A Most Unusual Puzzle
The original thread is entitled What Odds? and the poster was ThudandBlunder.
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 .
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
A finishes ahead of C in cases 1, 2, and 3, so in ordinary circumstances we would add these probabilities to arrive at the answer to the first question.
B wins the race in case 3 and 4.
The answer to the third question is case No. 3, but there is a catch. 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 ..."
The first question that arises is did ThudandBlunder know it was an impossible puzzle? One of those who submitted an answer was luskin77, to whom ThudandBlunder replied "Thanks it seems pretty sound reasoning to me. And no one else has disagreed" so apparently ThudandBlunder was not trying to trick us. He thought there really was an answer.
The next question is, is it possible to keep the original wording and merely adjust the probabilities so as to make the puzzle logically consistent? Instead of 2/3 and 3/4 let us use a and b.
1. ABC ab
2. ACB a(1-b)
3. BAC (1-a)b
4. BCA (1-a)b
5. CAB a(1-b)
6. CBA (1-a)(1-b)
Adding these, setting them equal to 1, and solving for a gives a = b/(2b-1). b cannot be equal to 1/2. Values less than 1/2 will make a negative and values greater than 1/2 will make a > 1, both of which are impermissible values for a probability, so unfortunately the puzzle cannot be fixed by altering the probabilities.
The next question is: how can the problem be reworded to preserve its literary style but make it possible to answer ThudandBlunder's three questions?
Quote: statmanIn all my years of reading puzzles I have never seen a puzzle as beautifully written as this one or with such an unusual answer.
The original thread is entitled What Odds? and the poster was ThudandBlunder.
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 .
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
A finishes ahead of C in cases 1, 2, and 3, so in ordinary circumstances we would add these probabilities to arrive at the answer to the first question.
B wins the race in case 3 and 4.
The answer to the third question is case No. 3, but there is a catch. 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 ..."
The first question that arises is did ThudandBlunder know it was an impossible puzzle? One of those who submitted an answer was luskin77, to whom ThudandBlunder replied "Thanks it seems pretty sound reasoning to me. And no one else has disagreed" so apparently ThudandBlunder was not trying to trick us. He thought there really was an answer.
The next question is, is it possible to keep the original wording and merely adjust the probabilities so as to make the puzzle logically consistent? Instead of 2/3 and 3/4 let us use a and b.
1. ABC ab
2. ACB a(1-b)
3. BAC (1-a)b
4. BCA (1-a)b
5. CAB a(1-b)
6.CBA (1-a)(1-b)
Adding these, setting them equal to 1, and solving for a gives a = b/(2b-1). b cannot be equal to 1/2. Values less than 1/2 will make a negative and values greater than 1/2 will make a > 1, both of which are impermissible values for a probability, so unfortunately the puzzle cannot be fixed by altering the probabilities.
The next question is: how can the problem be reworded to preserve its literary style but make it possible to answer ThudandBlunder's three questions?
Quoted for preservation.
Seriously? You are going to blame the conditions for your numbers not adding up to one? This is just priceless. :-)
You came up with the formula a = b/(2b-1). If you let b = 1 then a = 1. This will simply mean that the only possible order to the race is ABC. While I admit this is an edge case, it still does admit a viable answer.
Quote:weaselman: Seriously? You are going to blame the conditions for your numbers not adding up to one? This is just priceless. :-)
Weasel: a sneaky, untrustworthy, or insincere person (Merriam-Webster Dictionary, Definition No. 3). Whenever you are working a probability problem and you arrive at a probability greater that 1, you have made a mistake. In this case the mistake is in the problem itself.
I have a story for you:
Albert Einstein played the violin and he played in the Princeton Community Orchestra. During rehearsal he was having trouble making an entrance. The conductor stopped the music and remarked "What's the matter, Einstein, can't you count?"
If you go to the original thread you will see that the majority of respondents did not offer answers but said that there were insufficient data or something like that. They were right. I am just being a little more definite.
Quote: statman
The answer to the third question is case No. 3, but there is a catch. 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 ..."
Because the probabilities don't add to 1, it doesn't prove that the problem is impossible to solve, it merely proves that this answer is wrong. The conditions are not inconsistent, it is the logic used to arrive at an answer that is incorrect.
The problem is impossible to solve because you have 3 equations and 6 unknowns:
ABC + ACB + CAB = 2/3
ABC + BAC + BCA = 3/4
ABC + ACB + BAC + BCA + CAB + CBA = 1
Quote: CrystalMath...The problem is impossible to solve because you have 3 equations and 6 unknowns....
