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 2009-02-24, 13:17 #1 Mr. P-1     Jun 2003 116910 Posts The Colour of Money I'm not sure if this counts as a puzzle as I do not have a definitive answer. The Colour of Money is a gameshow currently being aired on British Television. Inevitably the show has a lot of choose-your-favourite-colour and suspenceful-count-downs and other woo which has no effect upon actual game strategy. Stripped of these elements, the game is as follows: The player is given a target score at the beginning of the game, which is stipulated as being between 55 and 155. (The actual values in the first two games were 62 and 64, and I suspect that in practice the targets will be around this level.) There are ten boxes, into which have been placed ten whole numbers chosen randomly from the range 1 - 20 inclusive, no two the same. The goal of the game is for the player to achieve his target score exactly. He scores by bidding on each of the boxes one at a time. If his bid is lower or equal than the number in the box, then he scores the amount bid. If the bid is greater, then he scores nothing. Regardless of whether the bid is successful, the box is opened and the player shown the actual number before he moves on to the next box. He does not need to bid on every box if he reaches his target before all have been bid. My questions are: 1. What's an optimal, or at least good strategy for playing this game? 2. For a given target, what is the initial probability of success given optimal strategy? 2. If, instead of being given his target, the player is allowed to choose it at the start of the game, what should he choose to maximise his expected score (i.e., to maximise the target multiplied by the probability of success)? The last two rounds of the game are easy to analyse. In the last round, the player has no choice, he must bid the exact amount to bring his score to his target, if this is achievable at all. In the penultimate round, the probability of success for each potential bid can be easily calculated, and will depend, not just on the amount left to score, but also upon the contents of the eight boxes already disclosed. Things are much more complex in the eighth and earlier rounds. In the very early rounds, a near optimal strategy will probably be to make the bid which maximises the expected score for that round. In the first round, this will be to bid 10 or 11, both of which yield an expected score of 5.5. In subsequent rounds, this will change the contents of successive boxes are disclosed. But at some point in the game, this will cease to be optimal. A player who has been unlucky (or who has a very high target) will need to increase his bids to have a chance of success, while one who has been lucky may be better off reducing them, in order to reduce his chance of failing to score at all. So, how best to play this game? Last fiddled with by Mr. P-1 on 2009-02-24 at 13:19
 2009-02-28, 08:40 #2 xorbe     Feb 2009 73 Posts Seems like a good strategy would be to bid the largest number initially, 20. Certainly you will quickly approach 64. Then, lower your bid by the amount as to not go over. Part of the problem is that the set of values in the 10 boxes are unknown, so some of it really comes down to hard luck. 1...10 is 55, 11...20 is 155, so 62-64 is really on the low side, sounds easy to get stuck with numbers that are too large to have an average winning strategy (if indeed the 10 numbers are picked at random).
2009-03-02, 09:13   #3
Mr. P-1

Jun 2003

7·167 Posts

Quote:
 Originally Posted by xorbe Seems like a good strategy would be to bid the largest number initially, 20. Certainly you will quickly approach 64.
I doubt that. If we ignore the rule that no two numbers in the boxes may be the same, your chance of success on a bid of 20 is 5% in each round. It's straight forward to calculate the probability of three or more successes in nine round, but because I'm lazy, I'll just calculate the odds of exactly three successes:

(.95^6)*(.05^3)*9!/(6!*3!) = 0.77%

Compare this with the probability of getting six successes in nine rounds with a bid of ten each time. That bid has a 55% success rate, so:

(.45^3)*(.55^6)*9!/(3!*6!) = 21.19%

Quote:
 Part of the problem is that the set of values in the 10 boxes are unknown, so some of it really comes down to hard luck. 1...10 is 55, 11...20 is 155, so 62-64 is really on the low side, sounds easy to get stuck with numbers that are too large to have an average winning strategy (if indeed the 10 numbers are picked at random).
Certainly luck will play a part. But are you sure you understand the game? It's to the player's advantage if the numbers in the boxes are large.

 2009-03-03, 02:34 #4 xorbe     Feb 2009 73 Posts "If his bid is lower or equal than the number in the box, then he scores the amount bid." Whoops! I read that as if the box number is lower or equal than the bid... so much for reading comprehension at 12:40 am!

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