## Another Probability Question, About Odds and betting

1. The problem statement, all variables and given/known data
I don't think the exact words are needed here, but let me describe the situation, basically you have 7 horses each with its own probability of coming first in a race. You place a "triactor" bet which is naming the exact order of the first 3 horses.

So you bet on the 3 horses with the highest probability of coming first, but since order matters, don't you need to probability of a horse coming second and third?

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 Quote by wk1989 1. The problem statement, all variables and given/known data I don't think the exact words are needed here, but let me describe the situation, basically you have 7 horses each with its own probability of coming first in a race. You place a "triactor" bet which is naming the exact order of the first 3 horses. So you bet on the 3 horses with the highest probability of coming first, but since order matters, don't you need to probability of a horse coming second and third?
Sure you do. Suppose the horses always finish in the same order, 1,2,3,... Then the probability of 1 coming in first is 1 and the probability of the rest is 0. But that still gives you no idea who to bet on for second. BTW I'm no gambler - but I think it's 'trifecta'.

 Quote by Dick Sure you do. Suppose the horses always finish in the same order, 1,2,3,... Then the probability of 1 coming in first is 1 and the probability of the rest is 0. But that still gives you no idea who to bet on for second.
Hmm this is an extreme case, but what if each horse has a non-zero probability of winning? I think the problem can be solved then, without knowing anything else. (Really, no one would want to watch/bet on a race when the winner is known in advance! )

In fact, I think if three horses have a non-zero chance of winning, it should be enough (but I might be naive)

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## Another Probability Question, About Odds and betting

All variations are 7!=7*6*5*4*3*2=5040 (variations, since position matters)
From 5040 variations, 120 are with 1,2,3 on first place (5!=5*4*3*2=120).
So the chance to win is 120/5040=0.0238 (around 2.4% if all horses have equal chance of winning). I think its pretty hard to match three horses in a row.

 Quote by Дьявол All variations are 7!=7*6*5*4*3*2=5040 (variations, since position matters) From 5040 variations, 120 are with 1,2,3 on first place (5!=5*4*3*2=120). So the chance to win is 120/5040=0.0238 (around 2.4% if all horses have equal chance of winning). I think its pretty hard to match three horses in a row.
Actually it should be 4!/7!, which is even lower.

Which is why you need those hot tips

 This is not really a purely mathematical, question - i.e., you're not really calculating any particular probability or trying to prove anything. In my view it's more a question of what would satisfy you. The example given (with all horses always coming in in a fixed order) is quite good and should convince that the given probabilities aren't enough. I would say that even if each horse has a positive probability of coming in first, it still wouldn't suffice: let's say horse 1 has a probability 1/2 of coming in first, and the remaining 1/2 is divided equally between all others. I would say that you then have more or less the same problem on who to bet on for the second and third places (and it could very well be that given the winning of the first horse, the second horse - gets really bummed and - comes in last with probability 1, so the answer "in that case there is no preference to any particular horse" is not convincing).

 Quote by Palindrom it could very well be that given the winning of the first horse, the second horse - gets really bummed and - comes in last with probability 1, so the answer "in that case there is no preference to any particular horse" is not convincing).
LOL I hadn't thought of this, it is a nice argument! If horses have complex emotions (which they probably do), I agree that it is not a mathematical problem.

However, if we assume that these probabilities only depend on the physical capabilities of the horse - and there is no reason to expect that these will change based on some other horse winning - the "in that case there is no preference" argument seems very convincing to me.

 Blog Entries: 1 If the first horse have probability of 50% gaining first and the other one "share" the probability the probability "jumps" for around 1/6=16%.

 Quote by naresh However, if we assume that these probabilities only depend on the physical capabilities of the horse - and there is no reason to expect that these will change based on some other horse winning - the "in that case there is no preference" argument seems very convincing to me.
Sorry, the english was perhaps unclear. I meant that the relative probabilities (of coming second, among the horses that did not come in first) will remain unchanged (from the probabilities of these horses winning). I guess this is still bad phrasing, but I suspect everyone is saying the same thing anyway.