Tenis Pro Problem
Poll
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14 members have voted
May 2nd, 2014 at 10:19:17 AM
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On a trivia game show you win $1,000,000 if you win 2 consecutive rounds of trivia played in alternating rounds against an audience member and the trivia champion. There are 3 rounds and you can choose which opponent to play first. So you will either play the champion twice or the audience member twice.
When is it advantageous to play against the champion first?
p(a) is the probability of beating the audience member
p(c) is the probability of beating the champion
the audience member is an easier opponent than the champion p(a)>p(c)
edit: if you are wondering why the title is called tennis pro problem it is because it was originally posed as a tennis match.. i changed the premise to trivia so the answer couldn't be googled as easily.. but i forgot to change the title. doh!!
When is it advantageous to play against the champion first?
p(a) is the probability of beating the audience member
p(c) is the probability of beating the champion
the audience member is an easier opponent than the champion p(a)>p(c)
answer in spoiler
edit: if you are wondering why the title is called tennis pro problem it is because it was originally posed as a tennis match.. i changed the premise to trivia so the answer couldn't be googled as easily.. but i forgot to change the title. doh!!
May 2nd, 2014 at 10:56:48 AM
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Looks to me as if it is always best to play the (weaker) audience member first and maybe twice.
What have I overlooked?
(Edit: see my comments below.)
For playing the audience member first:
p(win 2 straight) = p(a)*p(c) + (1-p(a))*p(c)*p(a) = p(a)*p(c)*(2-p(a))
For playing the champion first:
p(win 2 straight) = p(c)*p(a) + (1-p(c))*p(a)*p(c) = p(c)*p(a)*(2-p(c))
Playing the audience member first is better by a factor of (2-p(a))/(2-p(c)),
which is >1 since p(a)>p(c).
Of course, this assumes that the game stops once you have won two straight rounds or cannot do so, otherwise the formulae are for p(win 2 or more straight)
p(win 2 straight) = p(a)*p(c) + (1-p(a))*p(c)*p(a) = p(a)*p(c)*(2-p(a))
For playing the champion first:
p(win 2 straight) = p(c)*p(a) + (1-p(c))*p(a)*p(c) = p(c)*p(a)*(2-p(c))
Playing the audience member first is better by a factor of (2-p(a))/(2-p(c)),
which is >1 since p(a)>p(c).
Of course, this assumes that the game stops once you have won two straight rounds or cannot do so, otherwise the formulae are for p(win 2 or more straight)
What have I overlooked?
(Edit: see my comments below.)
May 2nd, 2014 at 11:05:45 AM
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Is the trivia themed? Do the questions get progressively harder?
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
May 2nd, 2014 at 11:54:25 AM
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I just noticed a problem with my previous answer. Guess I need to look at this a little deeper.
Suppose that the audience member is so weak that you are certain to win that competition; i.e., p(a)=1. In that case, your probability of winning two straight beginning with the audience member is:
p(win 2)=p(c) i.e., win your sole match against the champion.
Your probability of winning starting with the champion becomes:
p(win 2)=p(c)*(2-p(c)) i.e, win either of the two matches against the champion.
That suggests that you should begin with the champion if you are certain that you will beat the audience member. Maybe you should start with the champion if you are almost certain to beat the audience member.
More thinking is needed, I suppose.
p(win 2)=p(c) i.e., win your sole match against the champion.
Your probability of winning starting with the champion becomes:
p(win 2)=p(c)*(2-p(c)) i.e, win either of the two matches against the champion.
That suggests that you should begin with the champion if you are certain that you will beat the audience member. Maybe you should start with the champion if you are almost certain to beat the audience member.
More thinking is needed, I suppose.
May 2nd, 2014 at 12:26:27 PM
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Oh, I missed the 'two consecutive' part...
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
May 2nd, 2014 at 1:01:46 PM
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That is a pretty cool problem, since the answer might be seen as counter intuitive..
It is more intuitive when presented like this:
You must win Game 2 or you are out of luck. So it is always better to maximize the chances in game 2 - play the audience member, and have two cracks at NOT losing against the pro in game 1 and 3.
Total probability is therefore
P(Win against Audience) * (1- P(Losing against Pro)²)
As long as P(Win Audience) > P(Win Pro) this is always the best option. [Insert mathematical proof here, which I only have in messy longwinded format on an old receipt.]
Ergo: Always play the Pro first.
It is more intuitive when presented like this:
You must win Game 2 or you are out of luck. So it is always better to maximize the chances in game 2 - play the audience member, and have two cracks at NOT losing against the pro in game 1 and 3.
Total probability is therefore
P(Win against Audience) * (1- P(Losing against Pro)²)
As long as P(Win Audience) > P(Win Pro) this is always the best option. [Insert mathematical proof here, which I only have in messy longwinded format on an old receipt.]
Ergo: Always play the Pro first.
May 2nd, 2014 at 1:14:05 PM
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This is just a straightforward probability question.
