Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[timchen]

Remember some time ago we had a loan discussion going on? I just realized that the Saboteur is in fact equivalent to Loan.

Saboteur: Loan
Card with value >=3: Treasure

I seem to remember that the conclusion for loan is that it in average does not cycle your VP cards any faster than your treasure cards. So is it actually a misnomer that one of the Saboteur's disadvantages is that it cycles bad cards for your opponent?

[mith]

On average, you expect the Saboteur to cycle <3 cards proportionally to how many are in the deck, so that the only effect on the opponent's quality of hand is the trashing + getting to the next shuffle faster.

[pacovf]

Revealing cards until you reveal a card which meets certain criteria "usually" has the consequence of leaving you, on average, with a lower proportion of cards in your deck which meet that criteria.

So, on average, loan leaves you with a deck with proportionally less treasures, and saboteur leaves you with a deck with proportionally less cards costing 3$+. The effect, however, is pretty ignorable, on average.

This can be demonstrated (painfully), but if you want to visualize it, imagine the extreme case of two players that open saboteur/nothing, and assume that P1 plays saboteur turn 1, before P2 has a chance to draw his five cards because of some obscure rule that doesn't exist. So P2 is left with 0/n cards costing 3$+ in his deck (n being the number of cards not cycled, and averages 5), which is lower than the normal 1/11.

[carstimon]

Here's the discussion for venture, everything there should carry over.
http://forum.dominionstrategy.com/index.php?topic=1914.0
I want to self promote my experiment on page 4.

I would say no, there's no effect.

[RiemannZetaJones]

Here's the discussion for venture, everything there should carry over.
http://forum.dominionstrategy.com/index.php?topic=1914.0
I want to self promote my experiment on page 4.

I would say no, there's no effect.

The discussion about Venture asked about the next hand, which amounts to considering expected values before a reshuffle. So if it is known that being a victim of Saboteur will not empty your draw deck, then being a victim of Saboteur will not change the average value of your next card. It will however reduce the cumulative value of your entire deck (draw+hand+discard) and will bring the reshuffle closer.

edit: pacovf's example deals with a case where the draw deck could be exhausted by Saboteur, in which case the faster-reshuffle and worse-deck features of Saboteur are what bite, not some sort of mythical 'filter out cards' effect of Saboteur/Loan/Venture etc.

[timchen]

Revealing cards until you reveal a card which meets certain criteria "usually" has the consequence of leaving you, on average, with a lower proportion of cards in your deck which meet that criteria.

So, on average, loan leaves you with a deck with proportionally less treasures, and saboteur leaves you with a deck with proportionally less cards costing 3$+. The effect, however, is pretty ignorable, on average.

This can be demonstrated (painfully), but if you want to visualize it, imagine the extreme case of two players that open saboteur/nothing, and assume that P1 plays saboteur turn 1, before P2 has a chance to draw his five cards because of some obscure rule that doesn't exist. So P2 is left with 0/n cards costing 3$+ in his deck (n being the number of cards not cycled, and averages 5), which is lower than the normal 1/11.

Actually,  if you do this calculation, what you find is that this extreme case (with only 1 saboteurable card in the deck) is a singular one. For any other number of saboteurable cards in the deck the remaining deck quality does not change.

EDIT: to clarify, this has nothing to do with reshuffle. If you know you have exactly 1 saboteurable card in your remaining deck, then you know you will not trigger a reshuffle by the saboteur. As long as that card is not the last card in your deck, your average quality of the remaining deck will be 0. If that card is the last card, the remaining deck quality is undefined; but any sensible definition will not define that to be higher than the original average, so in this case the average quality of the next card is indeed lower.

It is quite curious why this case is just different from the cases for all other number of saboteurable cards in your deck though.

[RiemannZetaJones]

Revealing cards until you reveal a card which meets certain criteria "usually" has the consequence of leaving you, on average, with a lower proportion of cards in your deck which meet that criteria.

So, on average, loan leaves you with a deck with proportionally less treasures, and saboteur leaves you with a deck with proportionally less cards costing 3$+. The effect, however, is pretty ignorable, on average.

