## Actually infinite creatures

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billyh
Joined Feb 2006
404 Posts
I'm attempting to construct a scenario where a player gets an actually infinite number of creatures in a game of magic. (Amazingly, in Mirrodin Block only). I understand that there are several methods of attaining an arbitrarily large number of creatures, but because of the loop rules of magic these only produce finite numbers of creatures.

A caveat:
It might take an infinite amount of time to physically generate these creatures, but the fact remains that with high probability an infinite number of creatures are created.

Bob and Alice are playing a game of Magic. Bob has five life, a tapped Spikeshot Goblin, a Summoning Station, a Leonin Elder, and some tapped Mountains.

It's the Alice's turn (pre-combat main phase), and she has one life. She has a Wirefly Hive, a Gilded Lotus, a Voltaic Construct, and a March of the Machines.

Alice can use her Gilded Lotus, Voltaic Construct, and March of the Machines to produce an arbitrary amount of mana. She wants to create enough Wirefly tokens to fly over Bob's defenses and deal lethal damage to him. If she is unable to produce enough Wirefly tokens, Bob will be able to untap and shoot her with the Spikeshot Goblin.

If Alice gets lucky, she can win 5 flips in a row with her Wirefly Hive, which gives Bob 5 life from his Leonin Elder, and then fly over to kill him with her 5 2/2 Wirefly tokens. If Alice isn't quite as lucky and loses a flip after at least one Wirefly token is created, she gives Bob some life and then must be even more lucky to kill him.

Every time a Wirefly token leaves play, Bob will use his Summoning Station to create a colorless (and non-artifact) Pincher token. Bob is wary of Alice's tricks and wants as many creatures as possible, just in case.

With a high probability, no matter how many times Alice uses the Wirefly Hive, she will never create enough creatures to kill Bob. However, at any point in the process, she is not stalling, since she always has some (very small) chance of generating enough Wirefly tokens to win. Similarly, my understanding of magic loop rules leads me to believe that this scenario falls outside of them.

Bob creates a Pincher Token every time a Wirefly token goes to the graveyard. With high probability, Alice will continue creating Wirefly tokens forever, and Bob will have an infinite number of Pincher tokens (again, with the caveat that it will take Bob an infinite amount of time to generate these tokens).

I mentioned this loop in another post concerning infinity.

Does this work?

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cyphern
Joined Jan 2003
19354 Posts
If you were to repeat that process indefinitely, alice would eventually get the tokens she needs. The probability of her getting enough wireflies is non-zero, and thus will eventually happen. Even if it takes so many iterations that Knuth's up-arrow notation becomes combersome, you are no closer to infinity than if she got it after only 5 activations.

PS, you are correct that this scenario could not be shortened into a loop. And would need to be handled step by step.
billyh
Joined Feb 2006
404 Posts
If you were to repeat that process indefinitely, alice would eventually get the tokens she needs. The probability of her getting enough wireflies is non-zero, and thus will eventually happen. Even if it takes so many iterations that Knuth's up-arrow notation becomes combersome, you are no closer to infinity than if she got it after only 5 activations.

PS, you are correct that this scenario could not be shortened into a loop. And would need to be handled step by step.

Some infinite series converge to infinity, and some infinite sequences of actions have a probability of 1 of converging. This is not one of them. That is to say, even though Alice has an infinite number of chances to win, the chance of winning drops so fast that the probability that Alice will eventually win is NOT 1.

To give an easier example, suppose Alice is going through a loop that has a 1/3 chance of winning after one iteration, a 1/9th chance after 2, a 1/27th chance after 3, etc.

The overall chance that Alice will win is 1/3 + 1/9 + 1/27... If we sum this infinite series, the answer is 0.5. Even after an infinite number of attempts, Alice will have a 50% chance of winning, despite always having a chance to win.

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cyphern
Joined Jan 2003
19354 Posts
Some infinite series converge to infinity, and some infinite sequences of actions have a probability of 1 of converging. This is not one of them. That is to say, even though Alice has an infinite number of chances to win, the chance of winning drops so fast that the probability that Alice will eventually win is NOT 1.