When I read this thread prior to your post, I assumed that the way to fix the question would be to pick pseudo random probabilities that you liked for the six outcomes then calculate the P(A>B) and P(B>C). I was too lazy to pursue that.
Now that you point out that it is a matter of 3 equations in six unknowns, I don't think that implies that the problem is unsolvable but instead that it has multiple solutions (not uniquely solvable, I suppose). I'm still too lazy to pursue it.
Which I think it is as you are assuming that ABC is merely the chance of A over B times B over C. Don't you also need to know A over C to calculate this result?
CrystalMath: ABC, ACB, etc. What do these symbols mean in your statement? In my statement they are the order of finishing.
For someone who calls themselves statman you have a remarkably poor grasp of probability. Jufo81's post in the original thread makes it clear where your error is. How you could read that thread, then post something incorrect just baffles me.
P(BAC) does not equal 1/4
P(BAC) + P(BCA) = 1/4. That is the full set of outcomes where A loses to B (1/3) and B beats C (3/4).
The puzzle cannot be completely answered because you cannot differentiate between P(BAC) and P(BCA). Trying to add up incorrect probabilities, noting they don't add to 1, then declaring the puzzle broken is embarrassing.
Quote: statmanthecesspit: the project proposed is to fix the puzzle so that it can be solved. You're thinking about it; just dig a little deeper.
I disagree. You cannot have A beats be as 2/3rds as statement of the puzzle, then have your working show A beats B as 5/6, then declare the puzzle is broken.
Your workings ARE wrong in the first step. See the previous work I did on this puzzle, where I started with a similar approach. Also others have weighed in on this in a previous post. I don't understand why you'd ignore the results and work already given, even to cross check your own working.
Quote: dwheatleyP(BAC) does not equal 1/4
P(BAC) + P(BCA) = 1/4. That is the full set of outcomes where A loses to B (1/3) and B beats C (3/4).
Ta-dah! That's a far better way of saying what I was trying to say...
However P(ABC) is 1/2 ... only example where A beats B and B beats C.
This implies that P(ACB) + P(CAB) = 2/3 - 1/2 = 1/6, not 1/3 (1/6 + 1/6) as the original poster has put.
AND
P(BAC) + P(BCA) = 1/4
And we also have P(CBA) = 1/12
(this is because A beats B and B beats C are not independent events, and statman has assumed they are, from what I can tell).
This adds up to 1/2+1/6+1/4+1/12 = 6/12+2/12+3/12+1/12 = 1/1 => sums to 1, the problem is NOT intractable.
Quote:dwheatley:In CM's statement, they are also the order of finishing.
I don't think so. Since he adds them up and gets what looks like a probability, they probably mean the probabilities of those orders of finishing. Why don't you caclulate the probabilities of these, check against CM's totals, and let me know what you find. Remember that if a is the probability of A finishing before B, (1-a) is the probability of B finishing before A. In the finishing order BAC, horse A does not finish before horse B (1/3) but horse B does finish before horse C (3/4). If P(BAC) does not equal 1/4, what does it equal? You don't say. I can differentiate between A coming in second and coming in last. Why can't you?
Quote: statmanWhenever you are working a probability problem and you arrive at a probability greater that 1, you have made a mistake.
Exactly!
Quote:In this case the mistake is in the problem itself.
Nope. See your previous sentence. You made a mistake. It's not the problem, it's you.
I have a story for you:
Quote:
If you go to the original thread you will see that the majority of respondents did not offer answers but said that there were insufficient data or something like that. They were right. I am just being a little more definite.
Yes, they were right. The data is indeed insufficient, but that is not the reason your numbers do not add up. You are not being definitive, you are being wrong.
Quote: CrystalMathBecause the probabilities don't add to 1, it doesn't prove that the problem is impossible to solve, it merely proves that this answer is wrong. The conditions are not inconsistent, it is the logic used to arrive at an answer that is incorrect.
The problem is impossible to solve because you have 3 equations and 6 unknowns:
ABC + ACB + CAB = 2/3
ABC + BAC + BCA = 3/4
ABC + ACB + BAC + BCA + CAB + CBA = 1
This is what I get as well.
Well, I can reduce it to five unknowns, as CBA = ABC - 5/12:
Add the first two equations together: 2 ABC + ACB + BAC + BCA + CAB = 17/12
Add CBA to both sides: 2 ABC + ACB + BAC + BCA + CAB + CBA = 17/12 + CBA
Subtract equation 3: ABC = 17/12 - 1 + CBA = CBA + 5/12
You can't reduce it any further as the sums (ABC + CAB) and (BAC + BCA) always appear together - that is, whenever any of those four appears in an equation, it is added to its "partner".
