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Author Topic: Reasons for the 1st-player vs 2nd-player advantage  (Read 49791 times)

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kn1tt3r

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #25 on: January 18, 2012, 02:21:41 am »
0

Treasure map/wareouse is a typical opening that I aim for only as p2. The increased variance helps.
I don't get this. TM/Warehouse is such a hugh combo, you need very good reasons not to go for it.

I MIGHT agree on simple TMs on BMesque boards against a potentially better player, but with Warehouse support, 1st/2nd player shouldn't really play a big role in your decision making.
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Mean Mr Mustard

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #26 on: January 18, 2012, 07:30:42 am »
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I think that it is safe to say that even the best players, playing at top form, lose somewhere in the neighborhood of twenty-five percent of their games because of bad shuffle luck or by losing on turns, which is complicated by Isotropic's matching system that forces steady winners to play out of second player more often.  Games that are lost out of error in tactics or strategy do not effect that twenty-five percent, those losses subtract from the rest of the pie.

As to which is worse, variance or second player disadvantage?  Seems like a wash to me.  Both play a role regardless of skill differential.  In a close to evenly skilled match if the shuffles are equal the first player will usually win, and any subtle differences in builds reflect the skill differential.  A highly skilled player may win against a weaker opponent despite both bad shuffles and position, but if they do, again it is reflected in the seventy-five percent or so of games that it is possible to win.

This is the rub of Dominion.
« Last Edit: January 18, 2012, 07:34:51 am by Mean Mr Mustard »
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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #27 on: January 18, 2012, 07:46:23 am »
+1

In the long run the variance won't matter, but there will always be the second player disadvantage.
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Mean Mr Mustard

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #28 on: January 18, 2012, 07:47:52 am »
0

What do you mean?  Clarify?
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Empathy

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #29 on: January 18, 2012, 08:14:53 am »
0

Treasure map/wareouse is a typical opening that I aim for only as p2. The increased variance helps.
I don't get this. TM/Warehouse is such a hugh combo, you need very good reasons not to go for it.

I MIGHT agree on simple TMs on BMesque boards against a potentially better player, but with Warehouse support, 1st/2nd player shouldn't really play a big role in your decision making.

TM/warehouse is a good combo, but doesn't dominate any board its present on. I'd definitely open most ambassador openings rather than TM/warehouse on p1, but not on p2, though it depends on support for amb.

example: http://councilroom.com/game?game_id=game-20120116-065836-18fd44e7.html

That peddler deck my opponent played was *sweet*. Warehouse made sure that devel actually met a useful card to devel (rather than crappy copper).  I'm sure he could have made it explode T16 if it wasn't for me getting an early lead through TM luck, forcing him to find alternative green.

I still think my argument for 5/2 vs p1 advantage holds: if you are going to react to an opponent's 5/2 opening (changing your strategy because he got an unexpected HP/witch/mint), then so should you adapt to playing p2. The fact that any advantage gets "averaged out" after a high number of games doesn't change the fact that you can improve your play that way. And as Geronimoo mentioned, the p2 advantage will actually remain post-averaging if you play p2 more often than p1 (which is the case for a lot of us).
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WanderingWinder

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #30 on: January 18, 2012, 09:05:35 am »
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Treasure map/wareouse is a typical opening that I aim for only as p2. The increased variance helps.
I don't get this. TM/Warehouse is such a hugh combo, you need very good reasons not to go for it.

I MIGHT agree on simple TMs on BMesque boards against a potentially better player, but with Warehouse support, 1st/2nd player shouldn't really play a big role in your decision making.

TM/warehouse is a good combo, but doesn't dominate any board its present on. I'd definitely open most ambassador openings rather than TM/warehouse on p1, but not on p2, though it depends on support for amb.

Sure it does. Just because it doesn't dominate ALL boards it's on doesn't mean it's very very strong on many of them.
And yes, ambassador probably provides a reason to not go for it.

kn1tt3r

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #31 on: January 18, 2012, 09:12:07 am »
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I'm sure he could have made it explode T16 if it wasn't for me getting an early lead through TM luck, forcing him to find alternative green.
Turn 6 for a successful TM/Warehouse combo isn't extraordinary lucky.
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tlloyd

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #32 on: January 18, 2012, 09:35:31 am »
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I think that it is safe to say that even the best players, playing at top form, lose somewhere in the neighborhood of twenty-five percent of their games because of bad shuffle luck or by losing on turns, which is complicated by Isotropic's matching system that forces steady winners to play out of second player more often.  Games that are lost out of error in tactics or strategy do not effect that twenty-five percent, those losses subtract from the rest of the pie.

