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Author Topic: Events will have normal (non-randomizer) card backs due to missprint  (Read 30073 times)

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Rubby

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #50 on: April 10, 2015, 03:52:39 pm »
+2

IMO it would be far from perfect even if they had the randomizer backs. People who keep a randomizer deck would want an additional, non-randomizer copy of each Event so that if they occasionally use an app or just decide they want to play a game with Inheritance or something, they wouldn't have to fish stuff out of there.
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blueblimp

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #51 on: April 10, 2015, 05:08:55 pm »
0

Which app do most people use?  I have one I like but it doesn't have Prince.
The one I've used is D-Vault (for iOS). It doesn't seem to have Prince either though.
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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #52 on: April 10, 2015, 08:37:35 pm »
0

There's a relatively new Android app, Jack of All Dominion, that I think looks very nice. It's the only app I've found that includes all of the card text (not the art), which can come in handy when you're playing with people who haven't memorized all the cards and the Kingdom is facing away from them. It has a nifty little VP counter if you can't live without one of those. It has all the published recommended Kingdoms. It has all the Promos and even already includes 13 of the previewed Adventures cards.

Glad I asked.  Jack of All Dominion is cool.
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Deadlock39

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #53 on: April 11, 2015, 12:44:38 am »
0

This was discussed forever ago in this thread, but I have been sleeving my cards, and I come up with 12 extra cards with blue backs if you own everything.  I have 3 blanks with blue backs, 7 base cards from the original sets, and 2 Trash cards (Base cards Trash is grey). 

There are obviously much better solutions, so it doesn't matter.  If you absolutely need 20 proxy blue backed cards, and have those 12, you only need 8 sets of blanks from the BGG store to complete your set.

Throwaway_bicycling

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #54 on: April 11, 2015, 12:27:26 pm »
0


If you want to limit the number of events to 2, you can approximate these probabilities quite well by rolling d6+d8:

2-7: 21/48 = 43.75% (0 events)
8-10: 17/48 = 35.42% (1 event)
11-14: 10/48 = 20.83% (2 events)

So I don't usually have dice in my Dominion stuff, but I do have starter hands of 10 cards. I can approximate these probabilities by taking a hand of 10 cards, putting 3 copper aside, shuffling, then:

1) If top card is an Estate, (probability 3/7 = 9/21) no events
2) If top card is a Copper, keep that card on the table, put the other three coppers back, shuffle, then
3) If top card is now an Estate (probabilty 4/21), two events
4) Otherwise, one event

P(0 events) = 9/21 = .43
P(1 events) = 8/21 = .38
P(2 events) = 4/21 = .19

...but I would honestly probably just flip a coin, and put one event in on heads. :-)
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pacovf

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #55 on: April 11, 2015, 02:11:37 pm »
+1

Oh, I like convoluted probability approximations too!

Toss a coin 4 times. If you get two or four tails, don't use events.
Otherwise, toss the coin 4 more times. If you get a total of up to three tails or exactly five tails, use one event.
Otherwise, use 2 events.

 ;D
« Last Edit: April 11, 2015, 02:28:33 pm by pacovf »
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mith

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #56 on: April 11, 2015, 04:16:18 pm »
+1


If you want to limit the number of events to 2, you can approximate these probabilities quite well by rolling d6+d8:

2-7: 21/48 = 43.75% (0 events)
8-10: 17/48 = 35.42% (1 event)
11-14: 10/48 = 20.83% (2 events)

So I don't usually have dice in my Dominion stuff, but I do have starter hands of 10 cards. I can approximate these probabilities by taking a hand of 10 cards, putting 3 copper aside, shuffling, then:

1) If top card is an Estate, (probability 3/7 = 9/21) no events
2) If top card is a Copper, keep that card on the table, put the other three coppers back, shuffle, then
3) If top card is now an Estate (probabilty 4/21), two events
4) Otherwise, one event

P(0 events) = 9/21 = .43
P(1 events) = 8/21 = .38
P(2 events) = 4/21 = .19

...but I would honestly probably just flip a coin, and put one event in on heads. :-)

I like it... Only adding two coppers would actually be closer (5/14 is about 36%, 3/14 is about 21%).
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Throwaway_bicycling

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #57 on: April 11, 2015, 06:49:05 pm »
0


If you want to limit the number of events to 2, you can approximate these probabilities quite well by rolling d6+d8:

