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Old 2013-03-25, 15:29   #1
c10ck3r
 
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Default Price is Right Race Game

Okay, so I was watching The Price is Right today, and one of the games was "The Race Game". The gal that did it failed epically, but it got me thinking: What is the most efficient method to win?
I have my idea, but I'm curious if anyone on the forum has a better idea (or so much time on their hands that they feel like solving mathematically).
Pretty much: there are four prizes, described to the contestant. The contestant is given 45 seconds to affix the right tag to the right prize. Pulling a lever back at the start will tell you how many you got right. What's the best way to solve this puzzle, assuming you aren't certain on the prices?





See more: http://www.youtube.com/watch?v=2aw5sYeBlZk
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Old 2013-03-25, 19:30   #2
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Quote:
Originally Posted by c10ck3r View Post
Okay, so I was watching The Price is Right today, and one of the games was "The Race Game". The gal that did it failed epically, but it got me thinking: What is the most efficient method to win?
I have my idea, but I'm curious if anyone on the forum has a better idea (or so much time on their hands that they feel like solving mathematically).
Pretty much: there are four prizes, described to the contestant. The contestant is given 45 seconds to affix the right tag to the right prize. Pulling a lever back at the start will tell you how many you got right. What's the best way to solve this puzzle, assuming you aren't certain on the prices?





See more: http://www.youtube.com/watch?v=2aw5sYeBlZk
until you described it I thought you meant the one with the mice, I would randomly choose the prices hoping for as low as possible if not all of them, then switch 2 if you get 0 again, or 2 you then have a 1 in 4 ( or 1 in 1 respectively) chance of getting it on your next try instead of the usual 1/(4^4) to win all four.
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Old 2013-03-25, 20:22   #3
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The immediate, obvious generalisation is to consider the case with k different prizes.

There are numerous ways to interpret the problem. Are you trying to determine N, the least number of plays in which you can guarantee to find the fully correct answer, and a algorithm which will do this? Or are you trying to find an optimal solution for some n<N. If the latter, are you trying to maximise the probability of a perfect score? Your expected number of right answers? Your expected prize value, given that the prizes are not all the same value?

Are the k! different arrangements all equally likely? Or does your knowledge of how things are priced in general mean you assign different initial probabilities to different arrangements?

You might also model time differently, for example, it may take less time to swap two valuations over than to do a more complicated permutation.
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Old 2013-03-25, 21:07   #4
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Let's analyse the simplest interpretation, which is that you are trying to determine N the least number of plays in which you can guarantee to find the right answer, and an algorithm to do this.

Each play you make is either correct, in which case you're done, or it eliminates 1 possible arrangement, as well as giving you a score from 0 to k-2. (k-1 is not a possible score). In other words, you get a (k-1)-ary digit. After i plays, there will be k!-i untested arrangements, and (k-1)i information states. By the pigeon-hole principle, it is impossible for you to know for certain the correct arrangement for certain after i plays, unless (k-1)i >= k!-i. For k=4. the smallest i satisfying this inequality is 3, and your best hope is for an algorithm which allows you to get the right answer on the 4th play. This argument does not demonstrate that such an algorithm exists, only that no algorithm can solve the problem in fewer tries.

Last fiddled with by Mr. P-1 on 2013-03-25 at 21:11
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Old 2013-03-25, 21:13   #5
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Quote:
Originally Posted by science_man_88 View Post
until you described it I thought you meant the one with the mice, I would randomly choose the prices hoping for as low as possible if not all of them, then switch 2 if you get 0 again, or 2 you then have a 1 in 4 ( or 1 in 1 respectively) chance of getting it on your next try instead of the usual 1/(4^4) to win all four.
edit:sorry or all of them*
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Old 2013-03-25, 21:21   #6
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Clarification: you win all or none of the prizes. For the sake of the calculation, ignore time, focusing on least number of attempts. I've solved for a method that uses a maximum of 10 tries, with a very real possibility of 4 tries.
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Old 2013-03-26, 00:30   #7
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Quote:
Originally Posted by c10ck3r View Post
Clarification: you win all or none of the prizes. For the sake of the calculation, ignore time, focusing on least number of attempts. I've solved for a method that uses a maximum of 10 tries, with a very real possibility of 4 tries.
Do you mean you've found an algorithm which will give you the right answer in at most 10 tries, but which might (if you're lucky) give it in 4? Or that you're not sure what the maximum number of tries your algorithm requires, only that it's not more than 10 and may be as low as 4?

Of course, all algorithms might (if you're lucky) give the correct answer in 1 try.
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Old 2013-03-26, 00:47   #8
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Quote:
Originally Posted by Mr. P-1 View Post
Do you mean you've found an algorithm which will give you the right answer in at most 10 tries, but which might (if you're lucky) give it in 4? Or that you're not sure what the maximum number of tries your algorithm requires, only that it's not more than 10 and may be as low as 4?

Of course, all algorithms might (if you're lucky) give the correct answer in 1 try.
I mean I've found a method that will use somewhere between 4-10 attempts, dependent on the answer.
This algorithm, however, cannot give the correct answer in 1 attempt (by reason of the methodology).
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Old 2013-03-26, 01:02   #9
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Quote:
Originally Posted by c10ck3r View Post
I mean I've found a method that will use somewhere between 4-10 attempts, dependent on the answer.
This algorithm, however, cannot give the correct answer in 1 attempt (by reason of the methodology).
is it brute force ? because that's about how many it would take with brute force, 4 for the first one 3 for the next one 2 of the next one, oh wait the second one for that last one solves the last one allowing it to be solved by brute force in 9.
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Old 2013-03-26, 01:14   #10
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Quote:
Originally Posted by c10ck3r View Post
Clarification: you win all or none of the prizes. For the sake of the calculation, ignore time, focusing on least number of attempts. I've solved for a method that uses a maximum of 10 tries, with a very real possibility of 4 tries.
really? because she won one of them in the video.

Last fiddled with by science_man_88 on 2013-03-26 at 01:15
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Old 2013-03-26, 01:35   #11
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Quote:
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really? because she won one of them in the video.
c10ck3r is specifying the problem he wants us to solve, which is different from the one faced by the contestant.
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