You're right, I take it back.

It would still require an infinite amount of time to get an infinite amount of tokens (which was your goal with creating a situation like this). In a practical game, you would be limited by the duration of the round (if you are in a tournament), or the attention spans of alice and bob (if playing casually with sane people), or the lifespans of alice and bob (if they are insane), or the duration until the universe ends (if you built self-replicating robots with the ability to flee from a dieing sun to finish out the game for you, and assuming of course that the universe actually has an end).
condor
Joined Mar 2001
818 Posts
I'm attempting to construct a scenario where a player gets an actually infinite number of creatures in a game of magic.

All you did was create a situation where you could claim it isn't pointless to endlessly repeat a sequence of actions. What you don't seem to realize is that no rule of Magic says you can't do that anyway.

All the so-called "loop rules" do, is define an ending point to such a sequence, that will always be the same for a given setup. When you finish one set, you could do it again within the rules of Magic. But if the ending point is the same (or: predictable in advance), a judge can cite you for wasting time, since you knew that the result would be the same (or: you knew how to get to that predictable result by "naming a number").

What you didn't do, was actually create the time in which to get all those tokens. Whether your attempts stop because a judge tells you to stop stalling, or the tournament ends, or your friends all go home, or the world comes to and end - the results are the same. You will not have achieved an infinity of tokens.

Does this work?

Depends on what you mean. Does it allow Alice to draw the game, instead of lose? Yes. Does it get her infinite tokens, or somehow "prove" that Magic needs to recognize the concept? No.
billyh
Joined Feb 2006
404 Posts
How is a judge supposed to handle this? My understanding is that this example invokes neither the stalling nor the loop rules.

This situation is unlike any other loop I've seen in magic.

Consider:
1) Normal "arbitrarily large" loops, such as SquirrelCraft. These are well handled under the loop rules. You can't continue to create tokens.

2) Loops based on random conditions that will eventually be satisfied with probability 1. For example shuffling your deck and looking at the top three cards until you get the exact cards that you want. My understanding is that you must play out the scenario until the exact cards are found, and running the loop until that point is not stalling. Even in this case, the loop will eventually stop and you cannot get "infinite" iterations without stalling or breaking the loop rules.

3) This type of loop, which I contend is different than 1) or 2)

Can you construct a different loop (say, without Wirefly Hive) that with high probability will never stop, where neither player is stalling (and doesn't run afoul of the loop rules?).

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condor
Joined Mar 2001
818 Posts
How is a judge supposed to handle this? My understanding is that this example invokes neither the stalling nor the loop rules.

Correct. Which is why I said it was a draw. But without infinite time, you do not get infinite itteration of the loop. You can only get a finite number.

1) Normal "arbitrarily large" loops, such as SquirrelCraft. These are well handled under the loop rules. You can't continue to create tokens.

It isn't the rules of Magic that accomplish that. It is the fact that you can get any finite number you want by simply "naming" the number. So not naming a number is stalling.

Your situation is different - there is a point, albeit an increasingly hopeless one, in continuing. Just liek there is a point in continuing a Gaea's Blessing loop. Since it is random, you can't invoke the loop rules. But it is still finite. Why are you assuming otherwise?

3) This type of loop, which I contend is different than 1) or 2)

It is no different than 2).
Can you construct a different loop (say, without Wirefly Hive) that with high probability will never stop, where neither player is stalling (and doesn't run afoul of the loop rules?).

What would be the point? YOU DON'T HAVE INFINITE TIME.
Hayama
Joined Nov 2002
67 Posts
Based upon your original post, what is giving the tokens Haste, so they could then fly over & hit Bob?
diazona
Joined Dec 2007
82 Posts
Good point ;) Also Bob's Pincher tokens would have to have flying to block.