2-7: 21/48 = 43.75% (0 events)
8-10: 17/48 = 35.42% (1 event)
11-14: 10/48 = 20.83% (2 events)

So I don't usually have dice in my Dominion stuff, but I do have starter hands of 10 cards. I can approximate these probabilities by taking a hand of 10 cards, putting 3 copper aside, shuffling, then:

1) If top card is an Estate, (probability 3/7 = 9/21) no events
2) If top card is a Copper, keep that card on the table, put the other three coppers back, shuffle, then
3) If top card is now an Estate (probabilty 4/21), two events
4) Otherwise, one event

P(0 events) = 9/21 = .43
P(1 events) = 8/21 = .38
P(2 events) = 4/21 = .19

...but I would honestly probably just flip a coin, and put one event in on heads. :-)

I like it... Only adding two coppers would actually be closer (5/14 is about 36%, 3/14 is about 21%).

Excellent point. I can authoritatively state that would definitely be close enough for government work.
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Kirian

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #58 on: April 12, 2015, 01:09:00 pm »
+10


If you want to limit the number of events to 2, you can approximate these probabilities quite well by rolling d6+d8:

2-7: 21/48 = 43.75% (0 events)
8-10: 17/48 = 35.42% (1 event)
11-14: 10/48 = 20.83% (2 events)

So I don't usually have dice in my Dominion stuff, but I do have starter hands of 10 cards. I can approximate these probabilities by taking a hand of 10 cards, putting 3 copper aside, shuffling, then:

1) If top card is an Estate, (probability 3/7 = 9/21) no events
2) If top card is a Copper, keep that card on the table, put the other three coppers back, shuffle, then
3) If top card is now an Estate (probabilty 4/21), two events
4) Otherwise, one event

P(0 events) = 9/21 = .43
P(1 events) = 8/21 = .38
P(2 events) = 4/21 = .19

...but I would honestly probably just flip a coin, and put one event in on heads. :-)

I like it... Only adding two coppers would actually be closer (5/14 is about 36%, 3/14 is about 21%).

For even more precision, use 236 Coppers and 20 Estates, then reveal cards from the top until you have 10 Coppers or 2 Estates, whichever comes first.  Put in events equal to the number of Estates revealed.
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enfynet

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #59 on: April 12, 2015, 01:27:40 pm »
+1


If you want to limit the number of events to 2, you can approximate these probabilities quite well by rolling d6+d8:

2-7: 21/48 = 43.75% (0 events)
8-10: 17/48 = 35.42% (1 event)
11-14: 10/48 = 20.83% (2 events)

So I don't usually have dice in my Dominion stuff, but I do have starter hands of 10 cards. I can approximate these probabilities by taking a hand of 10 cards, putting 3 copper aside, shuffling, then:

1) If top card is an Estate, (probability 3/7 = 9/21) no events
2) If top card is a Copper, keep that card on the table, put the other three coppers back, shuffle, then
3) If top card is now an Estate (probabilty 4/21), two events
4) Otherwise, one event

P(0 events) = 9/21 = .43
P(1 events) = 8/21 = .38
P(2 events) = 4/21 = .19

...but I would honestly probably just flip a coin, and put one event in on heads. :-)

I like it... Only adding two coppers would actually be closer (5/14 is about 36%, 3/14 is about 21%).

For even more precision, use 236 Coppers and 20 Estates, then reveal cards from the top until you have 10 Coppers or 2 Estates, whichever comes first.  Put in events equal to the number of Estates revealed.
You could also use 118 Coppers, 5 Estates, and 5 Curses. Draw until you have 5 Coppers or 1 Estate or 1 Curse. If you draw a curse, use 2 Events. If you draw an Estate use 1. Otherwise use none.
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shmeur

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #60 on: April 12, 2015, 01:41:42 pm »
0

Or, you know, you could simply just go to https://www.random.org/ and type in 0-2.  Right?
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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #61 on: April 12, 2015, 04:27:32 pm »
+3

Oh, I like convoluted probability approximations too!

Toss a coin 4 times. If you get two or four tails, don't use events.
Otherwise, toss the coin 4 more times. If you get a total of up to three tails or exactly five tails, use one event.
Otherwise, use 2 events.

 ;D

I prefer convoluted non-approximations. Say we want a random event with a success chance of p. Then, flip a coin a bunch of times and write down a 1 for each head and a 0 for each tail. Add a decimal point in front of you list to make a binary number x. If x < p, success. If x > p, failure. If you aren't sure how x compares to p, flip the coin a few more times.

