5 hours ago, GrandSpleen said:

Oh and duh! Silly me. The situations where discarding from deck ceases to have any negative potential are 1) when you know what's going to be discarded [Imladris Stargazer, hero Gandalf] and 2) when there is no possible detrimental discard [you've built your deck to accommodate mining]

You're missing a key point: unless you know what you are discarding and revealing, you face the exact same 'negative potential' by not discarding too.

When you do know, you are clearly better off discarding by being able to clear out the rubbish cards.

When you don't know - you are getting no worse outcomes by discarding. You might get better outcomes if you have cards than benefit from stuff in your discard pile, and better decisions from knowing more about what is left in your deck.

Imagine you discarded the top 5 cards from your deck at the start of each game (prior to the Mulligan decision): are you any worse off?

(I think you'd actually be better off, despite a chance of discarding every copy of some key card!)

Ok, I've gone and written an actual computer program that plays the game 10,000 times under different scenarios and records the win percentage.

Each game starts with a 44 card deck with 3 SoG in it (we assume that the opening hand did not contain SoG). We make each game last 7 rounds. We call a game a "win" if we draw into SoG at any point in those 7 rounds. We also keep track of what turn the game was won.

1a) With no Thorin, just drawing 1 card per turn

Win percentage: 41.4
Average round won: 3.8

1b) Drawing 1 card per turn, with Thorin each round

Win percentage: 41.5
Average round won: 3.8

2a) Extra card each round, with no Thorin

Win percentage: 69.5
Average round won: 3.5

2b) Extra card each round, with Thorin each round

Win percentage: 69.7
Average round won: 3.6

3a) Master of the Forge each round, with no Thorin

Win percentage: 95.8
Average round won: 2.7

3b) Master of the Forge each round, with Thorin each round

Win percentage: 95.4
Average round won: 2.7

---

The small differences between Thorin/non-Thorin are statistical only. If I run the program again the results will be slightly different but will stay well within a percent of each other.

So there you have it. Even in the case of a deck running Master of the Forge, using Thorin every round did not impact the win percentage.

So Thorin is not a cost.

As an addendum:

Let's switch our key card to be 1x and make ourselves draw 7 cards per round over 7 rounds:

4a) 7 cards per round, looking for a 1x card, with no Thorin:

Win percentage: 100.0
Average round won: 3.6

4b) 7 cards per round, looking for a 1x card, with Thorin each round:

Win percentage: 88.4
Average round won: 3.3

So as I stated from the start, if you can draw your whole deck, then yes of course Thorin can have a downside. But this is not realistic for most decks.

59 minutes ago, Seastan said:

as I stated from the start, if you can draw your whole deck, then yes of course Thorin can have a downside. But this is not realistic for most decks.

I think you are being too kind - not only (1) do you need to draw through your deck, but (2) you have to lose after having drawn through your deck, (3) lose when you otherwise would have won, (4) not recognize that you are getting into danger in time to moderate your behaviour.

You might as well just say 'Mining is something you need to consider before doing in "Deadman's Dike"

6 hours ago, Seastan said:

Ok, I've gone and written an actual computer program that plays the game 10,000 times under different scenarios and records the win percentage.

Each game starts with a 44 card deck with 3 SoG in it (we assume that the opening hand did not contain SoG). We make each game last 7 rounds. We call a game a "win" if we draw into SoG at any point in those 7 rounds. We also keep track of what turn the game was won.

1a) With no Thorin, just drawing 1 card per turn

Win percentage: 41.4
Average round won: 3.8

1b) Drawing 1 card per turn, with Thorin each round

Win percentage: 41.5
Average round won: 3.8

2a) Extra card each round, with no Thorin

Win percentage: 69.5
Average round won: 3.5

2b) Extra card each round, with Thorin each round

Win percentage: 69.7
Average round won: 3.6

3a) Master of the Forge each round, with no Thorin

Win percentage: 95.8
Average round won: 2.7

3b) Master of the Forge each round, with Thorin each round

Win percentage: 95.4
Average round won: 2.7

---

The small differences between Thorin/non-Thorin are statistical only. If I run the program again the results will be slightly different but will stay well within a percent of each other.

So there you have it. Even in the case of a deck running Master of the Forge, using Thorin every round did not impact the win percentage.

So Thorin is not a cost.

I suppose the ridiculous boost from Master of the Forge only applied because we're looking for an attachment. I don't think these results are the same if we're looking for, say, a particular ally, or a powerful event, which is our must-win card.