I couldn't resist trying to figure it out... so assuming just for the sake of argument that Alice's tokens have haste and that Bob's tokens have flying, I think the odds are against Alice. The rough calculation I did shows Alice has less than a 4% chance of winning (although the operative word there is rough) . . . most of which comes from the 1/32 chance of winning the first 5 flips.
billyh
Joined Feb 2006
404 Posts
First of all, I forgot to put Mass Hysteria (also in Mirrodin Block) in my example. That fixes the haste issue. The Pincher Tokens cannot block the Wirefly Tokens (as they don't have flying).

C

Your situation is different - there is a point, albeit an increasingly hopeless one, in continuing. Just liek there is a point in continuing a Gaea's Blessing loop. Since it is random, you can't invoke the loop rules. But it is still finite. Why are you assuming otherwise?

It is no different than 2).

What would be the point? YOU DON'T HAVE INFINITE TIME.

This loop I've described and any loop involving the ordering of cards are fundamentally different.

Suppose I have the ability to run a loop where I can (as many times as I wish) shuffle my cards and look at the top 3. I want to run this loop until I get the exact cards I want on top of my deck, in the exact order I wish.

After any shuffle, I have a low probability of getting the exact cards I wish. After enough shuffles, with probability approaching 1, I will get exactly the cards I wished. Suppose that some other card allowed me to generate a creature every time my deck was shuffled. At some point, I will get the exact cards I want and I will have a very large, but finite number of creatures. The probability that I will eventually get the exact cards is 1.

Now consider the situation described. In this case, the probability that the loop ends is not 1. It's actually around 1/30. [Anecdotally, I asked a (currently level 4) judge this problem and off-the-cuff he said he'd roll a 30-sided die to determine who would win... turns out he was pretty close].

Suppose a judge in the shuffle/look case let the player stack the three cards instead of actually simulating each shuffle (I know, this isn't the current rules, but members of this message board have stated that they felt that the rules should allow this). If the judge allowed a similar simulation to occur in the Wirefly Token case then Bob would have infinite creatures.

I stated in my original post that it would take an infinite amount of time to physically generate all the creatures. Almost every response has also pointed this out. However, this brings an element of "action time" to magic that I don't think exists in the rules of the game (except in the stalling rules, which only puts effective upper limits on how much time something can take)

We often speed up actions when we play magic. We pass priority until something interesting happens. If Squirrelcraft happens, we skip millions of decisions in the space of a single demonstration. We could calculate what will eventually happen in the shuffle/look case and skip a number of random actions because we know what will eventually happen. [I know, current rules say you have to actually simulate]. What's the difference in skipping a million breakpoints and skipping an infinite number?

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carstenhaese
Joined Oct 2007
4039 Posts
What's the difference in skipping a million breakpoints and skipping an infinite number?

One is finite, the other one isn't. The "infinity" rules don't exist to let you short-cut an infinity of actions. They exist to let you short-cut a clearly defined finite number of actions.
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condor
Joined Mar 2001
818 Posts
This loop I've described and any loop involving the ordering of cards are fundamentally different.

Not with respect with anything having to do with this thread. There are only two points of significance in either:
• You don't know which will be the result of the next iteration.
This means that there is a strategically valid reason to repeat it until you hit the success state, and that you can't (that means "it's not possible to"; not "well, you can try if really really want to") use the looping rules and/or some other kind of shortcut.

The significant difference between the two - that only one has a probability of 100% of eventually hitting the success state - is irrelevant.

If the judge allowed a similar simulation to occur in the Wirefly Token case then Bob would have infinite creatures.

• Doing so means you end with a finite number of tokens. Not infinite. To get infinite, you have to continue to get the failure state INDEFINITELY.

I stated in my original post that it would take an infinite amount of time to physically generate all the creatures.

Which is why you won't get it. There can be only there possible outcomes:
• You repeat manually, and get a success. No infinite tokens.
• You repeat manually, and do not get success. The world ends (or maybe just the time alloted for the game) before you get infinite tokens.

