Dominion Strategy Forum
Dominion => Dominion General Discussion => Topic started by: Flip5ide on August 17, 2015, 04:10:49 pm
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If I have played 5000 games with all the cards (not counting Adventures), what are the odds that I have played the same kingdom twice?
I think for starters, I should try to recall how many kingdom cards there are altogether (not counting Adventures).
As an added bonus, at what point (after how many games) does it become a 50% chance that I have played 2 games with an identical setup?
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Pretty high, if you use the preset kingdoms.
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Pretty high, if you use the preset kingdoms.
So in other words, basically 100% after the first 2 games of playing the "First Game"? Dang I had no clue haha
I suppose I should add that you are picking a set of 10 from a set of 180 or however many kingdom cards there are. I guess you could throw in YW's bane card but I'm guessing it's negligible.
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I'd be willing to wager a lot that every single full random game played with all the pre-dark ages cards had a different kingdom. That is, NO kingdom has been generated more than once randomly.
Reason: 155 choose 10 is a lot. It has 15 digits, while the number of games that have been played has ~7 digits. (12 million were played on isotropic. How many on Goko/MF?). This puts the odds at around 10% that it's happened at all.
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As an added bonus, at what point (after how many games) does it become a 50% chance that I have played 2 games with an identical setup?
You would have to play on the order of 100 million games.
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If you're playing "full random" casual/unrated on Goko and you're not overly careful, you should expect multiple duplicate kingdoms per day.
(http://techzulu.com/wp-content/uploads/2012/08/Goko_Owl_Blue-252x300.png)
#DefyingTheOdds
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I feel like maybe this idea comes into play at some point?
https://en.wikipedia.org/wiki/Birthday_problem
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I feel like maybe this idea comes into play at some point?
https://en.wikipedia.org/wiki/Birthday_problem
This is exactly what Qmech and I are talking about, and where our estimates come from. Specifically, I used the estimate 1 - e-n2/2m, where n is the number of games you play and m is the number of possible kingdoms.
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I feel like maybe this idea comes into play at some point?
https://en.wikipedia.org/wiki/Birthday_problem
This is exactly what Qmech and I are talking about, and where our estimates come from. Specifically, I used the estimate 1 - e-n2/2m, where n is the number of games you play and m is the number of possible kingdoms.
-1 for ambiguous fraction notation.
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What, do you really think that the m might not be in the denominator?
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Never tell me the odds.
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Related thread:
http://forum.dominionstrategy.com/index.php?topic=13647
(I really do want to answer that question, just haven't had the time to sit down and do the math lately.)