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[GendoIkari]

http://boardgamegeek.com/thread/850260/randomizing-shelters-and-coloniesplatinum

What's the correct answer here? I believe that the discussion is actually missing a key option... I see 3 distinct possible ways to choose, each that would result in different calculations (even though the calculations may yield the same result).

1. Shuffle the 10 cards. If the top card is Dark Ages, use Shelters. If the bottom card is Prosperity, use Plat/Colony.

2. Shuffle the 10 cards. If the top card is Dark Ages, use Shelters. Shuffle the 10 cards again. If the top card is Prosperity, use Plat/Col.

3. Shuffle the 10 cards. If the top card is Dark Ages, use Shelters. Shuffle the remaining 9 cards. If the top card is Prosperity, use Plat/Col.

I am now leaning towards thinking that all 3 options yield the same, expected result, even though it's counter-intuitive at first.

(The expected result being that in a Kingdom with 5 cards from each, if you play that Kingdom 100 times, you will have 25 games with Colony but not Shelter, 25 games with Shelter but not Colony, 25 games with both, and 25 games with neither).

[LastFootnote]

Without thinking about it too long, I believe #1 and #3 will give identical results.

#2 is technically correct if you want the two decisions to be completely independent, but the other two methods are close enough that you might as well use #1 when using the real-life randomizer deck.

[shMerker]

Option 3 doesn't require a second shuffle, and therefore is the same as option 1.

And yeah I'm pretty sure all 3 give the same result.

The people in that other thread saying "draw two, if either is (dark ages/prosperity) use (shelters/colonies)" should double the chances of using either(or both).

[LastFootnote]

And yeah I'm pretty sure all 3 give the same result.

No, they don't. Let's say that you have a game with only 2 Kingdom cards. One is a Prosperity card and the other is a Dark Ages card. Using method #1, you either have both Colonies and Shelters or neither. Using method #2, you could have either, both, or neither.

[shMerker]

But you can't have a kingdom with only two cards?

Ok I see what you're getting at. If you have only dark ages and prosperity cards in the kingdom then it's not possible to have a kingdom that has colonies but not shelters or vice versa.

No wait that's not true either. It's that if all of your cards are prosperity or dark ages AND it's a 9/1 split then you won't be able to have a game with shelters, but not colonies or vice versa.

[rod-]

Go back to the original 9/10 dark ages 1/10 prosperity discussion:

Method 2 is the only one that allows all possibilities ; if you look at the top card and it's prosperity, the bottom card is not prosperity.  There is no way to get colonies w/o shelters using method 1(3 is, as far as i can tell, functionally identical). 

[LastFootnote]

But you can't have a kingdom with only two cards?

Do I have to write an entire induction proof for you? I was hoping you could extrapolate this for yourself.

If you have a game with 5 cards each from Prosperity and Dark Ages and you use method #1, you have a 50% chance of using Colonies and a 50% chance of using Shelters. However, the probabilities of having neither/justShelters/justColonies/both will not work out to exactly 25%/25%/25%/25%. You'll be more likely to have both or neither than to have just one or the other.

[HiveMindEmulator]

@shMerker: That's just a simplifying example. Say you have 10 cards, one prosperity and 9 DA. Using method 1, you will never use Colony and not Shelters, since if the top card is Prosperity, the bottom must be DA. However, by method 2, there is a 1/100 chance of playing with Colony and not Shelters (draw the prosperity card both times).

EDIT: wow triple ninja!

[jonts26]

LastFootnote is correct.

1 and 3 give the same result, but 2 is different because you are sampling with replacement vs without replacement. The larger the selection you choose however, the closer those become to the same. A 10 card kingdom might be large enough that it probably doesn't make a big difference.

[Jimmmmm]

The only difference between {1,3} and 2 is that 2 will result in slightly (10%?) fewer games with both Plat/Cols and Shelters since you will sometimes draw the same card twice.