Dang, I admire your dedication. Well it seems you are right, statistically speaking Thorin is not a cost!
Still I have a hard time seeing a card that I was desperate to draw go into the discard pile and thinking I have lost nothing. The next time this happens, I shall say to myself "this was not a cost, for the GrandSpleen in an equal number of alternate universes did not discard this."

7 hours ago, Seastan said:

As an addendum:

Let's switch our key card to be 1x and make ourselves draw 7 cards per round over 7 rounds:

4a) 7 cards per round, looking for a 1x card, with no Thorin:

Win percentage: 100.0
Average round won: 3.6

4b) 7 cards per round, looking for a 1x card, with Thorin each round:

Win percentage: 88.4
Average round won: 3.3

So as I stated from the start, if you can draw your whole deck, then yes of course Thorin can have a downside. But this is not realistic for most decks.

Ok Thorin is a cost if played in a noldor deck along Erestor and Galdor heroes full of additional card draw effect. I deem dwarves and elves are not a likely friendship but in few case...

Make me wonder how many cards can you actually draw without game breaking shenanigans (like invite loop or event recursion abuse).

15 hours ago, GrandSpleen said:

I like the shuffle example, it re-randomizes the deck, which is a nice complication. But if your deck is going to function well, you probably have added some card draw to it. You have Elven-light and Open the Armoury in your Everything Costs 1 deck. Let's use that repeatable card draw effect, and suddenly the thought experiment is biased toward "don't use Thorin." So let's imagine: Thorin will discard after you draw 2 cards (for example: in your 'everything costs 1' deck, this will be your Elven-light +Arwen combo being played each round before using Thorin).

Scenario A (SoG on top of deck), Draw SoG and win. Thorin: not a factor (result is the same whether or not you use him).

Scenario B (SoG 2nd card in deck), Draw X, draw SoG and win. Thorin or non-use of Thorin is not a factor in scenarios A and B. Thorin: not a factor.

Scenario C (SoG 3rd card in deck), Draw X, draw X, discard SoG and lose because you used Thorin. Thorin: caused a loss.

Scenario D (SoG is 4th card in deck), Draw X, Draw X, discard X, then on turn 2 draw SoG and win. But: if you don't use Thorin, turn 1 is Draw X, Draw X, and turn 2 is draw X, draw SoG. In scenario D, you win regardless of whether or not you use Thorin. Thorin: not a factor.

Scenario E (SoG is 5th card in deck), Draw X, Draw X, discard X on turn 1. Turn 2: draw X, draw X, discard SoG due to Thorin But without using Thorin, you were going to lose anyway (Game ends on turn 2, you were not going to draw SoG). Thorin: not a factor.

So with that addition, Thorin directly causes a loss in scenario C. For all the other scenarios, your win or loss is regardless of whether or not you use Thorin.

I made a scenario which suggests using Thorin carries more risk than reward, if we're staking a win-or-loss condition on drawing the desired card.

You made a scenario which suggests there is no difference between using or not using Thorin, and therefore you should just use him (and get +1 attack) as the win-or-loss is beyond your control (equal chance of either regardless of your actions).

If we add more complications, which way is this going to be biased? Would we see more and more scenarios in which using Thorin is equal to not using Thorin? Would we see more and more scenarios in which Thorin causes the loss? Would we ever see a set of scenarios in which Thorin directly causes more wins than losses?

Your Scenario E is wrong, it should be:

Scenario E (SoG is 5th card in deck), Draw X, Draw X, discard X on turn 1. Turn 2: draw X, draw SoG and win because you used Thorin, otherwise you would only have drawn the first four cards. Thorin: caused a win.

Scenario F would essentially be, what you describe in Scenario E.

Scenario F (SoG is 5th card in deck), Draw X, Draw X, discard X on turn 1. Turn 2: draw X, draw X, discard SoG due to Thorin But without using Thorin, you were going to lose anyway (Game ends on turn 2, you were not going to draw SoG). Thorin: not a factor.

So with that addition, Thorin directly causes a loss in scenario C and a win in Scenario E. For all the other scenarios, your win or loss is regardless of whether or not you use Thorin.

To conclude: Thorin changes the outcome of some games but in the end, he makes you win or lose with the same number of games.

[deleted for dumbness]

2 minutes ago, sappidus said:

This is an interesting claim. I am not sure if it's true. I will do some combinatorial calculations when I have time and report back.

(N.B.: this is different from what I think about Thorin or deck-discard effects in general. I generally agree with @Seastan et al. that there is essentially no functional "cost" for regular decks.)

This is in response to the example Grandspleen posted. For Rouxxor's example it does not work because of the fact that you can play Elven Light from your discard pile only. So if you get a desireable card there, Thorin can actually help. The only way to lose Rouxxor's example due to Thorin would be to discard SoG from the deck, before drawing it with Elven Light, which was against his rules. But then it does not matter what you discard, if you have no means afterwards to draw again.