Quote: statmanQuote:dwheatley:In CM's statement, they are also the order of finishing.
I don't think so. Since he adds them up and gets what looks like a probability, they probably mean the probabilities of those orders of finishing. Why don't you caclulate the probabilities of these, check against CM's totals, and let me know what you find. Remember that if a is the probability of A finishing before B, (1-a) is the probability of B finishing before A. In the finishing order BAC, horse A does not finish before horse B (1/3) but horse B does finish before horse C (3/4). If P(BAC) does not equal 1/4, what does it equal? You don't say. I can differentiate between A coming in second and coming in last. Why can't you?
ABC, ACB, BAC, BCA, CAB, and CBA are probabilities of the horses finishing in those orders. I don't understand the confusion.
Although, I forgot to state that we are assuming that two or more horses cannot tie.
Quote: statmanCheck against CM's totals
These are the conditions of the problem. I did no calculations here whatsoever.
No amount of flawed logic is going to solve this problem.
yet again, you just know he would go thru and erase all of them.
To be very specific, he assumed the events are independent so that he multiplied ... there is no way they can be independent since the horses are running in the same race ... that's as easy an explanation of statman's error as I can come up with ...Quote: weaselmanYes, they were right. The data is indeed insufficient, but that is not the reason your numbers do not add up. You are not being definitive, you are being wrong.
--Ms. D.
For example, BAC and BCA have a combined probability of 1/4 (B ahead of A times B ahead of C), not 1/4 each. Similarly, ACB and CAB are 1/6 together.
To little surprise, 1/2+1/4+1/6+1/12 = 1 :)
Quote:weaselman: BAC and BCA have a combined probability of 1/4 (B ahead of A times B ahead of C), not 1/4 each. Similarly, ACB and CAB are 1/6 together
Why are you combining these probabilities? The race can end in only one way.
Congratulations on 1/2+1/4+1/6+1/12 = 1. That's right, but what does it mean? Does it mean that the problem is consistent? Do you think that there are only four possible outcomes?
Quote: statman
Why are you combining these probabilities? The race can end in only one way.
The race can end in one way. But the probability of that ending is not what you think it is.
Quote:Congratulations on 1/2+1/4+1/6+1/12 = 1. That's right, but what does it mean?
That means that those are correct probabilities.
Quote:Does it mean that the problem is consistent? Do you think that there are only four possible outcomes?
Yes, the problem is consistent. It is underdefined, yes, but perfectly consistent.
And yes, these are the only four possible outcomes, if we exclude the possibility of a tie.
There are many ways to define "possible outcomes", BTW. For example, you can say, that B either wins the race or it does not, that those would be the only two possible outcomes with respective probabilities of 1/4 and 3/4.
Quote: statmanQuote:weaselman: BAC and BCA have a combined probability of 1/4 (B ahead of A times B ahead of C), not 1/4 each. Similarly, ACB and CAB are 1/6 together
Why are you combining these probabilities? The race can end in only one way.
Congratulations on 1/2+1/4+1/6+1/12 = 1. That's right, but what does it mean? Does it mean that the problem is consistent? Do you think that there are only four possible outcomes?
I thought this thread would be entertaining. It isn't. It's painful. Just stop...people *much* smarter than you are telling you that you're wrong. It's time to stop flailing and delete your posts.
Quote: statmanQuote:weaselman: BAC and BCA have a combined probability of 1/4 (B ahead of A times B ahead of C), not 1/4 each. Similarly, ACB and CAB are 1/6 together
Why are you combining these probabilities? The race can end in only one way.
Erm, yes... but there's six ways it can end. We can calculate the values of those, but we can only tell you that P(ACB) OR P(CAB) = 1/6.
Both of those possibilities add to 1/6th.
Quote:
Congratulations on 1/2+1/4+1/6+1/12 = 1. That's right, but what does it mean? Does it mean that the problem is consistent? Do you think that there are only four possible outcomes?
If you don't have the wit to work it out based on previous answers, I am not sure we can help you.
Please go back to my original answer : your workings cannot be right if the probability you calculate for A beating B is 5/6ths when the question states this as 2/3rds.
Simple as that. Your method is INCORRECT, go back, try again.
Or delete your answer for having too much "heat".
Quote:thecesspit: your workings cannot be right if the probability you calculate for A beating B is 5/6ths when the question states this as 2/3rds.
Where in my post do I calculate the probability of anything being 5/6 unless you mean that I declare the probability of A beating C should not be calculated as #1 + #2 + #3 because the problem is logically inconsistent?
