# Calculating Nth place from 1st place probability

• Feb 7th 2011, 08:06 AM
SumGuy
Calculating Nth place from 1st place probability
My apologies in advance, as I'm not sure how advanced this is, this might belong in basic Probability and Statistics.

I've done a lot of searching online, and I can't seem to find anything to say this problem has ever been attempted mathematically (which is the reason I'm guessing this would count as advanced).

The problem is this, lets say you have a race with 6 people in it. Given you know the chance of them to come first, can you mathematically calculate their chance to come 2nd (and 3rd, 4th, 5th, last)?

So for example:

Runner 1: 25% chance to win
Runner 2: 20% chance to win
Runner 3: 20% chance to win
Runner 4: 15% chance to win
Runner 5: 12% chance to win
Runner 6: 8% chance to win

If you are able to build a formula, would you be able to build it with a variable number of runners?

Thanks in advance to those of you who try. :-)
• Feb 7th 2011, 10:59 AM
theodds
Interesting question. The answer I think is no, no such formula exists (more accurately, the answer depends on other details that haven't been specified). You need to know a little bit more about the distribution of each runners time.
• Feb 7th 2011, 11:50 AM
SumGuy
I can understand how the distribution of time would come into play when calculating the probability in general, but if the probability of each one to win already factors this in, would those factors not cancel themselves out? I could well be wrong in saying that though. This question has plagued me for months!
• Feb 7th 2011, 02:36 PM
theodds
Suppose we add some new information.

(1) Whenever Racer 1 doesn't win, he gets last.

vs

(2) Whenever Racer 1 doesn't win, he gets 2nd.

In either (1) or (2), we can still meet the constraints you gave in the problem statement. However, in (1) Racer 1 has a 75% chance of coming in last while in (2) Racer 1 has a 75% chance of coming in 2nd. Thus we can't, on the basis of the constraints given in the initial problem, derive the formula you seek.

In fact, we haven't nailed things down very well at all. It seems to me that there are 25 free parameters in this problem, and you've specified 5.
• Feb 7th 2011, 03:33 PM
SumGuy
Well maybe it's not possible, I certainly don't know the answer. Well, to make things simpler, even for somewhere to start, let's say have 3 runners.

R1: 80%
R2: 10%
R3: 10%

Right so, R2 and R3 are of equal standard, therefore have equal chance of any given position, and they each have a 90% chance of not winning, so they would each have a 45% chance of 2nd and 3rd. The only real question here is how to split R1s chance across 2nd and 3rd. If the probability of [R2 OR R3 beating R1] or the probability of [R2 AND R3 beating R1] can be calculated, then we should be able to use the ratio to calculate the chance for R1 to come 2nd and third (sorry by the way if any of my logic is out). I haven't done this type of maths in years... Or at least to this degree anyway.
• Feb 7th 2011, 03:39 PM
theodds
Quote:

Originally Posted by SumGuy
Well maybe it's not possible, I certainly don't know the answer. Well, to make things simpler, even for somewhere to start, let's say have 3 runners.

R1: 80%
R2: 10%
R3: 10%

Right so, R2 and R3 are of equal standard, therefore have equal chance of any given position, and they each have a 90% chance of not winning, so they would each have a 45% chance of 2nd and 3rd. The only real question here is how to split R1s chance across 2nd and 3rd. If the probability of [R2 OR R3 beating R1] or the probability of [R2 AND R3 beating R1] can be calculated, then we should be able to use the ratio to calculate the chance for R1 to come 2nd and third (sorry by the way if any of my logic is out). I haven't done this type of maths in years... Or at least to this degree anyway.

The bold part is a problem. You are assuming something that isn't explicitly stated in the problem statement.

EDIT: The stuff following it is also not good. It makes sense that runners 2 and 3 should have the same probability of 2nd and 3rd, but there is no reason for it to be 45.

Think in terms of contingency tables. Here's one for the problem you stated:

80 ? ?
10 ? ?
10 ? ?

Each row and column needs to add to 1. If you add the constraint that runners 2 and 3 each need the same probability of 2nd and 3rd place, you get

80 1 - 2x 1 - 2y
10 x y
10 x y

so you still have a free parameter.
• Feb 7th 2011, 03:59 PM
SumGuy
I had a feeling that would be the part that was picked up on alright, but considering how their probability was calculated in the 1st place, I'm guessing it would be based on their mean average race times. I mean, do you think standard deviation would even come into play? If their mean average was the only factor in calculating their win probability, wouldn't the same hold true for the 2nd place probability. I know runners can have good and bad days, but given that their form on the day will be random, doesn't that factor out the standard deviation. I mean, even if 1 of them didn't deviate at all and the other one deviated by 10 seconds, if he was on bad form, then the steady runner would be at an advantage, but if he was on good form, the steady runner would be at the inversed disadvantage, and since he would have equal chance for good or bad form, doesn't it make sense that the law of random would cancel the 2 out? Maybe not if you take the number of times below average against the number of times above average as a ratio, granted, but how much of a margin of error would we really be talking about (realistically).

