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Default summoning cost

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questccg
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I am currently using a "default amount" of two (2) chips to summon a monster.

The value is FIXED because the monster cards are FACE DOWN (hidden from the opponent). You can summon many monsters, for example summoning four (4) requires eight (8) chips.

My initial intention was to be able to specify a COST for summoning each creature (some 1, others 2, maybe 3, etc.) But the problem is by doing so you REVEAL part of content of the hidden card (not exactly - but partially).

So currently (with the fixed amount), I am able to include bluffing and intimidation since the content of the cards remains hidden.

I wish there was a way to use variable summoning costs without the partial revealing of your hidden cards.

Yamahako
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When a monster is revealed,

When a monster is revealed, perhaps you have to pay more chips if its more powerful, or else get a refund if its less powerful.

ie. Low Power monster- played face down for 2 chips, when its revealed to the opponent, you receive 1 chip back

High power creature - played face down for 2 chips, when its revealed to the opponent, you have to pay 2 additional chips or discard the monster.

JHouse
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Agree

I agree with Yamahako in paying for the total costs when the cards are revealed. A further step that you can take is that a player must pay at least 2 chips for each monster played, but are able pay more (the chips paid can simply be placed on top of the card). This will allow players to bluff by overpaying for weaker monsters or underpaying for stronger monsters.

questccg
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Let me see if I understand

So are you saying that "by default" a player would play 2 chips for each monster of that type.

So let's say I have a weaker monster (with a summon cost of 1).

I summon four (4) of these monsters (so a total of 8 chips in the pot). Then my opponent can either call or raise. Assuming that he "calls". I would then REMOVE 4 chips from the pot because it is a weaker monster?

Am I understanding correctly???

Doesn't that remove the "bluffing" of playing a weaker card. You get back money you should have lost in the battle - because you played a weaker monster.

Yamahako
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questccg wrote:So are you

questccg wrote:
So are you saying that "by default" a player would play 2 chips for each monster of that type.

So let's say I have a weaker monster (with a summon cost of 1).

I summon four (4) of these monsters (so a total of 8 chips in the pot). Then my opponent can either call or raise. Assuming that he "calls". I would then REMOVE 4 chips from the pot because it is a weaker monster?

Am I understanding correctly???

Doesn't that remove the "bluffing" of playing a weaker card. You get back money you should have lost in the battle - because you played a weaker monster.

Well those other aspects of the game weren't explained in the original post...

I don't know how calling and raising work in this context.

Does the opponent raise or call by playing monsters of his own? Or is he betting against the success or failure of your monsters against a random threat? Need more information to help better on this I think.

questccg
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Sorry about that I will rectify.

Yamahako wrote:

Well those other aspects of the game weren't explained in the original post...

Sorry about that I will rectify.

Quote:
I don't know how calling and raising work in this context.

How it works is a player A will play a card face DOWN (hidden), he will then SUMMON a QUANTITY of monsters of that type. So let's say he has a dragon (type) and summons four (4) dragons. That would mean he needs to put eight (8) chips into the pot.

Quote:
Does the opponent raise or call by playing monsters of his own? Or is he betting against the success or failure of your monsters against a random threat? Need more information to help better on this I think.

Now, by default, the opponent is on the defensive. He may "call" by summoning four (4) of his own monsters OR he may summon MORE. By summoning more, he becomes the aggressor (so the attacker). When a player "calls", he becomes the defender. So Player B can then raise by putting into play six (6) unicorns (he puts 12 chips into the pot). During this time, neither player knows their opponent's card... All they know is how many monsters the other is summoning for the battle.

Once a player "calls" or "folds", the monsters are revealed (and then a formula is used to determine the winner - who will collect the entire amount of the pot).

So player A can then play two (2) more dragons to call or "re-raise".

I think your original idea would work if there was only one (1) monster in play for each player. In my context you can summon multiple monsters of the same type.

Yamahako
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It could still work. It

It could still work. It would work even well for bluffing (in favor of the bluffer).

The question is whether or not all Dragons are the same (let's assume they aren't)

I have 2 (1) cost Dragons, and 3 (2) cost dragons, and 1 (3) cost dragon.

I play 4 cards down, and my 8 chips. I played 2 (1) cost Dragons, and 1 (3) cost Dragon) and 1 (2) cost Dragon.

My opponent Raises me by 1 (playing 5 unicorns), so now he has 10 chips in the pot, and I have 8.

I Re-raise by playing my other 2 (2) cost dragons, leaving me with 12 chips in the pot.

My opponent plays another dragon to call me.

If either of us were to fold, the cards wouldn't be revealed and the last person in get the whole pot (with no cost reductions), and I would have underpaid for higher cost dragons, and overpaid for lower cost dragons. But we didn't.