Let p(A) be the probability of beating the first person you play, and p(B) be the probability of beating the 2nd person you play. Then your probability of winning 2 in a row is 2*p(A)*p(B) - p(A)*p(A)*p(B). (It's just the probability of winning the first 2, plus the probability of winning the second 2, minus the probability of winning all 3).
This can be reduced to (2-p(A)) * (p(A)*p(B)). Since these are all positive numbers, the value can be maximized by minimizing p(A), which means, play the one who is harder to beat (the pro) first. Always.
This can be reduced to (2-p(A)) * (p(A)*p(B)). Since these are all positive numbers, the value can be maximized by minimizing p(A), which means, play the one who is harder to beat (the pro) first. Always.
May 2nd, 2014 at 2:28:13 PM
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Quote: Doc
What have I overlooked?
The final expresion you came up with is always <= 1 since p(a)>p(c)Quote:which is >1 since p(a)>p(c).
May 2nd, 2014 at 2:51:51 PM
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Yes, chrisr, I got my inequalities reversed. At least I rather quickly noticed there was a problem, but I didn't get around to digging out the error.
Thanks for pointing it out.
Thanks for pointing it out.
May 2nd, 2014 at 4:34:54 PM
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If you play the audience member first, you win if either you beat the audience member and then the trivia champion, or lose to the audience member and then beat the trivia champion and audience member:
p(a) p(c) + (1 - p(a)) p(c) p(a) = p(a) p(c) (1 + (1 - p(a)) = p(a) p(c) (2 - p(a))
If you pick the trivia champion first, you win if either you beat the trivia champion and then the audience member, or lose to the trivia champion and then beat the audience member and the trivia champion:
p(c) p(a) + (1 - p(c)) p(a) p(c) = p(c) p(a) (1 + (1 - p(c)) = p(a) p(c) (2 - p(c))
p(a) > p(c) -> 2 - p(a) < 2 - p(c) -> p(c) p(a) (2 - p(a)) < p(c) p(a) (2 - p(c))
Play the champion first
It is a little counter-intuitive, but keep in mind that you get only one try to beat the person you play second, and have two to beat the person you play first.
May 2nd, 2014 at 5:01:00 PM
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The problem seem counter-intuitive, because we are tricked into believing we need to beat all 3 rounds.
If A is very easy and C is very hard to beat, we would definetly like a free-roll on C (and not on A).
Interestingly, if we could choose at any stage who to try next (with the win granted if we beat both A and C), our chance of success would not depend on the choice of the first round.
May 2nd, 2014 at 5:13:01 PM
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Quote: MangoJ
The problem seem counter-intuitive, because we are tricked into believing we need to beat all 3 rounds.
If A is very easy and C is very hard to beat, we would definetly like a free-roll on C (and not on A).
Interestingly, if we could choose at any stage who to try next (with the win granted if we beat both A and C), our chance of success would not depend on the choice of the first round.
Well, yeah. If the only requirement is that you beat them both then the only "wrong" choice would be to waste a round playing someone you have already beaten.
May 3rd, 2014 at 12:14:21 AM
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Here's my answer.
Always play the champion first.
To win two consecutive rounds, you always have to win the middle round. Therefore, you have to win the middle round AND (the first round OR the third round). Whichever you choose, the first and third rounds will be the same, so we'll call that p, and the middle round q. Assuming the three are independent, your chance of winning is:
q(1-(1-p)^2) = q(1-(1-2p+p^2)) = q(2p-p^2) = 2pq - p^2*q.
Since 2pq will be the same either way, the way to maximize this is to minimize p^2*q, and the way to do that is to make p less than q. This is counterintuitive since it means you'll be facing the harder opponent twice (which I suspect is why it's currently trailing 7-6 in the poll), but since you don't have to win all three matches, the middle one being the most important, I'm fairly sure it's right.
(Besides, it's always the counterintuitive answer in these riddles. Except when the teller botches the setup. But usually.)
It looks like everyone else who's given it any real thought has come to the same answer... but then what is with...
Always play the champion first.
To win two consecutive rounds, you always have to win the middle round. Therefore, you have to win the middle round AND (the first round OR the third round). Whichever you choose, the first and third rounds will be the same, so we'll call that p, and the middle round q. Assuming the three are independent, your chance of winning is:
q(1-(1-p)^2) = q(1-(1-2p+p^2)) = q(2p-p^2) = 2pq - p^2*q.
Since 2pq will be the same either way, the way to maximize this is to minimize p^2*q, and the way to do that is to make p less than q. This is counterintuitive since it means you'll be facing the harder opponent twice (which I suspect is why it's currently trailing 7-6 in the poll), but since you don't have to win all three matches, the middle one being the most important, I'm fairly sure it's right.
(Besides, it's always the counterintuitive answer in these riddles. Except when the teller botches the setup. But usually.)
It looks like everyone else who's given it any real thought has come to the same answer... but then what is with...
...the poll?
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