As to which is worse, variance or second player disadvantage?  Seems like a wash to me.  Both play a role regardless of skill differential.  In a close to evenly skilled match if the shuffles are equal the first player will usually win, and any subtle differences in builds reflect the skill differential.  A highly skilled player may win against a weaker opponent despite both bad shuffles and position, but if they do, again it is reflected in the seventy-five percent or so of games that it is possible to win.

This is the rub of Dominion.

But one of the points that has come out in our discussion so far is that first-player advantage (ignoring attacks for the moment) is essentially this: P1 is in a better position to benefit from relatively lucky shuffling, while P2 is more vulnerable to relatively unlucky shuffling. If P1 and P2 have similar shuffle luck, then P1 has no advantage when it comes to splitting key cards. I've mentioned before that attack cards can change this drastically, as can +buy to a lesser extent.
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WanderingWinder

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #33 on: January 18, 2012, 09:44:56 am »
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I think that it is safe to say that even the best players, playing at top form, lose somewhere in the neighborhood of twenty-five percent of their games because of bad shuffle luck or by losing on turns, which is complicated by Isotropic's matching system that forces steady winners to play out of second player more often.  Games that are lost out of error in tactics or strategy do not effect that twenty-five percent, those losses subtract from the rest of the pie.

As to which is worse, variance or second player disadvantage?  Seems like a wash to me.  Both play a role regardless of skill differential.  In a close to evenly skilled match if the shuffles are equal the first player will usually win, and any subtle differences in builds reflect the skill differential.  A highly skilled player may win against a weaker opponent despite both bad shuffles and position, but if they do, again it is reflected in the seventy-five percent or so of games that it is possible to win.

This is the rub of Dominion.

But one of the points that has come out in our discussion so far is that first-player advantage (ignoring attacks for the moment) is essentially this: P1 is in a better position to benefit from relatively lucky shuffling, while P2 is more vulnerable to relatively unlucky shuffling. If P1 and P2 have similar shuffle luck, then P1 has no advantage when it comes to splitting key cards. I've mentioned before that attack cards can change this drastically, as can +buy to a lesser extent.
Okay, but that means that there IS 1st player advantage, even without attacks, buys, etc........
Furthermore, if you're saying that it's easier for P1 to get lucky... isn't that less lucky somehow?

Mean Mr Mustard

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #34 on: January 18, 2012, 10:13:20 am »
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I think I am in over my head here, and should bow out.  Explain this, and I'll shove off: if a poor player has to get extremely lucky to beat an elite player, despite turn advantage, how exactly is variance getting cancelled out?
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rinkworks

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #35 on: January 18, 2012, 11:06:29 am »
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P1 is in a better position to benefit from relatively lucky shuffling, while P2 is more vulnerable to relatively unlucky shuffling. If P1 and P2 have similar shuffle luck, then P1 has no advantage when it comes to splitting key cards.

This is just not mathematically sound.  One player benefitting more from lucky shuffling and the other being more vulnerable to unlucky shuffling aren't effects that cancel each other out but in fact compound each other.

Let's assume that each player has an equal chance, each turn, of missing an opportunity to buy a key card.  We'll count the number of missed opportunities for each player over the course of, let's say, 8 turns.  (It doesn't matter how many, really.)  Let's say each player has a 20% chance of a turn being a missed opportunity.  (The exact percentage affects the magnitude of the resulting advantage but not whether or not there is one.)  Let's let M1 = the number of P1's missed opportunities.  Let M2 = the number of P2's missed opportunities.

If we simulate this, then the values of M1 and M2 we come up with will determine how the key card splits.  If M1 < M2, then P1 wins the split.  If M1 is equal to or one more than M2, then the split ties.  For P2 to win the split, M2 must be at most two less than M1.