Sorry by the way, I have a habit of rambling on when I try to explain my thought process. :-)

EDIT: my apologies, you're completely right, I factored out R1 from that all together.

So the contingency table should actually be as follows I'm guessing:

0.8, 1 - 2x, 1 - 2y
0.1, x, y
0.1, x, y
• Feb 7th 2011, 04:38 PM
theodds
Quote:

Originally Posted by SumGuy

EDIT: my apologies, you're completely right, I factored out R1 from that all together.

So the contingency table should actually be as follows I'm guessing:

0.8, 1 - 2x, 1 - 2y
0.1, x, y
0.1, x, y

Yeah, that one is correct if you make the assumption that runner 2 and 3 are equivalent (I came back to correct the error, but you beat me to it). It seems reasonable, but the point is that even if it seems like a reasonable assumption, it's still an assumption that needs stating. It isn't hard to cook up examples where it fails to hold and you still get the same probabilities of 1st.
• Feb 7th 2011, 04:45 PM
SumGuy
Imagine Runner 3 wasn't running, the probability ratio would stay the same wouldn't it? Meaning that R2has a 1/9 chance of beating R1. The same could be said about R3 if R2 were taken out. Now, since they are mutually exclusive (hoping the terminology is correct there), doesn't that mean there is a (1/9) * (1/9) chance of both R2 AND R3 beating R1? The event of them both beating R1 means that R1 must be 3rd. It would therefore happen 1/81 times (or 0.0123 in probability terms). So the probability of 2nd is 1 - 0.8 - 0.0123) = 0.1877

Now, that means the chances of R2 and R3 to come 2nd are each (1 - 0.1877) / 2 = 0.40615
and the probability of 3rd is (1 - 0.0123) / 2 = 0.49385

Making the table as follows:

R1: 0.8, 0.18770, 0.01230
R2: 0.1, 0.40165, 0.48395
R3: 0.1, 0.40615, 0.49385

That is, assuming you haven't corrected any of my logic in the time it took me to calculate and post that :-)

EDIT: that is if the error of assuming equivalence is neglected, but would it be so dangerous to neglect this. I mean, I completely understand that in maths, you can't make assumptions, but again, realistically, would you say this would bring that big of a margin of error into it? I mean, I'm after rounding 1/81 to 0.0123. There has to be some room for margin / rounding. Would you say that the margin of error from above would be that bad? Don't get me wrong, I'm still thankful you're pointing it out.
• Feb 7th 2011, 04:49 PM
theodds
Quote:

Originally Posted by SumGuy
Imagine Runner 3 wasn't running, the probability ratio would stay the same wouldn't it? Meaning that R2has a 1/9 chance of beating R1. The same could be said about R3 if R2 were taken out. Now, since they are mutually exclusive (hoping the terminology is correct there), doesn't that mean there is a (1/9) * (1/9) chance of both R2 AND R3 beating R1? The event of them both beating R1 means that R1 must be 3rd. It would therefore happen 1/81 times (or 0.0123 in probability terms). So the probability of 2nd is 1 - 0.8 - 0.0123) = 0.1877

Now, that means the chances of R2 and R3 to come 2nd are each (1 - 0.1877) / 2 = 0.40615
and the probability of 3rd is (1 - 0.0123) / 2 = 0.49385

Making the table as follows:

R1: 0.8, 0.18770, 0.01230
R2: 0.1, 0.40165, 0.48395
R3: 0.1, 0.40615, 0.49385

That is, assuming you haven't corrected any of my logic in the time it took me to calculate and post that :-)

I haven't looked much at this, but the bold part is wrong. If you know that Runner 2 beats Runner 1, then Runner 3 has a much higher chance of also beating Runner 1, because it means that Runner 1 has probably performed poorly relative to his capabilities. In technical terms, you are assuming that X, Y, and Z are independent, and then trying to use that to assert that X - Y and X - Z are independent - they aren't because if X is small then you have reason to think that X - Y and X - Z are both small as well.

Also, the term is independent. Mutually exclusive means that it would be impossible for both events to happen simultaneously.
• Feb 7th 2011, 05:24 PM
SumGuy
Oh, sorry about that, my terminology may be a little rusty alright. As for the bold part listed, you're right they're not independent alright, but could they not be balanced? I.e. R1 could have a bad run alright, but R2 or R3 might also be on a bad day and only one of them beats R1, R1 might be on an average day and either of R2 or R3 might be on an exceptionally good day and win that way.

Also, the original 80% chance would factor in the chance of a bad day. I know it doesn't put a number on it, but the 2nd and 3rd places are derived from a probability that already takes this into account. Having said that, I get what you're saying. If he doesn't win, there's a good chance he's on a bad day, increasing the probability or 3rd instead of 2nd.

Maybe you're right and it is impossible to calculate solely on 1st place probabilities, and I'm nowhere near the standard of people on this board, so I won't get it.

Even if I give up, I'll still come back to check other peoples opinions on this out of interest. I hope this survives as a discussion cause I am interested in this one.

You've earned my thanks good sir, and please don't take any of my arguments as attempts to be right, it was merely a mixture of hope that this was possible and a want to learn why I was wrong.