20130325, 15:29  #1 
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 
20130325, 19:30  #2  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:


20130325, 20:22  #3 
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<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. 
20130325, 21:07  #4 
Jun 2003
1169_{10} 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 k2. (k1 is not a possible score). In other words, you get a (k1)ary digit. After i plays, there will be k!i untested arrangements, and (k1)^{i} information states. By the pigeonhole principle, it is impossible for you to know for certain the correct arrangement for certain after i plays, unless (k1)^{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. P1 on 20130325 at 21:11 
20130325, 21:13  #5  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20130325, 21:21  #6 
Aug 2010
Kansas
547 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.

20130326, 00:30  #7  
Jun 2003
7×167 Posts 
Quote:
Of course, all algorithms might (if you're lucky) give the correct answer in 1 try. 

20130326, 00:47  #8  
Aug 2010
Kansas
1043_{8} Posts 
Quote:
This algorithm, however, cannot give the correct answer in 1 attempt (by reason of the methodology). 

20130326, 01:02  #9 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
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.

20130326, 01:14  #10 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
really? because she won one of them in the video.
Last fiddled with by science_man_88 on 20130326 at 01:15 
20130326, 01:35  #11 
Jun 2003
7×167 Posts 

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