There, now you can use a coin to create events with any probability of occurring!
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mith

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #62 on: April 13, 2015, 11:29:25 am »
0


If you want to limit the number of events to 2, you can approximate these probabilities quite well by rolling d6+d8:

2-7: 21/48 = 43.75% (0 events)
8-10: 17/48 = 35.42% (1 event)
11-14: 10/48 = 20.83% (2 events)

So I don't usually have dice in my Dominion stuff, but I do have starter hands of 10 cards. I can approximate these probabilities by taking a hand of 10 cards, putting 3 copper aside, shuffling, then:

1) If top card is an Estate, (probability 3/7 = 9/21) no events
2) If top card is a Copper, keep that card on the table, put the other three coppers back, shuffle, then
3) If top card is now an Estate (probabilty 4/21), two events
4) Otherwise, one event

P(0 events) = 9/21 = .43
P(1 events) = 8/21 = .38
P(2 events) = 4/21 = .19

...but I would honestly probably just flip a coin, and put one event in on heads. :-)

I like it... Only adding two coppers would actually be closer (5/14 is about 36%, 3/14 is about 21%).

For even more precision, use 236 Coppers and 20 Estates, then reveal cards from the top until you have 10 Coppers or 2 Estates, whichever comes first.  Put in events equal to the number of Estates revealed.

I'm sure someone with more meme-ing skills can come up with something for buying 4 extra copies of the base cards to make this happen...

(And also a Bad Luck Brian, trapped in an infinite coin flipping hell because he keeps hitting the probability boundary.)
« Last Edit: April 13, 2015, 11:33:41 am by mith »
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Witherweaver

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #63 on: April 13, 2015, 11:33:48 am »
+7

Consult your horoscope for that day.  If it is:

*Positive or uplifting, use 0 events
*Cautionary, use 1 event
*Hedging, use 2 events
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Deadlock39

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #64 on: April 13, 2015, 11:35:34 am »
0

With 60 Coppers, 40 Silvers and 30 Golds, you could do it with just two of three Base Set/Intrigue/Base Cards.

If you have all 3 you can use two for your event selection and not have to separate all your treasures back out before you play.  Perfect!

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #65 on: April 13, 2015, 11:38:33 am »
+1

I came up with another scheme the other night which only requires a starting hand and any other card:

Take 3 Estates, 4 Coppers, and 1 Harem (or whatever, but treat this card as if it were a treasure-victory card). Draw two at random.

If you draw 1 Copper and 1 Estate, use 0 events (24/56 = 3/7).
If you draw 2 Treasure, use 1 event (20/56 = 5/14).
If you draw 2 Victory, use 2 events (12/56 = 3/14).

Or change it to 1 Scheme (or any non-victory action card): If you don't draw Victory use 1 event, if you don't draw Treasure use 2 events.
« Last Edit: April 13, 2015, 11:41:19 am by mith »
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LastFootnote

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #66 on: April 13, 2015, 12:03:53 pm »
0

In my real-life games for the foreseeable future, I will be using 2 Events for every game with expert players and 0 Events for introducing novice players.
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mith

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #67 on: April 13, 2015, 12:33:38 pm »
+3

Yeah, I expect most of us will be totally ignoring the random method for a bit.

But if you want the exact probability distribution of the recommended method without the hassle of having to draw more than one card, you'll need 239607404704698 Overgrown Estates, 194802768052600 Estates, and 114398846157902 Islands... (Warning: a randomizer deck this size may reach the Sun.)
« Last Edit: April 13, 2015, 12:37:37 pm by mith »
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Witherweaver

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #68 on: April 13, 2015, 12:38:06 pm »
+2

Yeah, I expect most of us will be totally ignoring the random method for a bit.

But if you want the exact probability distribution of the recommended method without the hassle of having to draw more than one card, you'll need 239607404704698 Overgrown Estates, 194802768052600 Estates, and 114398846157902 Islands... (Warning: a randomizer deck this size may reach the Sun.)

However, these are too many calls to shuffle and draw a card from.  So you need to replicate this distribution somehow....
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jdgordon

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #69 on: April 14, 2015, 01:56:16 am »
0

The probabilities are slightly more than just doing random, but what I was going to do if events were blue backed was keep them seperate and grab 10 spare brownbacks in with the events, draw 0 on 4, 1 on 3, 2 on 2 and 3 on 1. shuffle those 10 cards to determine how many evenmt cards to throw in.