My calculation of P(ACB) is based on the following: A finished ahead of B (2/3) AND B did NOT finish ahead of C (1/4). I know of no reason why A's performance should influence B's. In probabilities, AND means multiply, so P(ACB) = 1/6. Based on similar reasoning, P(CAB) = 1/6. In probabilities OR means add if the events are mutually exclusive, so you are declaring that P(ACB) OR P(CAB) = 1/6; in other words that 1/6 + 1/6 = 1/6 and I suppose that's the reason you flunked grade school arithmetic.
I'm not looking for anyone to help me, I merely ask if someone can rewrite the problem so that answers are possible. Scorn + misquotes + bad math do not prove a point unless the jury is very naive.
Quote: statman
My calculation of P(ACB) is based on the following: A finished ahead of B (2/3) AND B did NOT finish ahead of C (1/4). I know of no reason why A's performance should influence B's. In probabilities, AND means multiply, so P(ACB) = 1/6. Based on similar reasoning, P(CAB) = 1/6. In probabilities OR means add if the events are mutually exclusive, so you are declaring that P(ACB) OR P(CAB) = 1/6; in other words that 1/6 + 1/6 = 1/6 and I suppose that's the reason you flunked grade school arithmetic.
You should try not to be rude. It makes you look especially pathetic, because you are not only wrong, but are also making an obvious and quite silly mistake in a very trivial matter. Your mentioning of grade school in this context is particularly laughable (I have been explaining the concept of probabilities to my 8 year old recently, and she was able to grasp it with a lot less struggle then it is causing you).
I don't hold your lack of mathematical skills against you, that is not a must have for a civilized person. But your arrogance, rudeness and lack of civility make you look like a ... well ... not a very good person. Stop it, and you may notice your encounters with others suddenly becoming much more pleasurable.
Your problem is that P(AB)*P(CB) is not P(ACB) as you stubbornly believe, but rather P(ACB)+P(CAB). The product gives you the probability of B being the last in the race, but does not tell you anything about the order of the other two horses. There are two outcomes that comprise this event - ACB, and CAB. Their combined probability is 1/6. You compute the probability of B being last and assign it to ACB, and then assign the same number to CAB. You end up adding these probabilities twice. No wonder the result is greater than 1.
Quote: statman
I'm not looking for anyone to help me, I merely ask if someone can rewrite the problem so that answers are possible.
Easy. Just add the missing condition. For example, you could specify the probability of A beating C. That should be enough to close the problem. It needs to be at least 1/2 though or the conditions will indeed become inconsistent.
ABC, ACB, BAC, BCA, CAB, CBA
A has 2/3 probability of finishing ahead of B.
ABC + ACB + CAB = 2/3.
or
BAC + BCA + CBA = 1/3
B has 3/4 probablity of finishing ahead of C.
ABC + BAC + BCA = 3/4.
or
ACB + CAB + CBA = 1/4
and
ABC + ACB + BAC + BCA + CAB + CBA = 1
(a) solve for ABC + ACB + BAC.
(b) solve for BAC + BCA
(c) solve for BAC.
None of these are possible as there are 6 unknowns and 5 statements. You need to state the odds of C finishing ahead of A.
I think you have fewer than five independent statements. You can add your first and second statements to get the fifth statement. Or, you can add your third and fourth statements to get the fifth statement. I think there are only four independent constraints here on the six unknowns, but I haven't looked into it thoroughly. I don't think this makes the problem unsolvable, rather that there are multiple, "correct" solutions.
Quote: statmanQuote:thecesspit: your workings cannot be right if the probability you calculate for A beating B is 5/6ths when the question states this as 2/3rds.
Where in my post do I calculate the probability of anything being 5/6 unless you mean that I declare the probability of A beating C should not be calculated as #1 + #2 + #3 because the problem is logically inconsistent?
My calculation of P(ACB) is based on the following: A finished ahead of B (2/3) AND B did NOT finish ahead of C (1/4). I know of no reason why A's performance should influence B's. In probabilities, AND means multiply, so P(ACB) = 1/6. Based on similar reasoning, P(CAB) = 1/6. In probabilities OR means add if the events are mutually exclusive, so you are declaring that P(ACB) OR P(CAB) = 1/6; in other words that 1/6 + 1/6 = 1/6 and I suppose that's the reason you flunked grade school arithmetic.
I'm not looking for anyone to help me, I merely ask if someone can rewrite the problem so that answers are possible. Scorn + misquotes + bad math do not prove a point unless the jury is very naive.
Well, I'll let you rewrite the problem so it fits your world of mathematics.
I'm not sure how you can state on one hand you can't add Your chances A betting C, but you can add all the chances together and the problem fails because those odds are greater than one.
A's performance obviously influences B's. It's a race. A beats B two thirds of the time.
