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This one's for Glen

Glen-

Last night we were playing Parthenon. If you haven't played it, it's a resource management game based in the Aegean Sea. But enough about that - it's good, and you should try it if you get the chance. I did want to hear your opinion on our last game last night.

Parthenon is played over three "years", each with four "seasons". Every time a new season starts, an event card is played. There's six event cards for each year - three good and three bad. The cards are randomized and two are discarded; four are played each year, one per season.

Our last game, last night was statistically lousy: played, in order, were the three bad cards, and then one good card. The odds on that are pretty low, yes? It gets worse: the cards have rankings (+3, +2, +1, -1, -2, -3, effectively). The beginning card of each year (which lasts all year) was the worst of the six cards. The "good" card played each year was usually useless by the time it was played. I don't remember if all three years were all as bad to be played in the worst order: -3, -2, -1, 1. It sure seemed that way. What are the odds?

Comments

( 5 comments — Leave a comment )
gingy
Mar. 5th, 2006 03:00 pm (UTC)
That sounds kinda like Diplomacy.

Passing by your neck of the woods today. I'll call if I get a chance to, have to be back for dinner at the in-laws' at 6.
ralphmelton
Mar. 5th, 2006 07:45 pm (UTC)
To keep Glen from giving me the hairy statistical eyeball, I will state that we are of course talking about the odds that this situation will come up in your next game.

If I understand correctly, the six event cards are shuffled between years, so the probabilities are independent within years, and then linked within years.

Within a season: the probability that you get three bad cards, then one good card, goes like this:
The chances of getting a bad card on the first draw are 3 in 6.
The chances of getting a bad card on the second draw after getting a bad card on the first draw are 2 in 5.
The chances of getting a bad card on the third draw (after the first two draws give bad cards) are 1 in 4.
The chances of getting a good card on the fourth draw after drawing three bad cards are 3 in 3, because all the bad cards are used up.
Multiplying those together gives a 1 in 20 chance of getting a year of three bad events then one good one.
Assuming all three years are independent, that's a 1 in 8000 chance of having all three years work out that way.

The probabilities of getting the order -3, -2, -1, 1 for a year: the seasonal probabilities are 1 in 6, 1 in 5, 1 in 4, 1 in 3, which multiply out to 1 in 360. The chances of having three years like that in a game are 1 in (360)^3, or 1 in 46,656,000.
glenbarnett
Mar. 5th, 2006 09:19 pm (UTC)
To keep Glen from giving me the hairy statistical eyeball, I will state that we are of course talking about the odds that this situation will come up in your next game.


Heh, yes. In retrospect, the chances of some circumstance occuring that would cause you to say "Wow, what are the odds of that?" are pretty high.

And of course, we're discussing probability rather than odds, though I can convert them to odds if anyone cares.

Ralph's answers look right to me, so I don't think there's much left to do, though it's not clear if you wanted a more general answer in the second part.
glenbarnett
Mar. 6th, 2006 01:44 am (UTC)

To be more specific,

The beginning card of each year (which lasts all year) was the worst of the six cards. The "good" card played each year was usually useless by the time it was played. I don't remember if all three years were all as bad to be played in the worst order: -3, -2, -1, 1. It sure seemed that way. What are the odds?

...so if you wanted say the odds of (i) "worst card first, then the other two bad cards and the least-good card in some order" or (ii) "worst card first, then two bad cards and some good card in some order" then those will be between the two probabilities Ralph gave.

If you want those (or some other probability still), post the specific event(s) you're after and I'll see what I can do.

sethcohen
Mar. 6th, 2006 05:24 pm (UTC)
Nope, thanks, the answers Ralph gave were more than sufficient. I would have thrown the question out more generally than to just you, but you were the usual suspect.
( 5 comments — Leave a comment )