The result of averaging this calculation over many simulations should be apparent, but just for fun I simulated this 10,000 times and got the following results:

* P1 wins the split: 3789 times
* Split is even: 4494 times
* P2 wins the split: 1717 times

In other words, P1 is more than twice as likely to win the split as P2 is.  Not that I recommend reading too much into these specific numbers!  Obviously I've made a lot of assumptions on how many turns it will take to exhaust the supply pile of the key card, and the likelihood of missing an opportunity to purchase one.  Moreover, the chance of missing any one opportunity is correlated with the chances of missing any of the others, because having a bad hand means possibly having a good hand next.  Still, you can finagle these numbers any way you want to:  ultimately, the odds MUST favor P1, simply because P1 can win the race after making more mistakes than P2 can make in order for him to.
« Last Edit: January 18, 2012, 12:02:45 pm by rinkworks »
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jomini

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #36 on: January 18, 2012, 11:12:06 am »
+1

Ttlyod:

Think of it this way for all cards in dominion except provinces there is an even number of cards. Say there is a dominant strat involving a dominant card, and each player has a 95% chance of acquiring that card (one copy) each turn. Statistical odds of P2 hitting that 5 times in a row are ~ 77.4%. Now what about P1? Well they also will hit 77.4% of the time on their first 5 runs, but they will also hit on their 6th shot 95% of the time. So of that 22.6% of games, 95% of those will also result in parity for P1 in other words P1 has  ~98.9% chance of reaching parity or better in the dominant strategy (I'm ignoring multiple misses as they are rather unlikely). P2 starts the game with a 21.5% disadvantage. Note - I'm using 95% odds which is remarkedly low variance (i.e. shuffle luck is close to non-existant) AND I'm treating each hand as independent (in reality if a player misses on one hand they generally are somewhat more likely to hit on the next one, which makes a P1 miss even less damaging relative to a P2 miss). Note that this scenario has no attacks, no +buys (or other card gain), no player interaction (aside from competition for scarce resources) and no preferential shuffling.

Now what about when there isn't a competition for the split on cards? Well say there is a card (like forge) where you want exactly one and where acquiring said card is the dominant strategy. Okay let's say we reach a parity point in the game where each player has a 75% chance of getting forge each turn (say with apothecaries & villages). Now P1 will hit it 75% of the time and 25% of the time P2 will not hit it. In 18.8% of games P1 has a solid lead. Now suppose P1 misses and P2 hits it, P2 doesn't get a solid lead unless P1 misses again. This will only occur in 4.7% of games. This leaves P1 with an advantage solely due to turn order in 14.1% of games.

There are just too many types of dominion games where it is a race to some critical threshold (e.g. 5) or some critical limited resource (e.g. curses, minions, cities, etc.), in either case there is a real P1 advantage. It gets tempered by shuffling (as we don't shuffle after every turn players have 2 or 3 turns to tie a race scenario), but shuffling dynamics also let P1 win out in a lot of interaction elements.

The long and the short of it is that the only scenarios where I want to be second player are those where I can adjust my play to interact with my opponent (e.g. going tunnel if he goes militia & going militia if he doesn't go tunnel). You can get rid of the +buys, the bonus card gains, all the attacks, and all the shuffle timing (e.g. by having a board where chancellor is dominant) and I will STILL take P1 even facing a tie-breaker penalty.
« Last Edit: January 18, 2012, 02:22:55 pm by jomini »
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tlloyd

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #37 on: January 18, 2012, 05:06:51 pm »
0

P1 is in a better position to benefit from relatively lucky shuffling, while P2 is more vulnerable to relatively unlucky shuffling. If P1 and P2 have similar shuffle luck, then P1 has no advantage when it comes to splitting key cards.

This is just not mathematically sound.  One player benefitting more from lucky shuffling and the other being more vulnerable to unlucky shuffling aren't effects that cancel each other out but in fact compound each other.

Let's assume that each player has an equal chance, each turn, of missing an opportunity to buy a key card.  We'll count the number of missed opportunities for each player over the course of, let's say, 8 turns.  (It doesn't matter how many, really.)  Let's say each player has a 20% chance of a turn being a missed opportunity.  (The exact percentage affects the magnitude of the resulting advantage but not whether or not there is one.)  Let's let M1 = the number of P1's missed opportunities.  Let M2 = the number of P2's missed opportunities.

If we simulate this, then the values of M1 and M2 we come up with will determine how the key card splits.  If M1 < M2, then P1 wins the split.  If M1 is equal to or one more than M2, then the split ties.  For P2 to win the split, M2 must be at most two less than M1.