My opponent had 1 (1) cost Unicorn, and 3 (2) cost Unicorns, and 2 (3) cost Unicorns. He has to put in 1 more chip [2 (3) costs, but 1 (1) cost, nets to +1] {If he couldn't pay he would have to discard one of the 3 cost Dragons}. I remove 1 chip from the pot, since I'm over invested. Leaving a 2 chip spread. Let's assume that his higher cost dragons are more powerful than mine, he then wins the pot.

If I want to bluff, I can play a lot of lower cost dragons to try and scare someone away from investing. Or if I want to downplay a strong hand, I can keep calling with powerful dragons (that I will have to invest extra money into the pot).

I don't know that its a perfect system, however I think it has enough interesting decision trees to be worth testing - if what you really want is a hidden variable cost mechanic.

questccg wrote:
Yamahako wrote:

Well those other aspects of the game weren't explained in the original post...

Sorry about that I will rectify.

Quote:
I don't know how calling and raising work in this context.

How it works is a player A will play a card face DOWN (hidden), he will then SUMMON a QUANTITY of monsters of that type. So let's say he has a dragon (type) and summons four (4) dragons. That would mean he needs to put eight (8) chips into the pot.

Quote:
Does the opponent raise or call by playing monsters of his own? Or is he betting against the success or failure of your monsters against a random threat? Need more information to help better on this I think.

Now, by default, the opponent is on the defensive. He may "call" by summoning four (4) of his own monsters OR he may summon MORE. By summoning more, he becomes the aggressor (so the attacker). When a player "calls", he becomes the defender. So Player B can then raise by putting into play six (6) unicorns (he puts 12 chips into the pot). During this time, neither player knows their opponent's card... All they know is how many monsters the other is summoning for the battle.

Once a player "calls" or "folds", the monsters are revealed (and then a formula is used to determine the winner - who will collect the entire amount of the pot).

So player A can then play two (2) more dragons to call or "re-raise".

I think your original idea would work if there was only one (1) monster in play for each player. In my context you can summon multiple monsters of the same type.

questccg
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Can you give another example...

Yamahako wrote:
The question is whether or not all Dragons are the same (let's assume they aren't)

You can only have one (1) dragon card (which is the creature type). Each 2 chips you put in is an identical dragon. So all dragons are the SAME. This is for easier management, one card is more than plenty...

Can you do an example with the cards being the SAME?

Yamahako
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questccg wrote:Yamahako

questccg wrote:
Yamahako wrote:
The question is whether or not all Dragons are the same (let's assume they aren't)

You can only have one (1) dragon card (which is the creature type). Each 2 chips you put in is an identical dragon. So all dragons are the SAME. This is for easier management, one card is more than plenty...

Can you do an example with the cards being the SAME?

Ahh ok I understand now.

So say there are 3 types of creatures. Dragon which costs 3, Unicorn which costs 2, and Squirrel which costs 1.

If I understand correctly, I play a card - and then pay 2 chips per "copy" of that creature I'm committing to the battle. This makes me the aggressor. I'm assuming being the "aggressor" changes the battle formula in some way - though if it doesn't that's fine.

My opponent then plays a card (might be the same, might be different, these are face down so we don't know), and then pays 2 chips per "copy" of that creature as well. If he out pays me for creatures then he becomes the "aggressor". Right?

Ok so how you could do a variable cost mechanic for that. When you play the original card face down, you put chips on it. These chips pay for the monster - but not the copies. You can over (but not under) pay for the monster. Maybe make it so you have to put 1 chip on the monster, but can pay up to 'X' chips. This means you can bluff a cost 1 monster is a cost 3 monster. These chips are lost to the pot if you fold similar to ante in a poker game. Additional monsters still all cost 2 chips (the extra cost would be absorbed in the initial card play). You could then balance out that first card - really powerful card might cost 5, really weak ones might cost 1, and then 2 per monster you summon.

Would that work?

questccg
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Wait a sec... I think I understand...

Yamahako wrote:
So say there are 3 types of creatures. Dragon which costs 3, Unicorn which costs 2, and Squirrel which costs 1.

If I understand correctly, I play a card - and then pay 2 chips per "copy" of that creature I'm committing to the battle. This makes me the aggressor. I'm assuming being the "aggressor" changes the battle formula in some way - though if it doesn't that's fine.

My opponent then plays a card (might be the same, might be different, these are face down so we don't know), and then pays 2 chips per "copy" of that creature as well. If he out pays me for creatures then he becomes the "aggressor". Right?

Yes, exactly correct.