This can be demonstrated (painfully), but if you want to visualize it, imagine the extreme case of two players that open saboteur/nothing, and assume that P1 plays saboteur turn 1, before P2 has a chance to draw his five cards because of some obscure rule that doesn't exist. So P2 is left with 0/n cards costing 3$+ in his deck (n being the number of cards not cycled, and averages 5), which is lower than the normal 1/11.

Actually,  if you do this calculation, what you find is that this extreme case (with only 1 saboteurable card in the deck) is a singular one. For any other number of saboteurable cards in the deck the remaining deck quality does not change.

EDIT: to clarify, this has nothing to do with reshuffle. If you know you have exactly 1 saboteurable card in your remaining deck, then you know you will not trigger a reshuffle by the saboteur. As long as that card is not the last card in your deck, your average quality of the remaining deck will be 0. If that card is the last card, the remaining deck quality is undefined; but any sensible definition will not define that to be higher than the original average, so in this case the average quality of the next card is indeed lower.

It is quite curious why this case is just different from the cases for all other number of saboteurable cards in your deck though.

I think it is a mistake even to think about the value of an empty draw deck.

Suppose the draw deck contains at least one card worth $3 or more, and Saboteur is played.
1. Saboteur does not worsen the expected value of the next card drawn provided the reshuffle is not triggered by drawing that card.
2. Saboteur may worsen the expected value of the next card drawn if the reshuffle is triggered by drawing that card (this depends on the contents of the discard pile)
3. Saboteur hastens the reshuffle.

In the case of exactly one card in the draw deck costing $3 or more and none in the discard, points 2. and 3. conspire to make the expected value of the next card worse.

[DG]

Suppose the draw deck contains at least one card worth $3 or more, and Saboteur is played.
1. Saboteur does not worsen the expected value of the next card drawn provided the reshuffle is not triggered by drawing that card.

This actually only applies with two cards worth 3 coins or more.

Having gone through a mathematical proof for ventures I do not want to do the same here. I would suggest that it is not certain that the values remain the same but for most practical purposes you can assume that they are close enough not to matter.

[timchen]

But your point 1. is just wrong, or I should say, not well defined. What do you mean by "provided the reshuffle is not triggered"?

For any given specific draw that leaves more than one card on your deck, the expected value may rise or fall depends on that specific draw.

If you average over all draws that do not trigger a reshuffle by drawing the next card, for a deck containing more than one $3 cards, this basically means average over all draws and yes, the expected value of the next card does not change.

However, in the specific case with exactly one $3 or more card in your deck, for all the draw that do not trigger a reshuffle by drawing the next card, the expected value of the next card is worsened. This has nothing to do with reshuffle.

[timchen]

Suppose the draw deck contains at least one card worth $3 or more, and Saboteur is played.
1. Saboteur does not worsen the expected value of the next card drawn provided the reshuffle is not triggered by drawing that card.

This actually only applies with two cards worth 3 coins or more.

Having gone through a mathematical proof for ventures I do not want to do the same here. I would suggest that it is not certain that the values remain the same but for most practical purposes you can assume that they are close enough not to matter.

Curious, I actually get this to work for any deck with two or more cards that worth 3 coins or more. The only singular case is with exactly 1 card worth 3 coins or more.

[cluckyb]

But your point 1. is just wrong, or I should say, not well defined. What do you mean by "provided the reshuffle is not triggered"?

For any given specific draw that leaves more than one card on your deck, the expected value may rise or fall depends on that specific draw.

If you average over all draws that do not trigger a reshuffle by drawing the next card, for a deck containing more than one $3 cards, this basically means average over all draws and yes, the expected value of the next card does not change.

However, in the specific case with exactly one $3 or more card in your deck, for all the draw that do not trigger a reshuffle by drawing the next card, the expected value of the next card is worsened. This has nothing to do with reshuffle.

True, but this isn't due to bad card cycling. It's due to lowering the overall value of their deck.

A Saboteur that ended with "After doing so, he puts the other revealed cards onto his deck and then shuffles his deck" (instead of the "reshuffle" line) would not change the expected value of the next drawn card in the don't reshuffle case.

[timchen]

?

