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 2013-03-25, 15:29 #1 c10ck3r     Aug 2010 Kansas 547 Posts 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
2013-03-25, 19:30   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by c10ck3r 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.

 2013-03-25, 20:22 #3 Mr. P-1     Jun 2003 7×167 Posts 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
 2013-03-25, 21:07 #4 Mr. P-1     Jun 2003 22218 Posts 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
2013-03-25, 21:13   #5
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

100000110000002 Posts

Quote:
 Originally Posted by science_man_88 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*

 2013-03-25, 21:21 #6 c10ck3r     Aug 2010 Kansas 10438 Posts 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.
2013-03-26, 00:30   #7
Mr. P-1

Jun 2003

7×167 Posts

Quote:
 Originally Posted by c10ck3r 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.

2013-03-26, 00:47   #8
c10ck3r

Aug 2010
Kansas

547 Posts

Quote:
 Originally Posted by Mr. P-1 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).

2013-03-26, 01:02   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by c10ck3r 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.

2013-03-26, 01:14   #10
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by c10ck3r 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

2013-03-26, 01:35   #11
Mr. P-1

Jun 2003

7×167 Posts

Quote:
 Originally Posted by science_man_88 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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