So what you mean is that the variable cost is ONLY on the card type - not each "copy"?! Did I understand properly? So you put in for each copy (2 chips each) PLUS a variable cost which can be bluffed...

I think the only thing is that you would never want to overpay (say you have a Dragon cost = 3, you never play 5, as an example) because that would lead you opponent to believe that you have a real strong creature - and therefore he would only oppose this creature with one that is very strong... However you can still bluff - but when a battle occurs you will probably lose.

Is my understanding correct???

Yamahako
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questccg wrote:Yamahako

questccg wrote:
Yamahako wrote:
So say there are 3 types of creatures. Dragon which costs 3, Unicorn which costs 2, and Squirrel which costs 1.

If I understand correctly, I play a card - and then pay 2 chips per "copy" of that creature I'm committing to the battle. This makes me the aggressor. I'm assuming being the "aggressor" changes the battle formula in some way - though if it doesn't that's fine.

My opponent then plays a card (might be the same, might be different, these are face down so we don't know), and then pays 2 chips per "copy" of that creature as well. If he out pays me for creatures then he becomes the "aggressor". Right?

Yes, exactly correct.

So what you mean is that the variable cost is ONLY on the card type - not each "copy"?! Did I understand properly? So you put in for each copy (2 chips each) PLUS a variable cost which can be bluffed...

I think the only thing is that you would never want to overpay (say you have a Dragon cost = 3, you never play 5, as an example) because that would lead you opponent to believe that you have a real strong creature - and therefore he would only oppose this creature with one that is very strong... However you can still bluff - but when a battle occurs you will probably lose.

Is my understanding correct???

Yup that's kind of what I was suggesting. You would want to overpay to bluff. You never want to bet up a weaker hand unless it's to bluff. I don't know how the calculations run to determine the winner - if it's a strictly "my number is bigger, I win" then you're right this won't be that simple. But if there's some interactivity in which monsters beat other monsters - then the bluffing mechanic could be interesting (overpaying for a weaker creature to get the opponent to fold or whatever).

genericm
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Not the money

What if you only put out 1 token for each monster summoned, this token represents the creature not the money to summon him. then resolve the money (or whatever) after the battle is revealed.

I put my creature down and 4 tokens on the card to summon 4 creatures
You put your card down and 3 tokens on the card to summon 3 creatures

I reveal a card with a value of 2, and I then put out 8 money into the Winners Pot
You reveal your card with a value of 3, and then put out 9 money into the Winners Pot

You win with 9 and claim the winners pot.

The tokens represent creatures not money.

questccg
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What do you think?

Yamahako wrote:
Yup that's kind of what I was suggesting. You would want to overpay to bluff. You never want to bet up a weaker hand unless it's to bluff.

Okay so I understood correctly. That could be interesting.

Quote:
I don't know how the calculations run to determine the winner - if it's a strictly "my number is bigger, I win" then you're right this won't be that simple.

Someone else suggest I look at an RPS (Rock-Paper-Scissors) system to resolve conflicts. I have been thinking about the formula... long and hard. In THEORY it works, BUT in practice I have a card which is defined as POWER. The POWER attribute (1-9) determines how many creatures can be summoned.

The problem is that when a player has a higher POWER, he can ALWAYS outbid the opponent to win the hand (intimidation). I DON'T like that...

And then there is having a creature with small STRENGTH opposing a stronger creature (one having STR = 3 and the other STR = 6). In such a case, you would require TWICE as many creatures to win.

So I THINK it has to be a comparison on one attribute: strength (STR). Multiply the number of creatures by strength. The higher of the two players is the victor.

The older formula had merit... but I don't think it will work in practice.

What do you think?

genericm
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One more thing to

One more thing to consider...

If it always costs me something to fight then I can be harassed by a weaker opponent and lose over the long run.

This would mean attacking with a weaker creature could have strategic value.

If I can make you lose something valuable while I lose something less valuable then I (numerically) have won even though I have lost.

(Also, yes, I think simplicity has great value, using only strength to resolve battle is the way to go)

questccg
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Divided into three (3) territories

Yamahako wrote:
But if there's some interactivity in which monsters beat other monsters - then the bluffing mechanic could be interesting (overpaying for a weaker creature to get the opponent to fold or whatever).

Well how the game is divided is into three (3) playing fields. Each field is sorta a "territory dispute" with its own pot. When a player lays down an original bet + summoning cost, he is sorta saying: "This is MY territory". As such he wins - by default - all antes to that pot (each turn you pay antes to all three (3) pots - so 3 chips).

It is not until the opponent counters, does a battle insue and therefore he can no longer collect the ante for that territory (or pot).

Yamahako
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questccg wrote:Yamahako

questccg wrote:
Yamahako wrote:
But if there's some interactivity in which monsters beat other monsters - then the bluffing mechanic could be interesting (overpaying for a weaker creature to get the opponent to fold or whatever).