[Schneau]

For those playing along at home, I think the percentages are computed exhaustively as follows. This is for a kingdom with 5 DA and 5 Prosperity (Pr) cards, where first card determines Shelters, second card drawn determines Plat/Col):

Option 1 and 3 (which are the same):

Code: [Select]

P(first card DA) = 5/10
P(first card PR) = 5/10

P(second card DA | first card DA) = 4/9
P(second card Pr | first card DA) = 5/9
P(second card DA | first card Pr) = 5/9
P(second card Pr | first card Pr) = 4/9

P(first card DA, second card Pr) = P(second card Pr | first card DA) * P(first card DA) = 5/9 * 5/10 = 25/90 = 27.8%
P(first card Pr, second card Pr) = P(second card Pr | first card Pr) * P(first card Pr) = 4/9 * 5/10 = 20/90 = 22.2%
P(first card DA, second card DA) = P(second card DA | first card DA) * P(first card DA) = 4/9 * 5/10 = 20/90 = 22.2%
P(first card Pr, second card DA) = P(second card DA | first card Pr) * P(first card Pr) = 5/9 * 5/10 = 25/90 = 27.8%Here, it is more likely to play with both Shelters and Plat/Col or neither than to play with one but not the other.


Option 2:

Code: [Select]

P(first card DA) = 5/10
P(first card PR) = 5/10

P(second card DA | first card DA) = P(first card DA) = 5/10
P(second card Pr | first card DA) = P(first card DA) = 5/10
P(second card DA | first card Pr) = P(first card Pr) = 5/10
P(second card Pr | first card Pr) = P(first card Pr) = 5/10

P(first card DA, second card Pr) = P(second card Pr | first card DA) * P(first card DA) = 5/10 * 5/10 = 25/100 = 25.0%
P(first card Pr, second card Pr) = P(second card Pr | first card Pr) * P(first card Pr) = 5/10 * 5/10 = 25/100 = 25.0%
P(first card DA, second card DA) = P(second card DA | first card DA) * P(first card DA) = 5/10 * 5/10 = 25/100 = 25.0%
P(first card Pr, second card DA) = P(second card DA | first card Pr) * P(first card Pr) = 5/10 * 5/10 = 25/100 = 25.0%Here, each of the four possible events are equally likely.


So, I think this backs up Jimmmmmm's claim that games with both Plat/Cols and Shelters will be less likely with options 1 and 3 than with 2. But also, games with neither will be more common.

[Mecherath]

I was just planning to look at the first 2 cards when I (eventually) play in real life.  If one of them is Prosperity, it's Colony/Platinum time.  If one is Dark Ages, it's Shelter time.  This is of course the same as the 1st and 3rd options.

The slightly higher chance of playing with Shelters, Platinum, and Colonies is a feature, not a bug if you ask me.  Especially since there are 192 cards if you play with all sets and Promos.

That means 1.8 Dark Ages, 1.3 Prosperity, and 6.9 Other cards.  I'm just fine with the chance of both being slightly boosted.

If Guilds has the expected 13 cards, that takes us to 205 where the chances expected values drop to 1.7 Dark Ages and 1.2 Prosperity, and 7.1 Other cards.

Easy enough to plug all those numbers into Schneau's formulas above, but meh... feeling lazy

[cayvie]

I was just planning to look at the first 2 cards when I (eventually) play in real life.  If one of them is Prosperity, it's Colony/Platinum time.  If one is Dark Ages, it's Shelter time.  This is of course the same as the 1st and 3rd options.

No, it's going to produce more c/p and shelter games than any of the 3 options.

Think about it, in 1, the top card has to be a prosperity card.

In your variation, either of the top 2 can be a prosperity card.

[ConMan]

In a game where there are d cards from Dark Ages and p cards from Prosperity, the probabilities are:

Method A (Draw 2 cards, 1st to check for Shelters and 2nd to check for Colonies):
P(Shelters + Colonies) = d*p/90
P(Shelters, no Colonies) = d*(9-p)/90
P(Colonies, no Shelters) = (9-d)*p/90
P(No Shelters or Colonies) = ((9-d)*(9-p)+9)/90

Method B (Draw 1 card for Shelters, reshuffle then draw 1 card for Colonies):
P(Shelters + Colonies) = d*p/100
P(Shelters, no Colonies) = d*(10-p)/100
P(Colonies, no Shelters) = (10-d)*p/100
P(No Shelters or Colonies) = (10-d)*(10-p)/100

