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I recently played two different card games with a friend: 500 Rum and Lost Cities (modified to use a regular deck of cards). We started arguing about which game was more subject to chance. I realized that I have no idea how to measure, compare, or even articulate the chance/strategy ratio of a game, and was wondering if anyone could point me in the right direction.

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Well, variance of distribution of results (with some normalization of course) when both players employ optimal strategies would be some measure, but it might be very hard to calculate. – dtldarek Jan 15 '13 at 23:36
@dtldarek, would you be able to apply variance of distribution of results to a game of no chance and complete information, eg chess? – alancalvitti Jan 16 '13 at 0:12
@alancalvitti Sure, it would be zero ;-) With optimal strategies, the outcome is deterministic, i.e. constant. – dtldarek Jan 16 '13 at 0:21
@Eric, you may be interested in this recent WSJ interview with Michael Mauboussin of Legg Mason "Is your Manger Skillful ...or Just Lucky":… ... and also in the Fundamental Theorem of Poker: – alancalvitti Jan 16 '13 at 0:22
@dtldarek, there are 3 possible outcomes: win, loss, draw, not 1. Obviously you know that, so I'm not understanding what you mean by "variance of distribution of results" – alancalvitti Jan 16 '13 at 0:24

Determine using absolutely no stratagy, (basically just playing the game without any particular aim) what is the probability of winning against a player using a stratagy.

Then determine the probability of winning against a player if using a stratagy.

That difference is the amount of "better" chance you have to win by using a particular stratagy. 1 - that number is the how much luck plays in the game.

You can test this out with different stratagies for your opponent and yourself to see which stratagies lend themselves to the most luck or least.

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you can consider this with a game of war with 2 cars. Lets say the stratagy in particular is just play the highest card out of the 2 you draw. Then: P1(winning) = P(drawing at least 1 card higher than opponents both cards) * 1/2 + P(drawing 2 cards higher than opponents both cards). Now compare that to P2(winning) = P(drawing 1 card higher than opponents both cards). P2(winning) - P1(winning) measures how much your probability of winning goes up by employing that particular stratagy. – frogeyedpeas Jan 15 '13 at 23:41
If you insert the 2 beats A rule then this becomes a lot more complicated all of a sudden. – frogeyedpeas Jan 15 '13 at 23:45
While this is a brilliant idea, a random play (i.e. with no strategy) is often too stupid to do anything (e.g. it often plays moves against itself). – dtldarek Jan 15 '13 at 23:47
So then what would be cool is to compare the probability of 2 different stratagies to measure how much does switching from one stratagy to the other improve the odds of winning. – frogeyedpeas Jan 15 '13 at 23:52
If you would like to have any way to compare two different games then you would need to find comparable strategies, which is impossible but for the best one (optimal) and the random one (the worst one would often have no sense). I don't know how to tell how much better one strategy is in one game, that other strategy in other game (measures such like "it wins 80% times against random strategy" depend on what you want to measure in the first place). – dtldarek Jan 16 '13 at 0:01

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