[Guest]

During my extensive search for new conumdrums I noticed a game often played in bars in South East Asia. There are 9 tiles numbered 1 to 9 on the back. Two dice are thrown and a tile is turned face up based on the following,

CHoose a number from either die and turn over the corresponding tile.

Turn over a tile that is the sum of the two numbers on the dice.

E.G. if a 2 and a 4 is thrown then the player has a choice of turning over tile number 2, 4, or 6. However if a double 6 is thrown, the player can only turn over tile number 6. The player keeps throwing until all tiles are turned over and he wins the game and gets a free beer, or until he cannot turn over a tile because it has been turned over already, in which event he looses.

WHat is the probibility of winning that free beer?

[bonanova]

Assuming the probability of success is affected by choices that are made,

what should we assume about how choices are made?

1. They are made at random
2. They are made following a best strategy

If, as seems likely, [2] is intended, then is part of the puzzle to find a best choosing strategy?

i.e. might the OP be stated: Assuming a player is experienced and well skilled in this game,

what is the best outcome he can achieve?

[Guest]

Assuming the probability of success is affected by choices that are made,

what should we assume about how choices are made?

1. They are made at random
2. They are made following a best strategy

If, as seems likely, [2] is intended, then is part of the puzzle to find a best choosing strategy?

i.e. might the OP be stated: Assuming a player is experienced and well skilled in this game,

what is the best outcome he can achieve?

The question should have stated that the player is skilled in the game and follows the best strategy. The answer shoud identify that strategy.

[Guest]

A good strategy for this game would be, when you get a choice you pick the tile with the least probability of achieving later...

For instance if you get 2 & 5 you need to choose between 2 5 and 7, the probability of getting a 2 is if you get 1-1, 2-x, x-2, which are 12, for 5 you'll need 1-4 2-3 3-2 4-1 5-x x-5 which are 15, for 7 you'll need a combo of 1-6 2-5 3-4 4-3 5-2 6-1 which are 6, so you'd pick 7, if 7 is picked you'd pick the 2, that would give you the best chances of winning...

I dunno how to calculate the probabilities yet, even if you do follow this strategy there are still cases in which it would have been better if you didn't (like if in above example you picked the 2 and the 7 but later on you never got a chance to do the 5 and you lose)

[Guest]

12/36 or 0.3333 of getting the option to flip one

13/36 or 0.3611 of getting the option to flip two

14/36 or 0.3889 of getting the option to flip three

15/36 or 0.4167 of getting the option to flip four

16/36 or 0.4444 of getting the option to flip five

17/36 or 0.4722 of getting the option to flip six

5/36  or 0.1667 of getting the option to flip seven

5/36  or 0.1389 of getting the option to flip eight

4/36  or 0.1111 of getting the option to flip nine

so in order 9,8,7,1,2,3,4,5,6 should be the order to flip for the stratagey

but I'm not sure how to add it all up ... 1/9th chance for getting the chance to flip a 9 and you only get nine flips anyways and it's the hardest one to get

so it obviously can't be more then a 11.11% chance and greater then 0.0% chance ... ha i narrowed it down

i'm currently writing a program that will go the this strategy to see what the percent is

[Guest]

Lost this many times: 9203606

Won this many times : 796394

Percent to win was: 7.96394% out of 10000000 games.

edit: for this question

Is the percent close and all I need to do is come up with the equation now or am I still way off?

Edited June 14, 2010 by PVRoot

[Guest]

Lost this many times: 92,016,755

Won this many times : 7,983,245

Percent to win was: 7.98325% out of 100,000,000 games.

(note): I add the commas just for readability

[Guest]

You're right, the answer is close to 8%. As far as I know there is no easy equation that can give you an exact solution. The only way is to run a simulation as you're done. I think the bar owners rely on the fact that most punters don't know about probability's and go for a random choice when choosing which tile to turn over. What % of wins would you expect if you adopted this approach?

Lost this many times: 92,016,755

Won this many times : 7,983,245

Percent to win was: 7.98325% out of 100,000,000 games.

(note): I add the commas just for readability