Dominion Strategy Forum
Dominion => Dominion General Discussion => Topic started by: suet63 on June 20, 2022, 09:14:33 am
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I want to hit $5 as soon as possible, but I start with a 4/3 split. Should I buy
2silvers OR 1wishing well 1silver OR 2wishing wells ?
I don't know how to calculate my expected money.
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One of each, if able to play WW on turn 4 you'll have perfect knowledge of the last card in your draw pile.
In the long run you might (ahem) wish you had more Wells and not the Silver but starting with exactly one Silver is the correct move on most boards.
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Yes. The main thing is, WW+Silver brings you to $9 total, and that means that as long as you can play the WW and guess correctly you are guaranteed a hand with at least $5 on the second shuffle. And you can always guess correctly if it's in your second hand.
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Yes. The main thing is, WW+Silver brings you to $9 total, and that mean that as long as you can play the WW and guess correctly you are guaranteed a hand with at least $5 on the second shuffle. And you can always guess correctly if it's in your second hand.
Even in your t3 hand, it's pretty easy to wish well because most of your deck is Coppers.
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WW+Silver:
Ways to get 5 on turn 3:
5C
4C + S
3C + S + E
3C + S + WW
(WW will guess copper cuz it’s more common than silver and is only way to get to 5)
4C + WW + (1C on top of deck)
4C + WW + (1S on top of deck)
4C + WW + (E + C on deck)
2C + 1S + WW + (1C top of deck)
2C + 1S + WW + (E + C top of deck)
3C + WW + E + (CC on deck)
3C + WW + E + (S on top of deck)
1C + 1S + 2E + (CC on deck)
3C + WW + E + (E + S on deck) **wish for silver**
Ah man this is gonna be tedious. Might come back later and fill these in.
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The better way to do this is via simulation. You can easily run enough simulations to converge on probailities down to a few decimal points, and it takes like 30 lines of code. Which I believe I've actually done before for ways to get to 5$. I don't remember the results, but I'd probably remember if wishing well+silver wasn't better than silver/silver.
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Actually, I was feeling like doing a programming task, so I did it again -- and it seems like we were all wrong. I'm getting 91.1% for hitting 5$ with silver/silver and 90.5% for hitting 5$ with silver/ww.
On the other hand, you could say the odds are basically the same, and wishing well is a much better card, so you'd rather open ww/silver. Still, your probability to hit 5$ is lower.
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(Link to script if anyone is interested (https://www.dropbox.com/s/l8ot6unwsf7von6/simulator.py?dl=1))
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The Wishing Well/Silver odds aren't too hard to compute though. We'll distinguish 3 cases:
WW in hand 3 (5/12 odds)
WW in hand 4 (5/12 odds)
WW misses shuffle (2/12 odds)
WW in hand 4 is $5 guaranteed.
WW in hand 3 - only way to miss $5 is to guess wrong. We'll look at what your hand 3 is after drawing your WW card.
A) SCCEE (3/22 odds)
B) CCCEE (5/22 odds)
C) CCCCE (5/22 odds)
are the only potential failstates.
For A), you only guess wrong if your single remaining E is on top; that will also immediately trigger a fail to hit $5 on T4. (1/6 odds)
For B) and C), guessing wrong can mean either hitting S or E.
If in B), if you hit S that guarantees $5 on T4, so disregard that. If you hit E, you fail only if you bottomdeck S. Overall odds: 1/30
For C), it is most complicated; both S or E can lead to failstates. With S, the failstate is bottomdecking C (3/5 odds), with E, the failstate is not bottomdecking E (4/5 odds) Overall odds: 11/30
All in all, WW in hand 3 has a failure rate of
3/22*1/6 + 5/22*1/30 + 5/22*11/30 = 5/44.
Finally, what if WW misses the shuffle? Let's go by what the other card missing the shuffle is.
If it's E, then you have $9 in 10 cards, so $5 is guaranteed.
If it's S (i.e. "Golden Sombrero"), then you have the same 10 cards as in the first shuffle, so the failure rate is 5/6.