The result of averaging this calculation over many simulations should be apparent, but just for fun I simulated this 10,000 times and got the following results:

* P1 wins the split: 3789 times
* Split is even: 4494 times
* P2 wins the split: 1717 times

In other words, P1 is more than twice as likely to win the split as P2 is.  Not that I recommend reading too much into these specific numbers!  Obviously I've made a lot of assumptions on how many turns it will take to exhaust the supply pile of the key card, and the likelihood of missing an opportunity to purchase one.  Moreover, the chance of missing any one opportunity is correlated with the chances of missing any of the others, because having a bad hand means possibly having a good hand next.  Still, you can finagle these numbers any way you want to:  ultimately, the odds MUST favor P1, simply because P1 can win the race after making more mistakes than P2 can make in order for him to.

That is a very impressive refutation of something I didn't say. My first comments in this discussion were to the effect that going first (aside from attack cards) isn't an advantage because the cards will split evenly. I was wrong, as has been explained, because my argument assumed similar shuffling between P1 and P2. Once you allow for different degrees of shuffle luck, you can see that P1 has a greater upside from shuffle luck while P2 has a greater downside. I never said these effects cancelled out. Claiming that they compound each other, however, is nonsensical -- the two effects are really just the two sides of the same coin. My point all along has been that P1's advantage is often overstated, since it really is dependent on differential shuffle luck. If P1 and P2 have similar shuffling outcomes, P1 will have no advantage in terms of splitting key cards. Given, however, that P1 and P2 are unlikely to have similar shuffling outcomes, on average P1 will be at an advantage, since P1 is more likely than P2 to benefit from better shuffle luck.
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Axxle

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #38 on: January 18, 2012, 07:10:19 pm »
0

P1 is in a better position to benefit from relatively lucky shuffling, while P2 is more vulnerable to relatively unlucky shuffling. If P1 and P2 have similar shuffle luck, then P1 has no advantage when it comes to splitting key cards.

This is just not mathematically sound.  One player benefitting more from lucky shuffling and the other being more vulnerable to unlucky shuffling aren't effects that cancel each other out but in fact compound each other.

Let's assume that each player has an equal chance, each turn, of missing an opportunity to buy a key card.  We'll count the number of missed opportunities for each player over the course of, let's say, 8 turns.  (It doesn't matter how many, really.)  Let's say each player has a 20% chance of a turn being a missed opportunity.  (The exact percentage affects the magnitude of the resulting advantage but not whether or not there is one.)  Let's let M1 = the number of P1's missed opportunities.  Let M2 = the number of P2's missed opportunities.

If we simulate this, then the values of M1 and M2 we come up with will determine how the key card splits.  If M1 < M2, then P1 wins the split.  If M1 is equal to or one more than M2, then the split ties.  For P2 to win the split, M2 must be at most two less than M1.

The result of averaging this calculation over many simulations should be apparent, but just for fun I simulated this 10,000 times and got the following results:

* P1 wins the split: 3789 times
* Split is even: 4494 times
* P2 wins the split: 1717 times

In other words, P1 is more than twice as likely to win the split as P2 is.  Not that I recommend reading too much into these specific numbers!  Obviously I've made a lot of assumptions on how many turns it will take to exhaust the supply pile of the key card, and the likelihood of missing an opportunity to purchase one.  Moreover, the chance of missing any one opportunity is correlated with the chances of missing any of the others, because having a bad hand means possibly having a good hand next.  Still, you can finagle these numbers any way you want to:  ultimately, the odds MUST favor P1, simply because P1 can win the race after making more mistakes than P2 can make in order for him to.

That is a very impressive refutation of something I didn't say. My first comments in this discussion were to the effect that going first (aside from attack cards) isn't an advantage because the cards will split evenly. I was wrong, as has been explained, because my argument assumed similar shuffling between P1 and P2. Once you allow for different degrees of shuffle luck, you can see that P1 has a greater upside from shuffle luck while P2 has a greater downside. I never said these effects cancelled out. Claiming that they compound each other, however, is nonsensical -- the two effects are really just the two sides of the same coin. My point all along has been that P1's advantage is often overstated, since it really is dependent on differential shuffle luck. If P1 and P2 have similar shuffling outcomes, P1 will have no advantage in terms of splitting key cards. Given, however, that P1 and P2 are unlikely to have similar shuffling outcomes, on average P1 will be at an advantage, since P1 is more likely than P2 to benefit from better shuffle luck.

edit: sorry, didn't read all of tlloyd's responses before this.

