constraints of the question!

Therefore you cannot prove anything from that model ... only suggest it as one of many solutions.

For instance in my model there are 2 balls labeled A and one B. One is drawn from the bag.

The ball drawn from the bag is the winner and second place is decided by the toss of a coin between the loser and C.

ABC = 2/3 * 1/2 = 2/6
ACB = 2/3 * 1/2 = 2/6
BAC = 1/3 * 1/2 = 1/6
BCA = 1/3 * 1/2 = 1/6
CAB = 0
CBA = 0

It proves nothing just as your model does.

Originally posted by wolfgang59
But your model is not the only one that will fulfil the constraints of the question!

Therefore you cannot prove anything from that model ... only suggest it as one of many solutions.

For instance in [b]my model there are 2 balls labeled A and one B. One is drawn from the bag.

The ball drawn from the bag is the winner and second place is dec ...[text shortened]... 1/2 = 1/6
BCA = 1/3 * 1/2 = 1/6
CAB = 0
CBA = 0

It proves nothing just as your model does.[/b]

Do I need to state my original answer again, or are we good now?

Originally posted by wolfgang59
But your model is not the only one that will fulfil the constraints of the question!

Therefore you cannot prove anything from that model ... only suggest it as one of many solutions.

For instance in [b]my model there are 2 balls labeled A and one B. One is drawn from the bag.

The ball drawn from the bag is the winner and second place is dec ...[text shortened]... 1/2 = 1/6
BCA = 1/3 * 1/2 = 1/6
CAB = 0
CBA = 0

It proves nothing just as your model does.[/b]

Very interesting! Here's what you said earlier in this thread:

"25% makes no sense.
There are 3 other outcomes equally likely (BCA, ACB, CAB) so they must be 25% also ... then there is the possiblity of ABC - the MOST likely.

Logically it cannot be more than 2/9 and with a few assumptions is greater
than 2/9"


So back then (only 3 days ago) ABC was the MOST likely; now it's no more likely than ACB according to your recently contrived model. Also back then BCA;ACB; and CAB were all equally likely; now they are all unequally likely in this new model. Though why you ever thought they WERE equally likely is another mystery to me. You seem very confused.

It IS of course very reasonable to infer that ABC is the MOST likely, but you've had to throw out that logic, and contradict your own words, in order to come up with this new forced model.
I doubt you will ever be able to cook up a 'solution' that's as coherent as mine, but keep trying if you must.

Originally posted by forkedknight
Do I need to state my original answer again, or are we good now?

'See above'...and then, yes, I think we will be done.

Originally posted by luskin
Very interesting! Here's what you said earlier in this thread:

"25% makes no sense.
There are 3 other outcomes equally likely (BCA, ACB, CAB) so they must be 25% also ... then there is the possiblity of ABC - the MOST likely.

Logically it cannot be more than 2/9 and with a few assumptions is greater
than 2/9"


So back then (only 3 days ago) ABC ...[text shortened]... be able to cook up a 'solution' that's as coherent as mine, but keep trying if you must.

In a model where
A has 3 balls in a bag: 3,6,9
B has: 2,5,8
C has: 1,4,7
Each player draws one of their 3 balls and compares whose is largest.

You get the following probabilities
ABC: 10/27
ACB: 4/27
BAC: 4/27
BCA: 4/27
CAB: 4/27
CBA: 1/27

No less contrived than your model, meets all requirements, completely different result
25% makes no sense: true
cannot be more than 2/9: true
ABC is most likely: true

I don't understand why you assert that your solution is the only/best one. It is one of many, and it's unreasonable to expect that anyone would match your exact solution.

Originally posted by luskin
Very interesting! Here's what you said earlier in this thread:

"25% makes no sense.
There are 3 other outcomes equally likely (BCA, ACB, CAB) so they must be 25% also ... then there is the possiblity of ABC - the MOST likely.

Logically it cannot be more than 2/9 and with a few assumptions is greater
than 2/9"


So back then (only 3 days ago) ABC ...[text shortened]... be able to cook up a 'solution' that's as coherent as mine, but keep trying if you must.

3 days ago I made 3 assumptions (that was clear in the post) to try and get a reasonable answer. The post clearly was tongue in cheek (note the smiley).

The model I gave recently was to demonstrate that there are many models that fit your initial conditions and they will of course give different answers.

Why should your model be the correct one?

Originally posted by forkedknight
In a model where
A has 3 balls in a bag: 3,6,9
B has: 2,5,8
C has: 1,4,7
Each player draws one of their 3 balls and compares whose is largest.

You get the following probabilities
ABC: 10/27
ACB: 4/27
BAC: 4/27
BCA: 4/27
CAB: 4/27
CBA: 1/27

No less contrived than your model, meets all requirements, completely different result
25% makes n ...[text shortened]... t is one of many, and it's unreasonable to expect that anyone would match your exact solution.

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!

Originally posted by wolfgang59
3 days ago I made 3 assumptions (that was clear in the post) to try and get a reasonable answer. The post clearly was tongue in cheek (note the smiley).

The model I gave recently was to demonstrate that there are many models that fit your initial conditions and they will of course give different answers.

Why should your model be the correct one?

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.