Well how the game is divided is into three (3) playing fields. Each field is sorta a "territory dispute" with its own pot. When a player lays down an original bet + summoning cost, he is sorta saying: "This is MY territory". As such he wins - by default - all antes to that pot (each turn you pay antes to all three (3) pots - so 3 chips).

It is not until the opponent counters, does a battle insue and therefore he can no longer collect the ante for that territory (or pot).

I thought of another way to do it. You could pay any amount of chips. The hidden card has a cost per monster, but you don't have to say how many monsters you're summoning. So I could have played a card that costs 3 per monster, and put 12 chips down, which would be 4 monsters. You may play a card with monsters costing 2, so you'd end up with 6 monsters. So then it would just be my 4 monsters vs. your 6 monsters. You could over pay to call, which would leave you with a partial payment - which could be as a bluff or whatever. For example, you could play a 2 chip monster, and play 15 chips, which would make it look like either 5 3 chip monsters, or 3 5 chip monsters instead of 7 2 chip monsters. The extra chip would be lost of course if you lost or folded.

Rock Paper Scissors is a good way to handle things - but maybe it's Rock is worth twice as much power against scissors,etc. You will need a system to handle ties if you don't already have one.

questccg
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Funny... I was doing math and...

Yamahako wrote:
I thought of another way to do it. You could pay any amount of chips. The hidden card has a cost per monster, but you don't have to say how many monsters you're summoning. So I could have played a card that costs 3 per monster, and put 12 chips down, which would be 4 monsters. You may play a card with monsters costing 2, so you'd end up with 6 monsters. So then it would just be my 4 monsters vs. your 6 monsters. You could over pay to call, which would leave you with a partial payment - which could be as a bluff or whatever. For example, you could play a 2 chip monster, and play 15 chips, which would make it look like either 5 3 chip monsters, or 3 5 chip monsters instead of 7 2 chip monsters. The extra chip would be lost of course if you lost or folded.

I was doing some math to figure out the value of creatures/armies. The thing is when you max out (9 STR with 3 QTY or 3 STR with 9 QTY) you can calculate the odds of winning. So "27" can beat how many hands (becomes the question). Same goes with any other hand - you do the math and get a value - from there you can pretty much figure out your odds of winning. However like I said, it does not make much sense to bet LESS than your maximum (because that will always be your BEST odds of winning). It does not guarantee you win - but it maximizes your odds...

So I still think there is some imperfection.

So the question is - how do we remove the condition that maximizing your army is the BEST way to go???

questccg
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Important note...

questccg wrote:
However like I said, it does not make much sense to bet LESS than your maximum (because that will always be your BEST odds of winning). It does not guarantee you win - but it maximizes your odds...

It does not mean you bet the MAX at the beginning... No you can bluff with less - to trick the other player in thinking you have less.

BUT it is when it comes to raising (the other player is willing to combat) that you would always want to MAX out... Sorta like - okay, now I'm all in...

questccg
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Clarification...

questccg wrote:
BUT it is when it comes to raising (the other player is willing to combat) that you would always want to MAX out... Sorta like - okay, now I'm all in...

By MAX out, I don't mean play all your chips. What I mean is play as MANY of the creature you are allowed. So if your POWER (PWR) = 6, you should raise up to 6 (if possible for that creature). That's what I mean by - "okay, now I'm all in". You might have only had 2 creatures before your opponent decides to do battle.

questccg
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It came to me in a dream...

questccg wrote:
So the question is - how do we remove the condition that maximizing your army is the BEST way to go???

I was up during the night and I came up with the following idea: your first three creatures cost 1 chip to summon, your next three cost 2 chips to summon and your last three cost 3 chips to summon.

So basically if you want to MAX your army, it's going to cost you 18 (1+1+1+2+2+2+3+3+3 = 18) chips! Game-wise that's a lot. It is equal to three turns of WEALTH = 9 (less antes = 3, so 3 x 6 chips = 18 chips). If your WEALTH = 6 (average), MAXing your army is worth 6 turns!

That means that you would not MAX your army each turn (too costly).

questccg
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Last minute update

questccg wrote:
I was up during the night and I came up with the following idea: your first three creatures cost 1 chip to summon, your next three cost 2 chips to summon and your last three cost 3 chips to summon.

This will allow the pot to be filled with the appropriate chips. BUT you can still CHOOSE to FOLD or CALL. If you call, you must TURN OVER your card and PAY the COST (your opponent does the same). This will reduce the amount of additional chips that need to be added to the pot... But it will allow for summoning costs to vary PER CREATURE.

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