threesome

In a certain three sided game, or any type of three way contest you care to imagine(but with no possibility of ties or draws), Alwyn, Bryce, and Cyril, (yes, at first I too thought they must be gay; but perhaps they're just Welsh), have competed against each other many hundreds of times over the years. The record shows that Alwyn has finished ahead of Bryce exactly two thirds of the time, and Bryce has finished ahead of Cyril exactly two thirds of the time. What proportion of these contests have resulted in B 1st; A 2nd ; C 3rd.
By the way, if there had been a fourth player, watchers of the TV comedy 'Little Britain' will have no trouble working out what his name would be.

Originally posted by luskin
In a certain three sided game, or any type of three way contest you care to imagine(but with no possibility of ties or draws), Alwyn, Bryce, and Cyril, (yes, at first I too thought they must be gay; but perhaps they're just Welsh), have competed against each other many hundreds of times over the years. The record shows that Alwyn has finished ahead of Bryce e ...[text shortened]... rs of the TV comedy 'Little Britain' will have no trouble working out what his name would be.

There are 4 possibilities for the game with regards to the constraints that you providied.
1) both A>B and B>C are met
ABC

2) A>B but not B>C
ACB
CAB

3) B>C but not A>B
BAC
BCA

4) neither constraint is met
CBA

There are a few possibilities then for the proportion of time that BAC occurs:
not at all)
(1) ABC occurs 2/3 of the time.
(4) CBA occurs 1/3 of the time.

BAC occurs 1/3 of the time:
(1) ABC occurs 1/3 of the time
(2) ACB + CAB occurs 1/3 of the time
(3) BAC occurs 1/3 of the time, BCA does not occur

Somewhere in between:
(1) ABC 1/3
(2) ACB + CAB 1/3
(3) 0 < BAC < 1/3, BCA is the rest

Therefore, I will say that there is not enough information to answer

apologies...answered the wrong question

Originally posted by forkedknight
Therefore, I will say that there is not enough information to answer[/b]

That is true in so far as there is no unique solution based solely on the explicit constraints. But what I'm really looking for is the solution which (I believe) is most logically consistent with the implications of the given data.
In other words, can anything be reasonably inferred regarding A versus C in terms of probability, while still keeping within the stated constraints? This way, I think, leads to a unique and plausible solution.

Originally posted by luskin
That is true in so far as there is no unique solution based solely on the explicit constraints. But what I'm really looking for is the solution which (I believe) is most logically consistent with the implications of the given data.
In other words, can anything be reasonably inferred regarding A versus C in terms of probability, while still keeping within the stated constraints? This way, I think, leads to a unique and plausible solution.

I think the answer you want is 2/13

However that is flawed.

Originally posted by luskin
That is true in so far as there is no unique solution based solely on the explicit constraints. But what I'm really looking for is the solution which (I believe) is most logically consistent with the implications of the given data.
In other words, can anything be reasonably inferred regarding A versus C in terms of probability, while still keeping within the stated constraints? This way, I think, leads to a unique and plausible solution.

If you calculate by straight probability, BCA + BAC should occur 2/9 of the time. If we presume that they are playing a skill game, and since A>B 2/3 of the time, and B>C 2/3 of the time, then A should beat C at least 2/3 of the time, then that puts the range for BAC between
4/27 <= BAC <= 6/27 (2/9)

2/13 falls into that range.

Originally posted by wolfgang59
I think the answer you want is 2/13

However that is flawed.

There is not enough information,
however if we make 3 'sensible' assumptions;
1. that the 4 possibilities where either A beats B or B beats C are all equally likely

and

2. CBA is less likely than any other result.

We find 1/9 > CBA > 0

Then with simple substitution we get

CBA 1/9 0/9
ABC 4/9 3/9
ACB 1/9 2/9
BAC 1/9 2/9
BCA 1/9 2/9
CAB 1/9 2/9

Our third assumption is to take the mid point
so BAC is probably, roughly about 1/6 ish

😉

Originally posted by wolfgang59
I think the answer you want is 2/13

However that is flawed.

No, my answer is much closer to 25%.

Originally posted by luskin
No, my answer is much closer to 25%.

It can't possibly be greater than 2/9, as I showed above. Why don't you just explain your thinking; I don't think we're on the same page.

Originally posted by luskin
No, my answer is much closer to 25%.

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

Originally posted by forkedknight
It can't possibly be greater than 2/9, as I showed above. Why don't you just explain your thinking; I don't think we're on the same page.

2/9 happens to be very close, but.....it IS slightly greater!

As it is apparently a game that combines skill with an element of chance, why not try to imagine a simple mechanical model, which would tend to reproduce the statistics quoted. See where that takes you. After that I'll explain my thinking.

Originally posted by luskin
2/9 happens to be very close, but.....it IS slightly greater!

As it is apparently a game that combines skill with an element of chance, why not try to imagine a simple mechanical model, which would tend to reproduce the statistics quoted. See where that takes you. After that I'll explain my thinking.

Different mechanical models could produce different results within the confines of forkedKnights analysis.

For instance a 'game' similar to a Mexican standoff would favour the weakest player of the three. A game like golf would favour the best player.

One could devise rules to produce any outcome within forkedknghts constraints.

Originally posted by wolfgang59
Different mechanical models could produce different results within the confines of forkedKnights analysis.

For instance a 'game' similar to a Mexican standoff would favour the weakest player of the three. A game like golf would favour the best player.

One could devise rules to produce any outcome within forkedknghts constraints.

Golf is a good example. In fact it 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 really consider the possibility that it's been ABC 2/3 of the time and CBA 1/3??? I don't think so, unless you know that B just hates being first or last.

Originally posted by luskin
2/9 happens to be very close, but.....it IS slightly greater!

As it is apparently a game that combines skill with an element of chance, why not try to imagine a simple mechanical model, which would tend to reproduce the statistics quoted. See where that takes you. After that I'll explain my thinking.

Please show us your math. I don't understand how it's possibly greater than 2/9.

B>A: that happens 1/3 of the time. It must be less than 1/3
B>C: that happens 2/3 of the time.
2/3 * 1/3 = 2/9

Now if you consider that in golf, cases where B>A mean that B>C is more likely than 2/3, then I could see where you're coming from, but from the problem, there is no indication that B vs C is dependent on B vs A in any way. Nor is it clear that A vs C is dependent on A vs B in any way.

There are other types of games where B>A would make B>C LESS likely.

I stand by my previous assertion that there is not enough information to provide an answer.

Originally posted by wolfgang59
There is not enough information,
however if we make 3 'sensible' assumptions;
1. that the 4 possibilities where [b]either A beats B or B beats C are all equally likely

and

2. CBA is less likely than any other result.

We find 1/9 > CBA > 0

Then with simple substitution we get

CBA 1/9 0/9
ABC 4/9 3/9
A ...[text shortened]... ur third assumption is to take the mid point
so BAC is probably, roughly about 1/6 ish

😉[/b]

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