You are assuming a fourth outcome, one that is not possible. That a judge lets you assume an infinite number of iterations, without success. The reason this is not possible, is that YOU WOULD CHOOSE TO REPEAT THE PROCESS AGAIN AT THIS POINT. You haven't reached "success," and it is still possible to, in theory.

Yes, I measn that even if you assume (as you can't) "infinity." I suggest you look up the "Grand Hotel Paradox" on wikipedia. It will explain some things about "infinity" to you, why you can't treat it as a number that you can compare to anything else in a normal way, and why you can still get a success after assuming it.

The point is, that by being boundless, there always will be more room, (or time, or whatever) to go through the process again, even after what you have assumed is already infinite.

What's the difference in skipping a million breakpoints and skipping an infinite number?

There is no such thing as an "infinite number," YOU WILL STILL WANT TO REPEAT THE PROCESS AFTER YOU SKIP WHATEVER IT IS YOU THINK YOU CAN SKIP, and what CarstenHaese said.
wprundv
Joined Aug 2007
156 Posts
guardian_phoenix
Joined Sep 2004
1727 Posts
supply+mox lotus

The Rules Q&A forum does not recognize the existence of Unglued or Unhinged. We cannot comment on silver-bordered cards.
billyh
Joined Feb 2006
404 Posts
To insert more formalism into this problem/question:

1) The limit of the number of Pincher Tokens that Bob controls approaches infinity as the turn approaches over.

I realize that "the turn approaches over" is not exactly a mathematical concept or a length of time, but I believe it's well defined and understood.

Furthermore, I contend that no other loop [say, no loop without Wirefly Hive] has this property. All other loops will not approach infinite creatures (mostly because of the loop/stalling rules). (But I'd be happy to be proved wrong, if there is another example.)

Also, if you are interested in the cool things you can do with infinite creatures (like make a possibly uncountable number of infinite creatures) see this thread.

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Ahlyis
Joined Aug 2003
11161 Posts
supply+mox lotus

The Rules Q&A forum does not recognize the existence of Unglued or Unhinged. We cannot comment on silver-bordered cards.

And if we were going to comment on silver bordered cards (beyond pointing out that they often violate the rules), I think most of us would point out that, even with infinite mana in your pool, you still have to choose a valid number for X in order to cast Supply. Infinity is NOT a valid number for X.
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Ahlyis
Joined Aug 2003
11161 Posts
Furthermore, I contend that no other loop [say, no loop without Wirefly Hive] has this property. All other loops will not approach infinite creatures (mostly because of the loop/stalling rules). (But I'd be happy to be proved wrong, if there is another example.)

Why does it matter whether there's another "similar" loop or not? Do you think special dispensation should be added to the Comp Rules just to handle the extremely unlikely loop that you've constructed?
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Brusko651
Joined Jan 2008
1656 Posts
I don't get it. When she tries it over and over again, she will get enough tokens after a while, won't she? Why are you talking about an infinite number of creatures?
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2goth4U
Joined Oct 2007
10383 Posts
I'm thinking Krark's Thumb would needlessly complicate things more?

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Ahlyis
Joined Aug 2003
11161 Posts
I don't get it. When she tries it over and over again, she will get enough tokens after a while, won't she? Why are you talking about an infinite number of creatures?

The opponent is gaining life each time she fails. So, as she tries longer and longer, the opponent gains more and more life, making the odds of her getting successful flips enough consecutive times less and less likely.
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2goth4U
Joined Oct 2007
10383 Posts
so to demonstrate:

Activate Wirefly Hive
success generates a 2/2 Insect Artifact Creature token
which gains a point of life for the opponent via Leonin's Elder
so a success nets her +1 more damage than opponent life gain
and she needs 5 straight successes to get the 5 tokens to kill him.

unfortunately, a fail destroys all accumulated insect tokens and stacks a bunch of Summoner station triggers which allows him to put Pincher tokens into play and gain even more life which makes the number of continuous successes required by the Wirefly Hive even higher and the chance of success even more remote.