P1: Buys one key card
P2: Buys one key card
P1: Buys two key cards
P2: Buys two key cards
P1: Buys three key cards
P2: Only one key card left, buys it.

Even if both players get the exact same shuffle luck, at the very least P1 gets an advantage in splitting any key card if there is a way to obtain more than one in a turn.
« Last Edit: January 19, 2012, 05:50:52 pm by Axxle »
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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #39 on: January 18, 2012, 07:45:13 pm »
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(e.g. going tunnel if he goes militia & going militia if he doesn't go tunnel).
I saw this and just had to point out http://councilroom.com/game?game_id=game-20120105-165629-8d87aaab.html.  He went tunnel because i went militia, and he actually continued with tunnel.  I just didn't play militia, so he bought a blank card on turn 2 and turn 3.  Admittedly, I also had a semi-blank card and he had a 4 vp advantage, but I wanted to bring it up because it was a one in a thousand kind of game involving tactical counterplay. 
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Anon79

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #40 on: January 18, 2012, 09:20:56 pm »
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Would have been interesting if your opponent had gotten Lookout earlier, say by opening Lookout/Tunnel in response to Militia. Now Tunnel isn't a blank card.
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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #41 on: January 18, 2012, 09:50:46 pm »
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My point all along has been that P1's advantage is often overstated, since it really is dependent on differential shuffle luck.
But that's like saying that King's Court isn't that good because you need good actions to multiply with it. King's Court IS of course good, because usually those actions are out there. Just as P1's advantage is... well I'm not sure who you're saying is overstating it, because as far as I can tell, all anybody's claimed is that it exists and it isn't negligible... but P1's advantage exists and is, well, not negligible, because there quite often is differential shuffle luck.
And there's usually something else to compound it, too, whether attacks, or gains, or three piles, or plus buy, or a mega-turn...
So... is your point just that it's dependent on differential shuffle luck? What I'm describing as type I in the article certainly is, sans extra gains or buys, which I don't think anyone's disputing. What I'm describing as type II is not, as I demonstrate in the article.
If that's not your point... I'm not sure what your point actually is.

Empathy

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #42 on: January 18, 2012, 10:07:06 pm »
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I'm sure he could have made it explode T16 if it wasn't for me getting an early lead through TM luck, forcing him to find alternative green.
Turn 6 for a successful TM/Warehouse combo isn't extraordinary lucky.

Agreed, but I felt somewhat more on the lucky end of my warehouse draw outcomes/mandarin plays.

My point was more that, if given the choice between his or my deck, I would prefer his. I think the odds are pretty close, probably almost 50-50, but biased towards the peddler deck. I also think I would have lost the peddler split if I had try to compete with him playing as p2, and that my odds playing his deck as p2 were easily 40-60.

The argument that this p1 advantage averages out as you play more games is true, but that also holds for 5/2 split advantages. There is no reason for you to not play as many 5/2 openings on advantageous boards than your opponent, given enough games. Actually, it is *more* the case than p1/p2 outcomes, given that these are biased by your skill level.

That doesn't mean you shouldn't micro-manage each single game according to the outcome "I have 3/4 split and my opponent has 5/2 split". Similarly, the outcome "I play second" should be weighted in your decision process.

The only way to nullify this argument is if the optimal p1 and p2 strategy always coincide. I am a firm believer that they don't, at least on some select boards (maybe 20% of them?). These boards are typically ones that favor p1.

My favorite example is chapel when there is no militia. Chapel without attacks makes for *very* deterministic games, especially if there is a solid engine involved. chapel/grand market, chapel/KC and chapel/tournament are so heavily biased towards p1 that I struggle to find ways to counter those openings.

If my opponent goes chapel/tournament, I will definitely not follow suitl. The only way I can see myself saving that game is by getting a lucky province => baron/coppersmith.
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gamesou

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #43 on: January 19, 2012, 03:12:43 am »
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Not quite the same as the current topic, but I wouldn't be surprised if, in a game between a good player and a mediocre player, the mediocre player had a 2nd-player advantage. In the following sense: he can simply copy the optimal strategy and win by luck, and that may be the most likely way for him to win.
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jomini

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #44 on: January 19, 2012, 10:59:04 am »
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P1 is in a better position to benefit from relatively lucky shuffling, while P2 is more vulnerable to relatively unlucky shuffling. If P1 and P2 have similar shuffle luck, then P1 has no advantage when it comes to splitting key cards.