Haha, my post was up for like a minute before I deleted it. You're too fast.

1 hour ago, GrandSpleen said:

Dang, I admire your dedication. Well it seems you are right, statistically speaking Thorin is not a cost!
Still I have a hard time seeing a card that I was desperate to draw go into the discard pile and thinking I have lost nothing. The next time this happens, I shall say to myself "this was not a cost, for the GrandSpleen in an equal number of alternate universes did not discard this."

I appreciate the concession. As for the psychological cost of using Thorin, I make no similar guarantee 😁.

1 hour ago, sappidus said:

Haha, my post was up for like a minute before I deleted it. You're too fast.

I see, you are not one of my workmates, they would never say that .

On 11/19/2018 at 1:22 AM, Seastan said:

You don't need to decide where to place undefended damage until after revealing the shadow card. So that's not how it works.

That doesn't contradict my statement or scenario. You decide to take the first attack undefended. You reveal *its* shadow card. You have to place the damage from the *first* undefended attack before you know what the shadow from the *second* undefended attack is. So if the second attack has Dol Goldur Orcs as a shadow, you won't know it until *after* you assign damage from the first undefended attack -- and since the whole point of taking two attacks undefended was to boost up Gimli, the damage from the first attack will be put on Gimli, not Thalin.

On 11/19/2018 at 2:37 AM, ColinEdwards said:

When you don't know - you are getting no worse outcomes by discarding. You might get better outcomes if you have cards than benefit from stuff in your discard pile, and better decisions from knowing more about what is left in your deck.

Imagine you discarded the top 5 cards from your deck at the start of each game (prior to the Mulligan decision): are you any worse off?

(I think you'd actually be better off, despite a chance of discarding every copy of some key card!)

Clarification -- would those five cards go back in your deck if you decided to Mulligan? If not, and you have a bad hand, the five you discarded are from a deck of greater average quality than your hand, so you are reducing the average quality of the deck by keeping your current hand but losing five cards from the (better) deck. If you mulligan because you are lacking a particular 3x card, and you just discarded two of them, you've just made your life *much* more difficult just for the informational advantage that mulliganing your current bad hand is not likely to get the key card you need.

If you rely on search effects to find key cards, discarding search cards will make your remaining deck less consistent. Worst case would be where you always use Thurindir with Gather Information to find a single copy of your key card -- and this initial discard tossed it. If you rely on search cards to find things that are *in* the deck, then discarding from the deck by any method can make those cards unavailable for search and represent a real cost.

On 11/19/2018 at 4:40 AM, ColinEdwards said:

I think you are being too kind - not only (1) do you need to draw through your deck, but (2) you have to lose after having drawn through your deck, (3) lose when you otherwise would have won, (4) not recognize that you are getting into danger in time to moderate your behaviour.

You might as well just say 'Mining is something you need to consider before doing in "Deadman's Dike"

I think you are being too harsh -- something can absolutely be a "cost" even if you end up winning anyways. Absent cards that care whether a particular card is in your deck or your discard, discarding a card from a randomized deck is not a cost if you never end up drawing your entire deck -- but if you do end up drawing your whole deck, it costs you future draws, and that's just as much a cost as exhausting a hero or paying resources or taking damage.

If you "recognize that you are getting into danger in time to moderate your behavior", then obviously the discard is a cost because you're changing your behavior *because* of the cost. If you stop using Stonehelm's ability because you're worried about running out of deck, than *its cost* is stopping you from using it.

1 hour ago, dalestephenson said:

If you stop using Stonehelm's ability because you're worried about running out of deck, than *its cost* is stopping you from using it.

Yeah, in the exact same way as Daeron's Runes or Cirdan has a 'cost' as you get close to running your deck down. (Unless you are playing a mission that plays off your discard pile or penalises you for running out of cards, it's a fairly theoretical one.)

On 11/19/2018 at 1:39 AM, Seastan said:

But most decks (nearly all?) do not get anywhere close to drawing themselves in a game. Therefore I maintain that in those decks Thorin is not a cost.

Just curious: do you trigger Thorin on essentially every single opportunity, even when his extra damage is not "needed", as long as you don't think you're going to reach that deep into your deck?

3 hours ago, dalestephenson said:

and since the whole point of taking two attacks undefended was to boost up Gimli, the damage from the first attack will be put on Gimli, not Thalin.