given enough time, Alice would eventually succeed in overcoming the odds against her (just like those monkeys would generate the complete works of Shakespeare). However in so doing, she would likely use up all the match time without achieving success, and the match would end following the procedures for time expiration in the Floor Rules. I'm thinking a judge may intervene and deem it stalling at some point once it become obvious that the chance of success is so remote as to be virtually unwinnable. There is an expected average time taken for each flip which can determine the maximum numbers of flips possible before the expiration of time in the match. Once you reach a point where the number of successive successes required exceeds the time available, I'm thinking the judge at this point would be right to call it stalling because a best case scenario at this point would be a draw and Alice would be playing only for the draw.

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darkartist
Joined Jul 2001
1418 Posts
@ the OP:

You've created an interesting situation, but I don't see how the concept of infinity plays into it at all.

So, the person is trying to win and is not stalling. How is this any different than any other situation where 2 players have some complicated and strategic stuff on the board, and eventually the clock runs out without anybody winning?

It just seems like you've created a voluntary loop that will either end in the player overcoming great odds and winning, or the game ending to time eventually. Whether that means the clock in a tournament, or her opponent conceding because he's sick of watching her flip coins, it's still going to stop eventually.

I don't think the concept of infinity even comes into this situation at all...

condor
Joined Mar 2001
818 Posts
1) The limit of the number of Pincher Tokens that Bob controls approaches infinity as the turn approaches over.

Certainly. And it NEVER reaches infinity. You are assuming that the game must provide you with a way to stop this process. Since the only way to do that is allow Bob to have "Actually infinite creatures," you conclude that you shoud be allowed to get there.

The game does not have to provide you with a way to end it. A forced draw is a an acceptable result, and in fact is the required result from your assumptions.
Furthermore, I contend that no other loop [say, no loop without Wirefly Hive] has this property.

So? It still doesn't require you to end with infinity.

Also, if you are interested in the cool things you can do with infinite creatures ....

Why would we be, since it can't happen? Besides, you are wrong in that thread, about what you "can do" with infinite creatures. Specifically, the answer to the question you asked, "Is your opponent able to block all of the creatures?" is "Yes." See the Grand Hotel Paradox. The situation there is an wexample of the same thing.
IronMagus
Joined Aug 2005
5667 Posts
Once you reach a point where the number of successive successes required exceeds the time available, I'm thinking the judge at this point would be right to call it stalling because a best case scenario at this point would be a draw and Alice would be playing only for the draw.

Can you provide a rules quote to back up your statement?
2goth4U
Joined Oct 2007
10383 Posts
Can you provide a rules quote to back up your statement?

I don't think I need one.

Ask yourself "What is the reasoning behind the stalling rule?"

it is to prevent a player from wasting time to prevent the result of a match which would likely end in their loss, ie. if they were not allowed to stall to get the draw.

now look at this scenario again.

Alice needs 5 straight successes to achieve her victory conditions at the outset. If she fails, she has sufferred a setback and increases the number of successful flips required to achieve victory. She may try again but each flip takes a certain amount of time to process and there is only so much time remaining in the match. At some point (assuming she has not arrived at the requisite number of successive successes for victory) the number of flips required for her to achieve victory will be greater than the amount of time remaining for her to do so. Once this point is reached, the intent can only be to stall for a draw as there is no possible way to complete the flips before time runs out and a draw is declared, and were she not to succeed, the sharpshooter would cause her to lose.

At this point, if Alice opts to try again the judge should interpret this as the same as Stalling and award the requisite penalty to Alice.

Wouldn't you agree?

Rule 151 Cheating — Stalling
Example D. A player intentionally exceeds the pregame time limit before the third game in an attempt to make it harder
for his opponent to win in time.
Philosophy If it is clear that a player is stalling, the integrity of the match is compromised and he or she will face a serious penalty.

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cyphern
Joined Jan 2003
19354 Posts
At this point, if Alice opts to try again the judge should interpret this as the same as Stalling and award the requisite penalty to Alice.