If it's C, you have to split $4/$4 to fail, so you first hand is one of
CCCCE (5/28 odds)
SCCEE (5/28 odds)
In total the failure rate is 5/14. All in all, the failure rate for WW missing the shuffle is
1/11*5/6 + 7/11*5/14 = 10/33.
That means that the chance of not hitting $5 when opening WW/Silver is
5/12*5/44 + 2/12*10/33 = 155/1584 = 9.7853...%
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Actually, I was feeling like doing a programming task, so I did it again -- and it seems like we were all wrong. I'm getting 91.1% for hitting 5$ with silver/silver and 90.5% for hitting 5$ with silver/ww.
On the other hand, you could say the odds are basically the same, and wishing well is a much better card, so you'd rather open ww/silver. Still, your probability to hit 5$ is lower.
I'm not surprised by this result, TBH. The only way playing a WW gives you more $ than playing a Silver does is in the rare case where you draw a Silver and a Copper with it. Usually it only gives $2 or less. Compared to Silver, WW has the advantage of better cycling, but also a substantial risk of being just a cantrip when you guess wrong.
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That means that the chance of not hitting $5 when opening WW/Silver is
5/12*1/12 + 2/12*3/10 = 61/720 = 8.47222...%
I bet 10 honor points that this number is incorrect since it doesn't align with mine.
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That means that the chance of not hitting $5 when opening WW/Silver is
5/12*1/12 + 2/12*3/10 = 61/720 = 8.47222...%
I bet 10 honor points that this number is incorrect since it doesn't align with mine.
Well I looked at your code and you do not always guess Copper on T3, which is the superior strategy.
EDIT: Nevermind, if your code is correct it should always guess Copper since there's no way that is not the most prominent card in your deck.
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If it's S (i.e. "Golden Sombrero"), then you have the same 10 cards as in the first shuffle, so the failure rate is 4/5.
Isn't it 5/6? Or have I missed something that raises the odds compared to normal?
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If it's S (i.e. "Golden Sombrero"), then you have the same 10 cards as in the first shuffle, so the failure rate is 4/5.
Isn't it 5/6? Or have I missed something that raises the odds compared to normal?
Ah, no I had an 80/20 split in my head but that's wrong. That explains the discrepancy hopefully.
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C) CCCCE (5/22 odds)
For C), it is most complicated; both S or E can lead to failstates. With S, the failstate is bottomdecking C (3/5 odds), with E, the failstate is not bottomdecking E (4/5 odds) Overall odds: 7/30
So, remaining cards here are CCCEES. Probability of silver-on-top is 1/6. Estate is 2/6. So I get 1/6*3/5 + 2/6*4/5 = 11/30. (I think you calculated 1/6 for estate instead of 2/6?)
And if I redo the other calculations with that, the result aligns with my simulation!
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(Link to script if anyone is interested (https://www.dropbox.com/s/l8ot6unwsf7von6/simulator.py?dl=1))
thanks, i find the peek behind the curtain very interesting. 8)
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If it's S (i.e. "Golden Sombrero"), then you have the same 10 cards as in the first shuffle, so the failure rate is 4/5.
Isn't it 5/6? Or have I missed something that raises the odds compared to normal?
Incorporating both changes, the results get even closer
3/22*1/6 + 5/22*1/30 + 5/22*11/30 = 5/44
1/11*5/6 + 7/11*5/14 = 10/33
5/12*5/44 + 2/12*10/33=155/1584
and 1 minus that as a decimal is 0.902, which is what I get. (I know I said 90.5, but I actually get 90.2 if I run more simulations.) So now it should be correct :-)
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One of each, if able to play WW on turn 4 you'll have perfect knowledge of the last card in your draw pile.
In the long run you might (ahem) wish you had more Wells and not the Silver but starting with exactly one Silver is the correct move on most boards.
I have imagined a board, that you may only want to buy WW, Witch, silver and gold. There is no village, and you don't want to draw WW when you play witch. Will you start with silver/silver in this board?