Method C (Draw 2 cards, check both for Shelters and Colonies):
P(Shelters + Colonies) = 2d*p/90 = d*p/45
P(Shelters, no Colonies) = (19d-d²-2dp)/90
P(Colonies, no Shelters) = (19p-p²-2dp)/90
P(No Shelters or Colonies) = (10-d-p)*(9-d-p)/90

So if we have 5 each of Dark Ages and Prosperity:

Method:	A	B	C
S+C	27.8%	25.0%	55.6%
S, no C	22.2%	25.0%	22.2%
C, no S	22.2%	25.0%	22.2%
No S or C	27.8%	25.0%	0.0%

And if we have 1 of each, and 8 of either:

Method:	A	B	C
S+C	1.1%	1.0%	2.2%
S, no C	8.9%	9.0%	17.8%
C, no S	8.9%	9.0%	17.8%
No S or C	81.1%	81%	62.2%

So Method C will give a much higher chance of at least one, and of both being in the game compared to the other two, whereas method B tends to smooth out the chances of those while giving a reasonable chance of neither being in. I'd say Method A is a decent compromise between fairness/accuracy to the rules and being easy to implement (only have to shuffle the randomisers once).

[Jimmmmm]

So, I think this backs up Jimmmmmm's claim that games with both Plat/Cols and Shelters will be less likely with options 1 and 3 than with 2. But also, games with neither will be more common.

Other way around, games with both are less likely with option 2, and yes you're right games with neither will also be more likely with option 2, since there's the possibility of pulling out the same non-Prosperity, non-DA card twice. So basically, option 2 is slightly worse if you want to see more games with Plat/Col and/or Shelters.

In other words, if you use option 2, 10% of the time you'll pull the same card out twice, which is equivalent to only pulling one card out, resulting in less Plat/Col/Shelters.

[Schneau]

In a game where there are d cards from Dark Ages and p cards from Prosperity, the probabilities are:

There's something wrong here. I don't have time to figure out what it is, but the columns in your last table and the first column of your first table don't add up to 100%.

[Davio]

Just use a d10 and throw it twice.

If the number thrown is equal to or lower than the number of cards from DA/Pros in the set, use the appropriate extra cards.

For instance:
3 DA Cards: 1-3 = Shelters, 4-10 = No Shelters, 3 in 10 chance of Shelters.
4 Pros Cards: 1-4 = Col/Plat, 5-10 = No Col/Plat, 4 in 10 chance of Col/Plat.

This follows the rules as close as possible, the same as option 2.

[Mecherath]

I was just planning to look at the first 2 cards when I (eventually) play in real life.  If one of them is Prosperity, it's Colony/Platinum time.  If one is Dark Ages, it's Shelter time.  This is of course the same as the 1st and 3rd options.

No, it's going to produce more c/p and shelter games than any of the 3 options.

Think about it, in 1, the top card has to be a prosperity card.

In your variation, either of the top 2 can be a prosperity card.

Oh, good call.  The version I was thinking of would largely increase the chance of Dark Ages and of Prosperity.  Which itself isn't a terrible idea as the relative probabilities of each decrease with each new set you add.  But not what I was going for when I typed that post yesterday.

Mine ends up with these numbers:
157/192 * 156/191 chance of no Dark Ages (66.79%) or 33.21% of Shelters
167/192 * 166/191 chance of no Prosperity (75.59%) or 24.41% of Colony & Platinum
132/192 * 131/191 chance of Neither (47.15%) or 52.85% of one or the other

I believe that means the chances of both are the overlap in those 3 possibilties, or about 4.77%.  I might have accidentally come across what I want.

Edit: My 2nd numerators were off the first time through.  And the 2nd time through I mis-interpreted that 52.85% number.  Been a few years outside of Stats classes.

[ConMan]

In a game where there are d cards from Dark Ages and p cards from Prosperity, the probabilities are:

There's something wrong here. I don't have time to figure out what it is, but the columns in your last table and the first column of your first table don't add up to 100%.

I found a mistake in the Method A probability of getting neither so I've adjusted those. And I think I made a colossal mistake in actually calculating the probability of getting both for Method B in Table 2, which I will correct after I make this post (I don't even know how I managed it).

[Watno]

Just use a d10 and throw it twice.

If the number thrown is equal to or lower than the number of cards from DA/Pros in the set, use the appropriate extra cards.

