## Probability of selecting the winner betting on 2 out of 5

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jacksmack
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### Probability of selecting the winner betting on 2 out of 5

Hi,

I have this exercise about probability. But I have not understand the reasoning followed.

This is the problem:
5 horses are in a race. John bets on 2 of them. Find the probability p that John picked the winner.

This is the solved example:
There are C(5,2)=10 ways to select 2 of the horses. Four of the pairs will contain the winner. Therefore, p = 4/10 = 2/5.

It's clear that the number of ways to select 2 out of 5 horses is 10,
but I don't understand the second part where states "Four of the pairs will contain the winner".

many thanks!

Sizik
Posts: 1215
Joined: Wed Aug 27, 2008 3:48 am UTC

### Re: Probability of selecting the winner betting on 2 out of

Say the horses are A, B, C, D, and E, and A is the winner.
The 10 possible pairs or horses are:
(A, B)
(A, C)
(A, D)
(A, E)
(B, C)
(B, D)
(B, E)
(C, D)
(C, D)
(D, E)

Out of these 10, four pairs contain A.
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Qaanol
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### Re: Probability of selecting the winner betting on 2 out of

A quick solution is to use the linearity of expectation:

For any particular horse I pick, my expected number of wins per play is 1/5. Therefore when I pick two horses my expected number of wins per play is 2/5.

Since only one horse can win at a time, my odds of picking the winner must also be 2/5.
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PeteP
What the peck?
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### Re: Probability of selecting the winner betting on 2 out of

You pair everyone with everyone and there are 4 losers=> Pairing the winner with all of them gives you 4 pairs which contain the winner.

@Qaanol I assume they wanted an example with obvious results to explain the concept. I think for an example like this it's intuitively clear for most that it could be done in the way you mention without having ever heard of the concept.

Qaanol
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### Re: Probability of selecting the winner betting on 2 out of

PeteP wrote:You pair everyone with everyone and there are 4 losers=> Pairing the winner with all of them gives you 4 pairs which contain the winner.

@Qaanol I assume they wanted an example with obvious results to explain the concept. I think for an example like this it's intuitively clear for most that it could be done in the way you mention without having ever heard of the concept.

Thank you, PeteP, for providing such a great segue into my number one rule about probabilities. And that is, “Never ever trust your intuition when it comes to probabilities, ever.”

For example, say you and I each pick a horse at random. What is the probability that at least one of us picked the winner? Hint: it’s not 2/5 anymore, even though together the expected number of wins for us is still 2/5.
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PeteP
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### Re: Probability of selecting the winner betting on 2 out of

Which is also intuitively obvious so entirely unconvincing to proof your point. (Yes there is much about probability where intuition is wrong and misleading, I was just amused by the example.)

Dopefish
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### Re: Probability of selecting the winner betting on 2 out of

PeteP wrote:Which is also intuitively obvious so entirely unconvincing to proof your point. (Yes there is much about probability where intuition is wrong and misleading, I was just amused by the example.)

Is it? I reckon it's
Spoiler:
1-(4/5)*(4/5)=0.36
but that's not a number that comes from intuition, but rather experience with that sort of problem. I think more people would conclude it's
Spoiler:
1/5+1/5=2/5
or something along those lines if they went by their intuitions if they hadn't dealt with this sort of problem before.

PeteP
What the peck?
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### Re: Probability of selecting the winner betting on 2 out of

I'm not talking about the exact result, but about it not being 2/5. That betting separately gives you lower chances than betting together because you could bet on the same thing.

Diemo
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### Re: Probability of selecting the winner betting on 2 out of

Yeah, but most people [citation needed] would say that it is 2/5, because most people would fail to take into account that you could bet on the same horse [citation needed].

I believe that most people would answer that as 2/5, but have no proof whatsoever
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