What Odds?

I have a question of abstract probability theory, but I will use horse racing to frame the problem. Suppose you have three horses in a race and the only information you are given is that: Horse A has 2/3 probability of finishing ahead of B; and B has 3/4 probability of finishing ahead of C.
Can you determine;
1) the probability that A will finish ahead of C
2) " " " B will win the race
3) " " " the result will be B 1st. A 2nd. C 3rd .

I'd like to get the Wizard's opinion on this, and any thoughts from others very welcome as well.

Giving this a quick once over, I think there is not enough information to determine all answers.

1) A will finish ahead of C at least 1/2 the time, but no more than 11/12 of the time. Since there is no information about how A and C fare against each other probability-wise, it cannot be determined whether the combination of CAB or ACB is more likely, or the combo of BCA or BAC.

2) B will win the race 1/4 of the time, as it will finish ahead of A 1/3 of the time, and ahead of C 3/4 of the time.

3) BAC is indeterminable. The answer to number 2 is the total chance of B winning, which covers the combinations BAC and BCA. There is no correlation between A and C that I can see.

Being abstract probability theory, of which I am not familiar, maybe there is a method to figuring out numbers 1 and 3 that I am unaware of. Anyone else have some insight?

Snip -- mistake made.

I looked at trying to calculate each possible result :

a - ABC
b - ACB
c - BAC
d - BCA
e - CAB
f - CBA

a+b+c+d+e+f = 1
a+b+e = 2/3
a+c+d = 3/4

There's 3 simulataneous equations given, but you need 6 to solve the problem, from what I can tell. It may be possible as all possibilities are between 0 and 1.

LET P(A)=the probability that A will win

LET P(A>B)= the probability that A will finish in front of B

Then if P(A>B)= p/q this implies that P(A)=p/(q-p)*(P(B))

e.g. if P(A>B)=1/2 then P(A)=P(B)
if P(A>B)=3/5 then P(A)=3/2*P(B)

So from the given information we have P(A)=2*P(B) and P(B)=3*P(C)

Combining the two equations we get P(A)=2*P(B)=6*P(C)

Since P(A)+P(B)+P(C)=1 then P(A)+1/2*P(A)+1/6*P(A)=1

And therefore P(A)=3/5 or 60%; P(B)=30%; P(C)=10%

So the answers to parts 1) and 2) are : 6/7 and 3/10 respectively.

As for part 3) P(B>A>C)= P(B)*P(A)/(1-P(B)*P(C)/(1-P(B)-P(A))=3/10*6/10*10/7*1=9/35

Can anyone find any flaw in the logic?

Thanks it seems pretty sound reasoning to me. And no one else has disagreed.

I don't think this is solvable, due to lack of information.
Knowing the probabiliy of who is ahead of whom is not enough, you also need to know the correlation between those events.

Say the run of C is rigged. C either wins or comes in last, never as second (i.e. by order). Then there are the only outcomes ABC, BAC, CAB, CBA.

Then ABC + CAB = 2/3 and ABC + BAC = 3/4.

You ask for A ahead of C, which is ABC + BAC = 3/4.

Honestly I couldn't follow Luskins calculations, but the rigged run of C is not excluded from your question (there is no paradox from C never 2nd).
Since Luskin came to a different result (6/7), it shows you that the original question is not solvable due to lack of information, and needs further assumptions.

Quote: luskin77

LET P(A)=the probability that A will win

LET P(A>B)= the probability that A will finish in front of B

Then if P(A>B)= p/q this implies that P(A)=p/(q-p)*(P(B))

e.g. if P(A>B)=1/2 then P(A)=P(B)
if P(A>B)=3/5 then P(A)=3/2*P(B)

So from the given information we have P(A)=2*P(B) and P(B)=3*P(C)

Combining the two equations we get P(A)=2*P(B)=6*P(C)

Since P(A)+P(B)+P(C)=1 then P(A)+1/2*P(A)+1/6*P(A)=1

And therefore P(A)=3/5 or 60%; P(B)=30%; P(C)=10%

So the answers to parts 1) and 2) are : 6/7 and 3/10 respectively.

As for part 3) P(B>A>C)= P(B)*P(A)/(1-P(B)*P(C)/(1-P(B)-P(A))=3/10*6/10*10/7*1=9/35

Can anyone find any flaw in the logic?

I don't think this solution is correct because I got P(B) = 1/4, the same as posted by Dween above. I think the flaw in your logic is that you consider that the probability of "A finishes ahead of B", P(A > B), is directly proportional to the ratio of how often A wins, P(A), compared to how often B wins, P(B). I don't think that the wording of the problem allows you to assume that.

To put it another way: the fact that A finishes before B twice as often as the opposite, doesn't necessarily mean that A is winner twice as often as B.