Having them bluebacked always seemed wierd to me so I'm glad the mistake happened :)
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mith

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #70 on: April 14, 2015, 03:11:47 pm »
+1

When we know the costs of all the event cards, we can come up with a scheme which involves drawing event cards to determine how many to use...
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Throwaway_bicycling

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #71 on: April 14, 2015, 09:22:32 pm »
+2


If you want to limit the number of events to 2, you can approximate these probabilities quite well by rolling d6+d8:

2-7: 21/48 = 43.75% (0 events)
8-10: 17/48 = 35.42% (1 event)
11-14: 10/48 = 20.83% (2 events)

So I don't usually have dice in my Dominion stuff, but I do have starter hands of 10 cards. I can approximate these probabilities by taking a hand of 10 cards, putting 3 copper aside, shuffling, then:

1) If top card is an Estate, (probability 3/7 = 9/21) no events
2) If top card is a Copper, keep that card on the table, put the other three coppers back, shuffle, then
3) If top card is now an Estate (probabilty 4/21), two events
4) Otherwise, one event

P(0 events) = 9/21 = .43
P(1 events) = 8/21 = .38
P(2 events) = 4/21 = .19

...but I would honestly probably just flip a coin, and put one event in on heads. :-)

I like it... Only adding two coppers would actually be closer (5/14 is about 36%, 3/14 is about 21%).

For even more precision, use 236 Coppers and 20 Estates, then reveal cards from the top until you have 10 Coppers or 2 Estates, whichever comes first.  Put in events equal to the number of Estates revealed.

So I have to admit: I'm now kind of glad they misprinted the cards. I haven't felt this mocked in weeks.  :)
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Throwaway_bicycling

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #72 on: April 14, 2015, 09:27:51 pm »
+3

It's pretty hard to shuffle a deck with 256 cards in it even when you're looking at it.
It's a problem I never expected to produce.
Eh...it's an even power of 2. There are plenty of sorting and shuffling algorithms that appreciate that, so I would claim this as a feature and not a problem.

The fact that there are 2^8 of these also gives you a nice compact way to express Kingdoms as strings of 10 extended ASCII characters...although some of those kingdoms would print out kind of funny these days. I personally haven't seen a bonafide vertical tab (^k in control character notation or \v in escape notation) in years.

Or just five Unicode characters!  Those would print out really funny though.

But one of those Unicode kingdoms is almost certainly a Perl program that can print its own source code!
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jdgordon

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #73 on: April 14, 2015, 09:43:14 pm »
0

It's pretty hard to shuffle a deck with 256 cards in it even when you're looking at it.
It's a problem I never expected to produce.
Eh...it's an even power of 2. There are plenty of sorting and shuffling algorithms that appreciate that, so I would claim this as a feature and not a problem.

The fact that there are 2^8 of these also gives you a nice compact way to express Kingdoms as strings of 10 extended ASCII characters...although some of those kingdoms would print out kind of funny these days. I personally haven't seen a bonafide vertical tab (^k in control character notation or \v in escape notation) in years.

Or just five Unicode characters!  Those would print out really funny though.

But one of those Unicode kingdoms is almost certainly a Perl program that can print its own source code!

ITs really not that difficult... Each card has a unique integer ID so randomly choose your 10(or however many). sort them (so the same cards are always in the same order) and base32/64 encode it.

You'll end up with a kingdomID something like 3245FAB4...
Chuck the algo on github and we can share them stupidly simply :) (OK OK.. I'll do it tonight if the wiki provides an easy enough way to get an id).


EDIT:

OK, each card is a 16bit value, 6 bits for the card id (card number is defined order (alphabetical for boxes and appearance order for promos) for each box). 4 bits for the box id (base == 0, intrigue 1, etc. for reprints and merged boxes cards would only be in the original appearance box). Then we are left with 6 bits to use as needed. Right now I can only think of using one to mark the BANE card, and one to mark a black market card, unused must be 0. Then you concat each card in the correct order, base32 it and boom, unique kingdom.

What other id markers are needed?
« Last Edit: April 15, 2015, 12:02:26 am by jdgordon »
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Beyond Awesome

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Re: Events will have normal (non-randomizer) card backs due to missprint
« Reply #74 on: April 15, 2015, 06:52:25 am »
0

Honestly, I feel that 1 year from now that no one will care that events have normal backs vs. randomizer backs.
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