Wouldn't you agree?

I do not agree.

Stalling is intentional slow play, and slow play is spelled out in the Penalty guide as follows:
133. Tournament Error — Slow Play
Definition
Players who take longer than is reasonably required to complete game actions are engaging in Slow Play. If a judge believes a player is intentionally playing slowly to take advantage of a time limit, the infraction is Cheating — Stalling.

The quantity of game actions you take is not what makes it stalling/cheating, it is the speed with which you perform individual game actions. Just because you have a lot of things to do doesn't mean you are playing slowly, so long as you do the actions in a timely manner.
Just_another_Wizard
Joined Jul 2008
968 Posts
Speaking of weird infinite things...I've come up with a combo that makes a Quillspike have infinite power/toughness and infinite -1/-1 counters...(not damage), but it isn't dead, and neither infinity is uncountable ;)
Joined Feb 2007
12590 Posts
No, you haven't.
Just_another_Wizard
Joined Jul 2008
968 Posts
...well technically not infinite, but you get the idea. I can give u a run down of how it works if you like....
2goth4U
Joined Oct 2007
10383 Posts
I do not agree.

Stalling is intentional slow play, and slow play is spelled out in the Penalty guide as follows:

The quantity of game actions you take is not what makes it stalling/cheating, it is the speed with which you perform individual game actions. Just because you have a lot of things to do doesn't mean you are playing slowly, so long as you do the actions in a timely manner.

read Example D in Rule 151

it is clearly stalling if I can not possibly achieve my victory in the time remaining and I am certain to lose if I stop, and I opt to continue the loop to wipe out the time in the game by following optional actions (even if that is a side effect of timely following the game actions).

How is this any different than continuously tapping and untapping a Silver Myr enchanted with Freed from the Real ?

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Ahlyis
Joined Aug 2003
11161 Posts
Sorry cyphern, but I'm leaning towards agreeing with 2goth4u on this one.

Once it becomes impossible for the game actions to produce a win, it stops being a valid sequence of actions and becomes nothing more than a stalling tactic. As long as there is still the possibility of winning the game, I don't see any way to call it stalling/cheating, no matter how remote the likelihood of winning actually is. But at some point, 2goth4u is correct, there is no longer enough time left in the match to achieve a winning condition via the "loop".
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Just_another_Wizard
Joined Jul 2008
968 Posts
Some infinite series converge to infinity, and some infinite sequences of actions have a probability of 1 of converging. This is not one of them. That is to say, even though Alice has an infinite number of chances to win, the chance of winning drops so fast that the probability that Alice will eventually win is NOT 1.

To give an easier example, suppose Alice is going through a loop that has a 1/3 chance of winning after one iteration, a 1/9th chance after 2, a 1/27th chance after 3, etc.

The overall chance that Alice will win is 1/3 + 1/9 + 1/27... If we sum this infinite series, the answer is 0.5. Even after an infinite number of attempts, Alice will have a 50% chance of winning, despite always having a chance to win.

Um, I don't agree with this conclusion, I am currently majoring in mathematics, and it is my understanding that if the probability of an outcome of an event is non-zero, then if that event occurs an infinite number of times, then it will occur at some point. Note that even though the probability is converging to 0, it is not zero at ANY iteration, thus at one of those iterations it will eventually produce enough tokens to give Alice the win. (Now probability is not my best area of math, but I'm pretty confident in this one.)

Your math is particularly what seems to be in error, what if instead of 1/(3^n), it was (3^n)/(4^n), then your sum is greater than 1 which means it couldn't possibly represent a probability.
cyphern
Joined Jan 2003
19354 Posts
Sorry cyphern, but I'm leaning towards agreeing with 2goth4u on this one.

Once it becomes impossible for the game actions to produce a win, it stops being a valid sequence of actions and becomes nothing more than a stalling tactic. As long as there is still the possibility of winning the game, I don't see any way to call it stalling/cheating, no matter how remote the likelihood of winning actually is. But at some point, 2goth4u is correct, there is no longer enough time left in the match to achieve a winning condition via the "loop".