I disagree that that was the point (my reason for doing it would be so that Gimli could be used to kill an enemy), but it doesn't matter. All I'm saying is the added risk of losing a hero by taking both those attacks undefended was around 5%, full stop. Certain decisions can be made to increase that risk, but I based my statement on the decisions I would make. Other decisions lead to different risks, of course, as someone who decides to put the damage on Eowyn is going to face 100% chance of losing a hero.

After placing the initial 2 undefended damage on Gimli, the risk of losing a hero by taking the second attack undefended was high. Higher than I'd be comfortable with personally as a conservative player. This is why I would've placed the first 2 damage differently.

53 minutes ago, ColinEdwards said:

Yeah, in the exact same way as Daeron's Runes or Cirdan has a 'cost' as you get close to running your deck down. (Unless you are playing a mission that plays off your discard pile or penalises you for running out of cards, it's a fairly theoretical one.)

It's hardly the exact same way. Cirdan's cost isn't optional -- if you draw, you discard, and the discard certainly is a cost. Discarding is a cost of playing Daeron's Runes, and unless you want to discard a card currently in your hand, it's a dead card with an empty deck.

If you stop using Stonehelm's ability because you're worried about the deck, his cost isn't theoretical, it's *real*. You're forgoing a point of direct damage because of the cost.

Outside quests that penalize you for running out of cards or penalize you for cards in your discard, the game itself penalizes you for having an empty deck -- you don't get to draw a new card. If it's impossible for you run out your deck without winning or losing, then discarding cards from your deck does not affect the likelihood of this game-induced penalty coming into play. If it's possible, then discarding from your deck is an actual cost; and the likelier it is to run out of deck the higher the cost becomes (assuming you're not running a Noldor deck that wants to recur Lords of the Eldar every turn).

There are 34 pages of decks that include Will of the West, and they aren't *all* wasting their card slots on a useless card. Will of the West is included in close to a third of the mining (Zigil Miner) decks, so clearly the possibility of mining through a deck prematurely is sufficiently compelling to justify deck space for a strong minority of those decks.

21 minutes ago, sappidus said:

Just curious: do you trigger Thorin on essentially every single opportunity, even when his extra damage is not "needed", as long as you don't think you're going to reach that deep into your deck?

Since I play him in a mining deck, yeah absolutely. I've yet to experiment with him outside of one, but given the math, I don't see why not. At worst its a neutral effect and at best he gives you increased information about what you're going to draw.

On 11/19/2018 at 9:24 AM, Seastan said:

Ok, I've gone and written an actual computer program that plays the game 10,000 times under different scenarios and records the win percentage.

Each game starts with a 44 card deck with 3 SoG in it (we assume that the opening hand did not contain SoG). We make each game last 7 rounds. We call a game a "win" if we draw into SoG at any point in those 7 rounds. We also keep track of what turn the game was won.

1a) With no Thorin, just drawing 1 card per turn

Win percentage: 41.4
Average round won: 3.8

1b) Drawing 1 card per turn, with Thorin each round

Win percentage: 41.5
Average round won: 3.8

2a) Extra card each round, with no Thorin

Win percentage: 69.5
Average round won: 3.5

2b) Extra card each round, with Thorin each round

Win percentage: 69.7
Average round won: 3.6

3a) Master of the Forge each round, with no Thorin

Win percentage: 95.8
Average round won: 2.7

3b) Master of the Forge each round, with Thorin each round

Win percentage: 95.4
Average round won: 2.7

---

The small differences between Thorin/non-Thorin are statistical only. If I run the program again the results will be slightly different but will stay well within a percent of each other.

So there you have it. Even in the case of a deck running Master of the Forge, using Thorin every round did not impact the win percentage.

So Thorin is not a cost.

I figured out, this should also be calculable, so I gave it a try.

1a) With no Thorin, just drawing 1 card per turn

I will lose when there is no copy of Steward in the first seven cards. The number of combinations for 7 cards out of 44 is C(44, 7) = 44! / 7! / (44-7)! = 38 320 568.

Of course C(n, k) = n! / k! / (n - k)! applies here.

In order to get the number of combinations for not having Steward in the first 7 cards means I have to multiply the number of combinations for 0 cards out of 3 (for Steward) and the number of combinations for 7 cards out of 41 (everything save Steward): C(3, 0) * C(41, 7) = 3! / 0! / (3-0)! * 41! / 7! / (41 - 7)! = 22 481 940.

Dividing the second result (2) by the first (1) gives us the probality to lose the game:

22 481 940 / 38 320 568 = 58.7%

Thus the win percentage is the difference to 100% which is close enough to Seastan's result: 1 - 22 481 940 / 38 320 568 = 41.3%. As he already said

Quote

the results will be slightly different but will stay well within a percent of each other.