This is what I got by continuing from thecesspit's post:

Initial info given:
P(A > B) = 2/3
P(B > C) = 3/4

Questions:

1. P(A>C) = ?
2. P(B) = ?
3. P(BAC) = ?

Now, P(ABC) = P(A > B) & P(B > C) = 2/3*3/4 = 1/2
And, P(CBA) = P(C > B) & P(B > A) = (1-3/4)*(1-2/3) = 1/12

where symbol '&' refers to the intersection of the probability sets
(see image: http://en.wikipedia.org/wiki/File:Venn_A_intersect_B.svg ).

From the initial info given we also know that:

P(A>B) = P(ABC) + P(ACB) + P(CAB) = 2/3
P(B > C) = P(ABC) + P(BAC) + P(BCA) = 3/4

Substituting the known value for P(ABC) = 1/2 leaves two equations with four unknowns:

P(ACB) + P(CAB) = 2/3 - 1/2 = 1/6
P(BAC) + P(BCA) = 3/4 - 1/2 = 1/4

The latter expression is P(B), because P(B) = P(BAC) + P(BCA), so P(B) = 1/4 (agrees with Dween).

This can be, of course, confirmed by: P(B) = P(B > A) & P(B > C) = (1-P(A>B))*P(B>C) = 1/3*3/4 = 1/4

It seems that the questions 1. P(A>C) and 3. P(BAC) cannot be solved from the information given, but it was possible to solve question 2. for P(B).

Quote:

I have a question of abstract probability theory, but I will use horse racing to frame the problem. Suppose you have three horses in a race and the only information you are given is that: Horse A has 2/3 probability of finishing ahead of B; and B has 3/4 probability of finishing ahead of C.
Can you determine;
1) the probability that A will finish ahead of C
2) " " " B will win the race
3) " " " the result will be B 1st. A 2nd. C 3rd .

Good puzzle! The number of possible outcomes is 3! = 6. They are, with their probabilities:

1. ABC (2/3)(3/4) = 1/2

2. ACB (2/3)(1/4) = 1/6

3. BAC (1/3)(3/4) = 1/4

4. BCA (1/3)(3/4) = 1/4

5. CAB (2/3)(1/4) = 1/6

6. CBA (1/3)(1/4) = 1/12

Since these outcomes are exhaustive, their probabilities must add to 1. Since they do not, the given conditions are inconsistent and the answer is "No, you can't determine ..."

This is like an ancient Chinese puzzle:

Duke Wen has three horses, W1, W2, and W3. Duke Chang also has three horses, C1, C2, and C3. W1 is faster than C1, W2 is faster than C2, and W3 is faster than C3. They agree to a contest of three heats, two horses each. Although Duke Wen has the faster horses, Duke Chang manages to win the contest. How?

Quote: ThudandBlunder

I have a question of abstract probability theory, but I will use horse racing to frame the problem. Suppose you have three horses in a race and the only information you are given is that: Horse A has 2/3 probability of finishing ahead of B; and B has 3/4 probability of finishing ahead of C.
Can you determine;
1) the probability that A will finish ahead of C
2) " " " B will win the race
3) " " " the result will be B 1st. A 2nd. C 3rd .

I'd like to get the Wizard's opinion on this, and any thoughts from others very welcome as well.

I took another stab at this.

1 - P(A beats C) = 11/12
2 - P(B wins) = 1/4
3 - P(BCA) = 0, and P(BAC) = 1/4

Here's how I get there. I imagined instead of 3 race horses, lets have 3 12 sided dice, labelled A, B and C. You roll these dice, and order them based on the values rolled. These could be considered to be the possible times the horses run the races in.

Dice A = 8 sides labelled "6", 4 sides labelled "1"
Dice B = 12 sides labelled 2.
Dice C = 9 sides labelled 0, 3 sides labelled 3.

(You could imagined them as 1 minute + x seconds, as rolled on the die. And even the seconds could be different within each range... So A is labelled 1m 6.1s, 1m 6.2s and so forth... the idea of would hold fine).

Thus
P(A beats B) = 2/3 (2/3rds of the time dice A rolls higher than 2).
P(B beats C) = 3/4 (only 3 times out of 12 dice C rolls higher than a 2).

=>

P(ABC) = 6,2,0 on the dice = 2/3 * 3/4 = 6/12
P(BAC) = 2, 1, 0 on the dice = 1/3 * 3/4 = 3/12 = 1/4
P(BCA) = Not possible, For C to beat A, it has to roll a 3, which means it beats B.

=> B wins = 1/4

Carrying on:

P(ACB) = 6,3,2 on the dice = 2/3*1/4 = 2/12

=> A wins = 8/12 = 2/3

P(CBA) = 3,2,1 = 1/4 * 1/3 = 1/12
P(CAB) = 0 (C rolls 3, which to win A has to roll a 1, which means it loses to B).