[pacovf]

Revealing cards until you reveal a card which meets certain criteria "usually" has the consequence of leaving you, on average, with a lower proportion of cards in your deck which meet that criteria.

So, on average, loan leaves you with a deck with proportionally less treasures, and saboteur leaves you with a deck with proportionally less cards costing 3$+. The effect, however, is pretty ignorable, on average.

This can be demonstrated (painfully), but if you want to visualize it, imagine the extreme case of two players that open saboteur/nothing, and assume that P1 plays saboteur turn 1, before P2 has a chance to draw his five cards because of some obscure rule that doesn't exist. So P2 is left with 0/n cards costing 3$+ in his deck (n being the number of cards not cycled, and averages 5), which is lower than the normal 1/11.

Actually,  if you do this calculation, what you find is that this extreme case (with only 1 saboteurable card in the deck) is a singular one. For any other number of saboteurable cards in the deck the remaining deck quality does not change.

EDIT: to clarify, this has nothing to do with reshuffle. If you know you have exactly 1 saboteurable card in your remaining deck, then you know you will not trigger a reshuffle by the saboteur. As long as that card is not the last card in your deck, your average quality of the remaining deck will be 0. If that card is the last card, the remaining deck quality is undefined; but any sensible definition will not define that to be higher than the original average, so in this case the average quality of the next card is indeed lower.

It is quite curious why this case is just different from the cases for all other number of saboteurable cards in your deck though.

I've read blueblimp's proof, and he's absolutely correct, so I guess my demonstration is wrong.

The reason his demonstration doesn't work for only one special card is that he assumes the draw deck won't be empty by the end of the reveal, which is only true if you have at least two special cards in your draw deck. That assumption makes the initiation of his recursion, and since we cannot define a density for zero cards decks, dropping it would invalidate his demonstration.

[RiemannZetaJones]

But your point 1. is just wrong, or I should say, not well defined. What do you mean by "provided the reshuffle is not triggered"?

For any given specific draw that leaves more than one card on your deck, the expected value may rise or fall depends on that specific draw.

If you average over all draws that do not trigger a reshuffle by drawing the next card, for a deck containing more than one $3 cards, this basically means average over all draws and yes, the expected value of the next card does not change.

However, in the specific case with exactly one $3 or more card in your deck, for all the draw that do not trigger a reshuffle by drawing the next card, the expected value of the next card is worsened. This has nothing to do with reshuffle.

I think my point 1 is stupid, it can only refer to the case of 2 $3 cards or more.

In general, blueblimp's argument shows that a card in the style of Venture/Loan/Saboteur will necessarily not change the expected values of the next n cards precisely when it is guaranteed that drawing those n cards will not trigger the reshuffle. So although you're right that you can explain the change in expected value without referring to the reshuffle, the changing of the expected values and the possible triggering of the reshuffle must be bound up with one another.

[blueblimp]

Looks like the important points from the Venture discussion have already been covered, so I'll chime in with something interesting I've realized since then.

In general, in Dominion, there is a lot of luck of the form: a high probability of something mildly good happening for you, and a low probability of something fairly bad happening for you. Because the expected value arguments sidestep analyzing the actual distribution, they don't say much about this.

For example, when you play Saboteur, the most likely case is skipping a small number of bad cards, which is mildly good for you, because your opponent's draw deck becomes a little worse, but there's a small chance of skipping a large number of bad cards, which is fairly bad for you, because your opponent's draw deck becomes a lot better.

When you play Loan/Venture, in many games the most likely case is that you only skip bad cards (because you have more bad cards than good cards), which is mildly good for you. But there is a small chance of getting unlucky and skipping a good card, which can be very bad for you.

Cards missing the reshuffle is another example. Usually doesn't happen, which is mildly good, but when it happens, it's pretty bad, and if both opening buys miss the reshuffle, it's very bad. Similar idea with opening double terminal, where it's more likely that they don't collide, but you're in big trouble when they do.

There is some luck that goes the other way (e.g. Treasure Map collision, which is unlikely without enablers), and well this is all symmetric if you consider the effect on your opponent, but I still think it's part of the flavour of the game.