Dominion Strategy Forum
Dominion => Dominion General Discussion => Topic started by: Jeebus on April 09, 2015, 03:02:33 pm
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In a recent game I opened Spice Merchant + Silver, and then I drew SM + Silver + 3 Estates. I wondered about the odds of that.
I don't know a lot about calculating probabilities, but I tried to figure it out. Maybe someone here can correct me if I'm wrong.
Odds of getting it on turn 3: 0.13 %.
Odds of getting it on turn 4: 0.13 %.
Odds of getting it on turn 3 or 4: 0.25 %
Odds of getting it on turn 3, 4 or 5 (assuming two more buys on turn 3 and 4): 0.32 %
Calculations:
Turn 3: 1/(12 choose 5) = 1/792 = 0.13 %
Turn 4:
Odds of coppers only turn 3: (7 choose 5)/(12 choose 5) = 21/792
If so: Odds of drawing SM+S+3E on turn 4: 1/(7 choose 5) = 1/21
Total odds of this happening turn 4: 21/792 * 1/21 = 1/792 = 0.13 %
(Are the odds of drawing a specific hand, for instance 5 coppers, always the same for turn 3 and 4?)
Turn 5:
Odds of drawing two of the needed cards at the bottom: (5 choose 2)/(12 choose 2) = 10/66
If so: Odds of drawing the last three needed cards from the next shuffle: 1/(12 choose 3) = 1/220
Total odds of this happening turn 5: 10/66 * 1/220 = 1/1452 = 0.07 %
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Well the first problem that I see is that you are using the "choose" function which gives you the total number of combinations of cards possible. This gives you faulty percentages because the order matters in that calculation. Your hand, however, does not care what order the 5 cards are in.
In other words:
SM/S/E/E/E = SM/E/S/E/E = E/S/SM/E/E = ...
So yes, the total number of possible draws is 792, but a significant number of them are exactly the same.
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So yes, the total number of possible draws is 792, but a significant number of them are exactly the same.
No, 1/792 is correct. The choice function accounts for the number of same choices.
(12 choose 5) = 5/12 * 4/11 * 3/10 * 2/9 * 1/8 = 1/792 or 12! / (5! (12-5)!)
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Well the first problem that I see is that you are using the "choose" function which gives you the total number of combinations of cards possible. This gives you faulty percentages because the order matters in that calculation.
This is not correct. The choose function gives you the number of combinations in which order does not matter. Permutations are for situations where the order does matter.
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Turn 4:
Odds of coppers only turn 3: (7 choose 5)/(12 choose 5) = 21/792
If so: Odds of drawing SM+S+3E on turn 4: 1/(7 choose 5) = 1/21
Total odds of this happening turn 4: 21/792 * 1/21 = 1/792 = 0.13 %
(Are the odds of drawing a specific hand, for instance 5 coppers, always the same for turn 3 and 4?)
Yes, they are.*
You are looking for two hands XXXXX / ?????, where X is a specified card and ? is a random card. The order of the hands does not matter.
*exception: card draw that causes a reshuffle (e.g. Smithy)
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Ninja'd... but
(Are the odds of drawing a specific hand, for instance 5 coppers, always the same for turn 3 and 4?)
Yes, the odds of a specific 5 cards ending up in positions 1-5 of a 12 card shuffle are identical to the odds of them ending up in positions 6-10 of that same shuffle. (If you don't effect the shuffle with cards 1-5)
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Turn 5:
Odds of drawing two of the needed cards at the bottom: (5 choose 2)/(12 choose 2) = 10/66
If so: Odds of drawing the last three needed cards from the next shuffle: 1/(12 choose 3) = 1/220
Total odds of this happening turn 5: 10/66 * 1/220 = 1/1452 = 0.07 %
The odds of the two needed cards at the bottom is just (12 choose 2) = 1/66, the rest is correct.
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Turn 5:
Odds of drawing two of the needed cards at the bottom: (5 choose 2)/(12 choose 2) = 10/66
If so: Odds of drawing the last three needed cards from the next shuffle: 1/(12 choose 3) = 1/220
Total odds of this happening turn 5: 10/66 * 1/220 = 1/1452 = 0.07 %
The odds of the two needed cards at the bottom is just (12 choose 2) = 1/66, the rest is correct.
But you're not taking into account that there are 5 "needed cards", just that you draw 2 cards...
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Sorry, perhaps we should first make clear what probability we are talking about.
I gave the probability for SM+Silver missing the reshuffle, so that you draw them for the first time in turn 5 (12 choose 2). Then you shuffle the deck and draw 3 estates (12 choose 3). You cannot draw estates as your first 2 cards, if SM+Silver missed the reshuffle.
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Sorry, perhaps we should first make clear what probability we are talking about.
I gave the probability for SM+Silver missing the reshuffle, so that you draw them for the first time in turn 5 (12 choose 2). Then you shuffle the deck and draw 3 estates (12 choose 3). You cannot draw estates as your first 2 cards, if SM+Silver missed the reshuffle.
To be clear, assuming you buy two cards in the first shuffle, this gives 1/(12 choose 2) * 1/(12 choose 3) = 1/66 * 1/220 = 1/14520
If you bought one card (because EEESM is usually a dud), then 1/(12 choose 2) * 1/(11 choose 3) = 1/66 * 1/165 = 1/10890
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Sorry, perhaps we should first make clear what probability we are talking about.
I gave the probability for SM+Silver missing the reshuffle, so that you draw them for the first time in turn 5 (12 choose 2). Then you shuffle the deck and draw 3 estates (12 choose 3). You cannot draw estates as your first 2 cards, if SM+Silver missed the reshuffle.
Okay, I see now. Yeah, if I draw SM on turn 3 or 4, I would play it, so that would change everything. So my calculation for turn 5 was wrong. I could draw Silver though, and then draw Estate + SM at the bottom, shuffle and draw Silver again + 2 Estates. That would make SM fail again. Anyway, now we're talking about several different scenarios.
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I did manage to once get that disastrous series of draws on turns 3, 4 and 5, in a Gokodom match, against Marin.