For instance:
3 DA Cards: 1-3 = Shelters, 4-10 = No Shelters, 3 in 10 chance of Shelters.
4 Pros Cards: 1-4 = Col/Plat, 5-10 = No Col/Plat, 4 in 10 chance of Col/Plat.

This follows the rules as close as possible, the same as option 2.

It doesn't. option 1/3 would be what the rules say.

[michaeljb]

It doesn't. option 1/3 would be what the rules say.

No, option 2 is what the rules say.

They just say Platinum and Colony should be used based on the proportion of Prosperity cards in use, and Shelters should be used based on the proportion of Dark Ages cards in use. Options 1 and 3 don't quite follow this. The only rule it mentions for games with both Prosperity and Dark Ages cards is to not use the same card to determine both at once.

Just cuz no one else has posted them here yet, here are the relevant paragraphs from the rulebooks.

If only Kingdom cards from Prosperity are being used this game, then the Platinum and Colony piles are added to the Basic cards in the Supply for the game. If a mix of Kingdom cards from Prosperity and other sets are being used, then the inclusion of Platinum and Colony in the Supply should be determined randomly, based on the proportion of Prosperity and non-Prosperity cards in use. For example, choose a random Kingdom card being used - such as the first card dealt out from the Randomizer deck - and if it is from Prosperity, add Platinum and Colony to the Supply. Platinum and Colony are not Kingdom cards; when those are included, there are 10 Kingdom cards, plus Copper, Silver, Gold, Platinum, Estate, Duchy, Province, Colony, and Curse, in the Supply. Use 8 Colonies for a 2-player game, or 12 Colonies for a game with 3 or more players.

If only Kingdom cards from Dark Ages are being used this game, the Shelter cards replace starting Estates - each player's starting deck is seven Coppers, a Hovel, a Necropolis, and an Overgrown Estate. If a mix of Kingdom cards from Dark Ages and other sets is being used, then the use of Shelters should be determined randomly, based on the proportion of Dark Ages cards in use. For example, choose a random Kingdom card being used - such as the last card dealt from the Randomizer deck - and if it is from Dark Ages, use Shelters in place of starting Estates. Do not use the same card to choose whether or not to use Shelters as you use to choose whether or not to use Platinum and Colony (from Prosperity). Using Shelters does not change the Estate Supply pile; it still contains 8 Estates for 2 players and 12 Estates for 3 or more players.

[Mecherath]

The only difference between Options 2 and 3 is that option 2 puts the Prosperity-decider card back in the pile and Option 3 does not.  Since the main thing you want to avoid is using the same card for Prosperity-deciding and Dark-Ages-deciding, Option 3 seems closer to the intent than Option 2.  Obviously if you take that first card out when you reshuffle the remaining 9, it will never come up again.

In the paragraphs you quote it mentions top card for Prosperity and bottom card for Dark Ages.  These are of course only "for example" suggestions, but as the choice the means the least shuffling, it works for me.

Heck, even mentioning to not use the same card for both seems like a "well duh" to me.  Of course the same card can't be from two sets.  Obviously there are people who want to be by the book and use the Donald X sanctioned method, but just do whatever you want.

[RiemannZetaJones]

Since the main thing you want to avoid is using the same card for Prosperity-deciding and Dark-Ages-deciding, Option 3 seems closer to the intent than Option 2.

No, the thing you want to do is have the chance of using Colonies be proportional to the number of Prosperity supply cards and the chance of using Shelters be proportional to the number of Dark Ages supply cards. Option 2 is correct (and different from Options 1 & 3 which are, as reported, the same)

[HiveMindEmulator]

All of these options obey the rules, and give you the appropriate probability of using Platinum/Colony and shelters. The difference between options 1/3 and option 2/dice is about dependence. In option 2/dice, the events {use Platinum/Colony} and {use shelters} are independent given the set of kingdom cards. Option 1/3 have a slight dependence.

EDIT: I switched the meanings of 2 and 1/3...

[michaeljb]

Options 1 and 3 don't quite follow this.

Woops, just worked through the math for myself and realized what has already been pointed out--that it's only the odds of using both Shelters and Platinum/Colony or using neither is affected by Options 1 and 3. So my post added nothing of value...