Harvest /Library / Merchant Ship /Sea Witch /Witch
Fortuneteller / Sea Chart / Wishing Well / Bureaucrat / Duke
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Actually, I was feeling like doing a programming task, so I did it again -- and it seems like we were all wrong. I'm getting 91.1% for hitting 5$ with silver/silver and 90.5% for hitting 5$ with silver/ww.
On the other hand, you could say the odds are basically the same, and wishing well is a much better card, so you'd rather open ww/silver. Still, your probability to hit 5$ is lower.
I have a Maths question. How do I calculate my expected money in turn3, if I start with silver/silver? I suppose this is easy probability, but I forgot all I have learn in high school! 🤬🤬
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The Wishing Well/Silver odds aren't too hard to compute though. We'll distinguish 3 cases:
WW in hand 3 (5/12 odds)
WW in hand 4 (5/12 odds)
WW misses shuffle (2/12 odds)
WW in hand 4 is $5 guaranteed.
WW in hand 3 - only way to miss $5 is to guess wrong. We'll look at what your hand 3 is after drawing your WW card.
A) SCCEE (3/22 odds)
B) CCCEE (5/22 odds)
C) CCCCE (5/22 odds)
are the only potential failstates.
For A), you only guess wrong if your single remaining E is on top; that will also immediately trigger a fail to hit $5 on T4. (1/6 odds)
For B) and C), guessing wrong can mean either hitting S or E.
If in B), if you hit S that guarantees $5 on T4, so disregard that. If you hit E, you fail only if you bottomdeck S. Overall odds: 1/30
For C), it is most complicated; both S or E can lead to failstates. With S, the failstate is bottomdecking C (3/5 odds), with E, the failstate is not bottomdecking E (4/5 odds) Overall odds: 7/30
All in all, WW in hand 3 has a failure rate of
3/22*1/6 + 5/22*1/30 + 5/22*7/30 = 1/12.
Finally, what if WW misses the shuffle? Let's go by what the other card missing the shuffle is.
If it's E, then you have $9 in 10 cards, so $5 is guaranteed.
If it's S (i.e. "Golden Sombrero"), then you have the same 10 cards as in the first shuffle, so the failure rate is 4/5.
If it's C, you have to split $4/$4 to fail, so you first hand is one of
CCCCE (5/28 odds)
SCCEE (5/28 odds)
In total the failure rate is 5/14. All in all, the failure rate for WW missing the shuffle is
1/11*4/5 + 7/11*5/14 = 3/10.
That means that the chance of not hitting $5 when opening WW/Silver is
5/12*1/12 + 2/12*3/10 = 61/720 = 8.47222...%
nice analysis!! 😊
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Actually, I was feeling like doing a programming task, so I did it again -- and it seems like we were all wrong. I'm getting 91.1% for hitting 5$ with silver/silver and 90.5% for hitting 5$ with silver/ww.
On the other hand, you could say the odds are basically the same, and wishing well is a much better card, so you'd rather open ww/silver. Still, your probability to hit 5$ is lower.
I have a Maths question. How do I calculate my expected money in turn3, if I start with silver/silver? I suppose this is easy probability, but I forgot all I have learn in high school! 🤬🤬
Expected money is easy: you have $11 among 12 cards, so the expected value for your money on T3 (or T4) is 5*11/12=55/12=$4.6.
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yes. (Though expected money doesn't allow you to conclude anything about the probability of hitting a certain amount.)
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If it's S (i.e. "Golden Sombrero"), then you have the same 10 cards as in the first shuffle, so the failure rate is 4/5.
Isn't it 5/6? Or have I missed something that raises the odds compared to normal?
Incorporating both changes, the results get even closer
3/22*1/6 + 5/22*1/30 + 5/22*11/30 = 5/44
1/11*5/6 + 7/11*5/14 = 10/33
5/12*5/44 + 2/12*10/33=155/1584
and 1 minus that as a decimal is 0.902, which is what I get. (I know I said 90.5, but I actually get 90.2 if I run more simulations.) So now it should be correct :-)
Thanks, I made the changes in my post so the correct information is available directly!