This is just not mathematically sound.  One player benefitting more from lucky shuffling and the other being more vulnerable to unlucky shuffling aren't effects that cancel each other out but in fact compound each other.

Let's assume that each player has an equal chance, each turn, of missing an opportunity to buy a key card.  We'll count the number of missed opportunities for each player over the course of, let's say, 8 turns.  (It doesn't matter how many, really.)  Let's say each player has a 20% chance of a turn being a missed opportunity.  (The exact percentage affects the magnitude of the resulting advantage but not whether or not there is one.)  Let's let M1 = the number of P1's missed opportunities.  Let M2 = the number of P2's missed opportunities.

If we simulate this, then the values of M1 and M2 we come up with will determine how the key card splits.  If M1 < M2, then P1 wins the split.  If M1 is equal to or one more than M2, then the split ties.  For P2 to win the split, M2 must be at most two less than M1.

The result of averaging this calculation over many simulations should be apparent, but just for fun I simulated this 10,000 times and got the following results:

* P1 wins the split: 3789 times
* Split is even: 4494 times
* P2 wins the split: 1717 times

In other words, P1 is more than twice as likely to win the split as P2 is.  Not that I recommend reading too much into these specific numbers!  Obviously I've made a lot of assumptions on how many turns it will take to exhaust the supply pile of the key card, and the likelihood of missing an opportunity to purchase one.  Moreover, the chance of missing any one opportunity is correlated with the chances of missing any of the others, because having a bad hand means possibly having a good hand next.  Still, you can finagle these numbers any way you want to:  ultimately, the odds MUST favor P1, simply because P1 can win the race after making more mistakes than P2 can make in order for him to.

That is a very impressive refutation of something I didn't say. My first comments in this discussion were to the effect that going first (aside from attack cards) isn't an advantage because the cards will split evenly. I was wrong, as has been explained, because my argument assumed similar shuffling between P1 and P2. Once you allow for different degrees of shuffle luck, you can see that P1 has a greater upside from shuffle luck while P2 has a greater downside. I never said these effects cancelled out. Claiming that they compound each other, however, is nonsensical -- the two effects are really just the two sides of the same coin. My point all along has been that P1's advantage is often overstated, since it really is dependent on differential shuffle luck. If P1 and P2 have similar shuffling outcomes, P1 will have no advantage in terms of splitting key cards. Given, however, that P1 and P2 are unlikely to have similar shuffling outcomes, on average P1 will be at an advantage, since P1 is more likely than P2 to benefit from better shuffle luck.

You are still wrong. As I showed, P1, solely by dint of turn position also has an advantage in a race - first to get a card that dominates. Even when you aren't fighting to split cards (say in in forge/KC x2 /Monument x3  game - nothing will pile), you still are fighting to be the first one to a forge & the first one to start scoring megapoints regularly. This happens whenever you have a key card (like goons, golem, possession, forge, witch, followers, mint, KC, etc.) that greatly benefits playing it first but has high cost.

Further, you are mistaking what "shuffle luck" actually entails here. When we do these probabilistic analyses we are giving the players the EXACT SAME distribution of luck. The distribution of hands in my analysis is IDENTICAL for both players. P1 is just a likely as P2 to luck out and get the good cards on any given hand, the ONLY difference is he gets to play an extra hand a statistically significant percent of the time.

This as actually an unrealistic assumption. Once a player breaks parity and pulls ahead, they tend to have better odds of getting whatever it is they want (key kingdom card, province, etc.). So take a minion race. Both open chancellor/silver (everything else is crap or too pricey). If P1 enters T3 (newly shuffled from chancellor) with 2 minions and P2 enters T3 with 1 minion, then P1 has better odds of acquiring a minion on T3 (if he has to minion for 4 he has better odds of hitting 5 coin the next 4 card hand or trying his luck again). When the card you buy makes buying the next card you want more likely, we witness a (normally slight) compounding of odds.

