Is this game beatable? (weird rock-paper-scissor thing)

[Fryguy]

Ok, you are heads-up with an opponent, and each of you has 6 numbers, 1-6, and you use each number exactly once. There is also a "set of prizes" labeled 1-6 that neither of you are in control over.

One of the prizes is determined at random, and revealed to both players. Each player then picks one of their numbers, and whoever picks the higher number gets the value of all 3 numbers.

Example: The prize comes out "3" Player 1 picks 1, and player 2 picks 6. Player 2 wins 10 (3 + 1 + 6).

At this point another prize comes out, and you repeat this process for each of the 6 numbers. Each prize number comes out exactly once.

Ties: If both players pick the same number, it gets added to the prize and nobody wins that round, and the next rounds prize is added on as well.

Example: The prize is 3. Both players pick 3. The prize becomes 9 + whatever the value of the next prize is. So if the next prize is a 1, the prize is now 10.

The object of the game is to have a higher score than your opponent at the end of the 6 rounds.

From experience it seems intuitive that you cannot beat this game 100%, however is there an optimal strategy for it so that you can win a majority of the time?

[Thremp]

Yeah, play against really dumb people.

[ughaulkghalugh]

moola?

[Fryguy]

yes moola. I played for a while and it doesn't seem like it's remotely beatable for even $100 at a time. I eventually gave up, claiming it's basically rock-paper-scissors with too much luck factor to be worth trying to make money from.

[AlphaWice]

It's weighted rock paper scissors. Consider a strategy P of playing the same card as the prize card. It beats most strategies, except a strategy to play one higher than the prize card.. etc.

[SheetWise]

If I understand your example --

[ QUOTE ]
Example: The prize comes out "3" Player 1 picks 1, and player 2 picks 6. Player 2 wins 10 (3 + 1 + 6).


[/ QUOTE ]

You then play all 6 rounds?

[AlphaWice]

<u>What happens if it is tied on the last round?</u>


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I think an optimal strategy is really close to the following:

play the same as the prize card, except if you can force a win by playing your highest, second highest, etc.

[Siegmund]

Sounds like John Conway's "Gops" (Game of Pure Strategy) for n=6. (Actually, not sure Conway invented it - I just first read about it in his big book on strategies of games.) There exist strategies to ensure a 50-50 chance of winning for each player. If you know your opponent is playing non-optimally it is possible to take advantage.

Don't forget to specify a rule for what happens if you tie on the last round of the game.

The game is solvable with pencil and paper if you start with only 3 or 4 cards, and enough to give you a headache if you start with a full suit of 13. See http://home.netcom.com/~goldkerr/gops.htm for solutions up to size 5. Of course, you're playing with six

[Fryguy]

[ QUOTE ]
<u>What happens if it is tied on the last round?</u>


--
I think an optimal strategy is really close to the following:

play the same as the prize card, except if you can force a win by playing your highest, second highest, etc.

[/ QUOTE ]

Sorry I forgot to mention this. If it ties the last round, nobody gets any of those points (so if you are leading, you win).

I have no idea what happens if the game score is tied on the last round though.

My first instinct was to play the same value as the prize too, and then try to force wins in the later rounds, I never really hardcore stuck to the strategy though, as it doesn't feel like it works.

[Fryguy]

I just read about gops VERY briefly, and it doesn't seem to match this game. In gops, you only score from the middle. In the game I described, you get points for what each player plays as well.