To get the average round, in which the game was won, we have some additional work: We need to figure out, how often will we win in the first round, the second round and so on until the seventh round. This means, we have to calculate the probability of the first Steward being the first, second, ... , seventh card in the deck.

The first card is easy, as 3 cards out of the deck with 44 cards will let us win, thus the probability is p1 = 3 / 44 = 6.8%.

For the second card we take the percentage of the first one being no Steward (1 - p1 = 41 / 44) und multiply it with the second card being Steward (3 / 43): p2 = 41 / 44 * 3 / 43 = 6.5%

For the third card we take the percentage of the first two being no Steward (1 - p1 - p2) and multiply it with the third card being Steward (3 / 42): p3 = (1 - p1 - p2) * 3 / 42 = 6.2%

This goes analogously for the next four cards:

p4 = (1 - p1 - p2 - p3) * 3 / 41 = 5.9%

p5 = (1 - p1 - p2 - p3 - p4) * 3 / 40 = 5.6%

p6 = (1 - p1 - p2 - p3 - p4 - p5) * 3 / 39 = 5.3%

p7 = (1 - p1 - p2 - p3 - p4 - p5 - p6) * 3 / 38 = 5.0%

Of course these 7 percentages sum up to 41.3% again.

In order to calculate the expected value we need to multiply every percentage with its corresponding card's position in the deck and add everything up. But as our total win percentage is only 41.3%, we also need to divide the result by this value, otherwise we would regard losses as wins in round 0.

(6.8% * 1 + 6.5% * 2 + 6.2% * 3 + 5.9% * 4 + 5.6% * 5 + 5.3% * 6 + 5.0% * 7) / 41.3% = 3.8

This time the result is the same as from Seastan's simulation, which proves that at least the first part of it is correct.

Now there are only 5 cases left, which need the same treatment, just this time they will be way more complicated, due to extra cards being drawn, cards being discarded and decks being shuffled. This post alone took me about 3 hours to compute and write down, so I will be done for today. And of course at least some of you need to understand my calculations, otherwise there would be no point in posting them here.

4 minutes ago, Amicus Draconis said:

Now there are only 5 cases left, which need the same treatment, just this time they will be way more complicated, due to extra cards being drawn, cards being discarded and decks being shuffled. This post alone took me about 3 hours to compute and write down, so I will be done for today. And of course at least some of you need to understand my calculations, otherwise there would be no point in posting them here.

I suspect analysis of the remaining cases would be best done using Markov chains, the mere appearance of which would probably narrow the interested public to statistically 0%, heh.

9 minutes ago, sappidus said:

I suspect analysis of the remaining cases would be best done using Markov chains, the mere appearance of which would probably narrow the interested public to statistically 0%, heh.

Given this crowd? I'm not sure I would take that bet.

5 minutes ago, Seastan said:

I disagree that that was the point (my reason for doing it would be so that Gimli could be used to kill an enemy), but it doesn't matter. All I'm saying is the added risk of losing a hero by taking both those attacks undefended was around 5%, full stop. Certain decisions can be made to increase that risk, but I based my statement on the decisions I would make. Other decisions lead to different risks, of course, as someone who decides to put the damage on Eowyn is going to face 100% chance of losing a hero.

After placing the initial 2 undefended damage on Gimli, the risk of losing a hero by taking the second attack undefended was high. Higher than I'd be comfortable with personally as a conservative player. This is why I would've placed the first 2 damage differently.

If you place the first 2 damage on Thalin, you reduce the number of hero killing shadows from an undefended attack to 3 instead of 5. It's still higher than 5% (3/26 rounds to 12%). In order to get below 5% you need to place the first 2 damage on Thalin and *defend* with Gimli, which reduces the number of killing Shadows to 1 (1/26). But in that case, Gimli increases his effective attack *at all* only if one of the two Dol Goldur Orcs shadows come out, and is not available to kill an enemy. Clearly not practical at all under the circumstances.

However, you make a good point that in this particular playthrough, it was not necessary to place the first undefended attack's damage on Gimli. While Gimli wants the damage eventually for the Nazgul, he doesn't need the extra two points of damage *this* turn, and since he will still have a 2-strength attack enemy engaged after the combat phase he could easily take that damage *next* turn instead with another undefended attack (in fact, in Beorn's playthrough Gimli did exactly this, after being healed by Daughter of Nimrodel). The only practical downside to putting the first undefended on Thalin is that it put him in range of Necromancer's Reach killing him during the next questing phase -- but since it would also kill Theodred and Eowyn, that would hardly matter.