This is not to say that the P1 advantage is overwhelming, but that it will account for a good percentage of the odds of winning a game.
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DStu

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #45 on: January 19, 2012, 11:14:25 am »
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@jomini:
The point is you are talking about something else the he is:

He said: Given both player at the current game get "equally lucky" (in the sense of: missing their target equally often), then, given no +buy (which I think was not mentioned in this posts but was explicitly mentioned in all posts before, so we can savely assume that this assumption is still valid), player 1 has no advantage in this split. But p1 profits more or earlier for a lucky outcome than p2 does.
Which is exactly what's happening.

You say that obviously both player have the same distribution of luck, which is of course save to assume.  But that's not what he is talking about. He is not talking about what's happening when the random variable $LUCK has the same distribution for both players, but what is happening when the actual realization of $LUCK is the same (or different) for both players. To understand in which of this cases p1 has an advantage over p2, and in which not.
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rrenaud

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #46 on: January 19, 2012, 11:38:57 am »
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It's not that interesting to consider only games in which p1 and p2 have the same exact "luck"  outcome, because that will almost never happen.  This is setting aside the matter of measuring luck given divergent strategies.

It's like saying.  First, let's restrict our statement to the less than 1% of games in which the sequence of multiple random outcomes is the same for both players, then X.  Sure, I'll agree with you on that.  But it says very little because it rarely applies.

OTOH, considering comparable levels of luck is interesting.  The 6-4 split when p1 gets a little luckier than p2 getting matched to a 5-5 split when p2 is a little luckier than p1 is indeed insightful.
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DStu

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #47 on: January 19, 2012, 11:53:35 am »
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I anyway think that is getting a bit offtopic:
We started with the thesis that for racing for a pile a firstplayer advantage does not exists, because ... I don't know anymore .... We explained to whomever that this is wrong, because if both player does not have the same shuffleluck, than there really is an advantage. Then whoever realized what was the mistake, namely that his analysis only holds in the case with equal shuffleluck, and wrote a post where he for my understanding explained exactly that. Then there was a post that critics exactly that you may not only consider the cases with equal shuffleluck, but whatever.

I think that is clear to everybody since at least yesterday, leading to my last post, which obviously did exactly the opposite of what it should do. m(
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Empathy

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #48 on: January 19, 2012, 11:58:05 am »
+1

How about the following toy model:

Take two Gaussian r.v. for the outcome of "luck". All you affect is therefore the mean and variance of your "luck"/deck.

Now every turn, you draw a random sample from your Gaussian, and add it up. Whoever gets first to some threshold number (say 40ish points) wins.

It makes sense that, to minimize the average number of turns to attain the threshold, you simply maximize the mean of your deck. To *win* however, you need to do something more subtle: maximize the probability of getting there before your opponent.

Now it's pretty clear that the first player has an edge in games where the variance of the deck is small: at the limit where sigma ->0, he always wins! This advantage decreases drastically as sigma becomes large.

Now, I am sure someone can come up with a pair of 2 means and variances such that p1 goes for a high mean, low variance deck, and p2 increases his odds by going with a lower mean, higher variance.

Actually, just take mean = 39, variance = 0, mean = 0 , variance = 100000000000000.

p1 wins deterministically on T2, but p2 has 1/2 of winning on t1. Best odds p2 can hope to get in this toy model.

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jomini

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Re: Reasons for the 1st-player vs 2nd-player advantage
« Reply #49 on: January 19, 2012, 12:12:08 pm »
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Dtsu:

As I noted very early on, parity rules only apply in a deterministic setting. E.g. where each player gets equivalent hands at all times. By this I mean if P1 and P2 will both always hit or miss at the same time then parity is maintained and there can be no other way to split cards than 5:5. However if P1 and P2 merely hit or miss at the same rate, then P1 has an advantage.

The first condition is generally not applicable, players will have differing hands from time to time. The second condition is a good first order approximation. P1 is not getting any luckier than P2, he has no deviation from expected hand performance, the "luck" component is indeed identical. P1 simply has an extra turn.

Now in a lot of games, P1 has additional advantages that WW was pointing out (like being able to witch on T3 before P2 shuffles) and in some games he has fewer (e.g. smugglers with something like IGG). Nonetheless, P1 retains an inherent advantage in winning the split on limited cards and in winning the race to a key card thanks to his extra turn (or half turn if you want an expectation value). In some very simple games (like big money/smithy) this is mitigated by P2 winning ties ... but in a lot of games (those involving card gainers, +buys, interactive cards, those involving dominant high value cards, etc.) P1's advantage is real and explains a good bit of win distribution in matches.
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