As for the crack about flunking grade school arithmetic, I never went grade school or studied arithmetic. I went to secondary school and studied both Mathematics and Statistics, which I suppose might have been different from what you do in Harvard, but evidence of other people suggest maths is the same the world over. Fancy that.
Quote: statman
I'm not looking for anyone to help me, I merely ask if someone can rewrite the problem so that answers are possible. Scorn + misquotes + bad math do not prove a point unless the jury is very naive.
You're the statman, who claims to know everything about this stuff; why don't YOU do it?
I'm a bystander in this discussion, but it looks to me like you're an imposter standing in front of a room full of professionals and treating them like 6th graders... "Now, class, who can show me the solution?" You're Jack Black singing the math song in School of Rock. But whatever, it's fun to read. keep doing it, I need to waste more time today.
Quote:Mosca: You're the statman, who claims to know everything about this stuff; why don't YOU do it?
I have completely rewritten my answer on seeing that some of those who submitted answers in the original post have gone back to do more work on them. Although some of the responses here are obviously malicious, I think that some are motivated by personal pride. They want me to be wrong so that they will not be seen as having been on a fool's errand.
Go to Marilyn Vos Savant's personal web site and read her correspondence on the Monty Hall Problem. She received thousands of letters saying she was wrong, many on academic stationery and signed by Ph.D.'s. Under similar circumstances I wouldn't have the patience to resolve it the way she did. Since her explanation of the logic of the problem was not understood, she asked the grade school teachers of America to have their classes do simulations, all of which confirmed her result.
I have submitted this puzzle to Marilyn and I hope she reads it. If she sends a reply, I shall share it with you.
Quote: statmanI have been teaching math for thirty years
Ouch. I pray that you are lying about this. Because, if not, that must mean that our schools are in an even more dire state than it is generally recognised.
Quote:I have read many puzzles but never before have encountered one that has no solution.
That's just fantastic.
Here is another "unusual" one to add to your collection.
Jonny buys some amount of candy, and a bottle of beer. How much does he pay if candy costs $5 per pound?
Quote: statmanQuote:Mosca: You're the statman, who claims to know everything about this stuff; why don't YOU do it?
I would if I could, but I find it too tough. That's why I came to the celebrated gaming professionals; however they find it more fun to bash statman than to get to work on the problem. I attempted to make the puzzle solvable by adjusting the two given probabilities but found that there are no values of these that do the trick. I'm sorry I can't go through all of the failed attempts at solution. I merely have offered an explanation of why they failed. I may not be right, but at least it should be obvious that there is work to be done.
Many problems are presented to this group and are met with responsive answers, but just because some people think I have been "proven wrong" on some point they concentrate on trying to prove me wrong again. If you accuse me of being pedantic I would have to plead guilty because I have been teaching math for thirty years, however I never have had to do so in a tough inner city neighborhood, at least not until now. Since I have no experience in such a situation I don't quite know how to deal with it. Perhaps the answer is to get some of the most offensive posters banned from the forum, and I am working on how to accomplish this.
I have read many puzzles but never before have encountered one that has no solution. That is why I call it "A Most Unusual Puzzle."
If you look at the original thread there are now two solutions from me and a explanation from weaselman on why there is n infinite number of solutions.
What are the odds
You aren't being bashed because you were wrong before.nyou were being bashed for being wrong this time.
Quote: statmanThe 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
A finishes ahead of C in cases 1, 2, and 3, so in ordinary circumstances we would add these probabilities to arrive at the answer to the first question.
Let a = ABC, b = ACB, c = CAB, d = CBA, e = BAC, f = BCA
1. a+b+c = 2/3
2. a+e+f = 3/4
3. d+e+f = 1/3
4. b+c+d = 1/4
5. a+b+c+d+e+f = 1
where a,b,c,d,e,f >= 0
Subtract 4 from 1 to get a-d = 5/12. Therefore a= 5/12 + d
Let's test this. If P(ABC) = 2/3.
P(CBA) = 1/4.
P(ACB) = 0
P(CAB) = 0
P(BCA) + P(BAC) = 1/12
If P(ABC) = 5/12
P(CBA) = 0
P(BCA) + P(BAC) = 1/3
P(ACB) + P(CAB) = 1/4
If P(ABC) = 1/2
P(CBA) = 1/12
P(BCA) + P(BAC) = 1/4
P(ACB) + P(CAB) = 1/6.
All conditions are satisified in each of these results. B beat C 3/4 of the time and A beat Bs 2/3rds of the time, and the probabilities add up to 1.
P(ABC) is NOT P(A beats B) * P(B beats C). That caught me for a bit.