=> C wins = 1/12

And there we go, there's one set of solutions, where two of the 6 possible results have a zero chance.

Quote:

Dice A = 8 sides labelled "6", 4 sides labelled "1"
Dice B = 12 sides labelled 2.
Dice C = 9 sides labelled 0, 3 sides labelled 3.
...
P(BCA) = Not possible, For C to beat A, it has to roll a 3, which means it beats B.

What prevents you from labelling them differently though?
<never mind ... I am still thinking> :)

Nothing... it was just giving one solution (there may be more).

Ah, ok then ...
I thought you were saying that it was possible to find a single solution ...

Quote: weaselman

Ah, ok then ...
I thought you were saying that it was possible to find a single solution ...

That may also be the case. All I know is that there is at least one solution. P x beats y is non-transitive. I thought about this after thinking of the horses as dice, and thinking about non-transitive dice (If you've seen them, it's pretty cool.. A beats B 2/3rds of the time, B beats C 2/3rds of the time and C beats A 2/3rds of the time when used heads up).

Here's another set, that I think meets the criteria, but with 3 six -sided dice.

A : 17,16,15,13,11 and 9
B : 18,14,12,10,4 and 0
C : 8,7,5,3,2 and 1

Note that C never beats A.

=> P(CBA), (CAB) and P(BCA) = 0

=> P(ABC) = 5/12
=> P(BAC) = 4/12
=> P(ACB) = 3/12

=> P(A) = 8/12
=> P(B) = 4/12
=> P(C) = 0/12

There we go, two answers are possible :) There's likely to be more.... :)

There is an infinite number of solutions, within certain boundaries.
I am a horrible drawer, but here goes an illustration...

Click this, I could not figure out how to embed it


The yellow area represents C beating B. Green is A beating B.

Grey sector is where B beats both A and C. It represents BAC and BCA. Note, that it can be split between these two outcomes in any way, and will still satisfy the conditions. Note also, that you can make that area larger and smaller by rotating the yellow sector around. The probability of B winning the race can have any value between 1/12 and 1/3.

The overlap between green and yellow is where C and A both beat B - CAB and ACB. Again it can be split in an infinite number of ways between the two outcomes, and can be made larger and smaller. The probability of B coming last is any value between 0 and 1/4.

The green part, not overlapping yellow is A beats B and B beats C - ABC. In the picture it is about 1/2, but note that, again, you can rotate the yellow sector around the center, and increase or decrease the overlap, consequently changing the "pure green" area. You can make the overlap disappear completely for example (because 1/4 < 1/3), making P(ABC) = 2/3, or you can make the whole yellow thingy be inside the green area, thus making P(ABC)=2/3-1/4 = 5/12.
So, P(ABC) can have any value between 5/12 and 2/3.

Nice... I was thinking you could easily have continuous probability functions for A, B and C's times for the 1 mile race, which would then produce the possible results.

That's that finished.

Quote: thecesspit

...
=> B wins = 1/4...
=> A wins = 8/12 = 2/3...
=> C wins = 1/12...

Call the probabilities of the order of finish ABC, ACB, BAC, BCA, CAB, and CBA. And call the probabilities of a horse beating another horse AB, AC, and BC. We have 6 unknowns in terms of independent variables AB, AC, and BC. Here is a system of six equations:

ABC + ACB + BAC + BCA + CAB + CBA = 1
ABC + ACB + CAB = AB
ABC + ACB + BAC = AC
ABC + BAC + BCA = BC
ABC + BAC = AC * BC (C is in last place. This equation and the following one were revealed by weaselman.)
ACB +CAB = AB * CB = AB * (1-BC) (B is in last place.)

These six equations can be solved on an Excel sheet. Set AB=2/3 and BC=3/4. Then vary AC from 0 to 1 to see how ABC, ACB, BAC, BCA, CAB, and CBA behave. When AC is less than 2/3, BAC is always negative, but when AC is greater than 2/3, CAB is always negative. When AC=2/3, both BAC and CAB are 0. Since probabilies must be greater than or equal to zero, then AC=2/3.

Then ABC=1/2, ACB=1/6, BAC=0, BCA=1/4, CAB=0, and CBA=1/12. So, I agree with thecesspit that P(B wins) = BCA = 1/4; P(A wins)= ABC + ACB = 2/3; and P(C wins)= CBA = 1/12.

So, the values of AB and BC were adjusted by the puzzle writer to force AC to one possible value.

Quote: ChesterDog

ABC + BAC = AC * BC (C is in last place. This equation and the following one were revealed by weaselman.)

Actually, I am not sure this necessarily is true. AC and BC do not have to be independent (maybe, C is having a really bad day), and if they aren't the equation does not hold.