Ok, i can agree that if it is impossible for the game actions to produce a win, then it is stalling. However, at what point does it become impossible? 2goth4u proposed the point at which there is no longer enough game time in which to do all the actions that are needed to win, but how does one determine that point?
Brusko651
Joined Jan 2008
1656 Posts
Some infinite series converge to infinity, and some infinite sequences of actions have a probability of 1 of converging. This is not one of them. That is to say, even though Alice has an infinite number of chances to win, the chance of winning drops so fast that the probability that Alice will eventually win is NOT 1.

To give an easier example, suppose Alice is going through a loop that has a 1/3 chance of winning after one iteration, a 1/9th chance after 2, a 1/27th chance after 3, etc.

The overall chance that Alice will win is 1/3 + 1/9 + 1/27... If we sum this infinite series, the answer is 0.5. Even after an infinite number of attempts, Alice will have a 50% chance of winning, despite always having a chance to win.

but, staying in this simple example, wouldn't her chance of still winning after those first three tries be at 0,5-(1/3+1/9+1/27) = about 0,018%?

so, after a few tries, wouldn't it be safe for her to give up?

edit: now that I think about it, shouldn't it be 1/9 of the remaining 2/3 when you want to determine the total chance?
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sebastian1968
Joined Jul 2008
5 Posts
Initially her chance is 1/32. She must win 5 in a row.

If she wins 4 and then loses, his life will be at 13.
She will now need to win 13 in a row which is about 1/8000 chance.

Lets just say she hits 12 in a row and then loses. His life is 37.
Her chance is now 1/128000000000.

At some point it has to become stalling. But her chances will never be zero.
Ahlyis
Joined Aug 2003
11161 Posts
Your math is particularly what seems to be in error, what if instead of 1/(3^n), it was 3/(4^n), then your sum is greater than 1 which means it couldn't possibly represent a probability.

huh? How does that summation total more than 1? 3/4 + 3/16 + 3/64 + ... looks like it converges to 1 as you go to infinity. But I don't see how it exceeds 1, as you claim.

Besides, if you DO construct such a summation that has each piece less than 1 (so it could be a probability) but the summation is greater than one, then I'd like to see you construct an example, even a hypothetical one with made up cards, where that summation would apply. I'll give you a hint, you couldn't.

The only way for your summation to exceed 1 is for at least one of the entries to be larger than the remaining difference between the summation so far and 1. But that's impossible if the scenario is constructed properly. It would be claiming that the given event has greater than 100% probability of occurring.
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Just_another_Wizard
Joined Jul 2008
968 Posts
Sorry, I should have written (3^n)/(4^n) ( which equals 3 according to my TI-200 Voyage calculator)

But in any case summing the chances she will win does not represent her total chance to win, in fact it has no meaning w/ regard to probability, a correct way to calculate her chance to win is calculate the chance she ALWAYS fails to win and then calculate 1 minus that number, This is done by taking the probability that she loses the first iteration TIMES the probability she loses the second TIMES the third etc. etc.... This is a product of decimals between 0 and 1 (not including them), and thus the limit as the number of iterations goes to infinity the probability she will lose all of the iterations goes to zero, thus she will eventually win.
Kedar
Joined Sep 2007
6823 Posts
Sorry, I should have written (3^n)/(4^n)

Which can be written in a more simple form: (3/4)^n. >.> And yes, summing that up would total more than 1. An impossibility in probability--nothing can have more than a chance of 1 of happening.

Ahlyis
Joined Aug 2003
11161 Posts
Sorry, I should have written (3^n)/(4^n) ( which equals 3 according to my TI-200 Voyage calculator)

Fine. Now, show me an example where that would be the sequence of probabilities.

You can't because it's an impossible sequence for probabilities.

Your second iteration has probability of 9/16, meaning 9 out of 16 times you run the sequence, you will succeed on EXACTLY the second try. But since 12 of 16 times you try, you'll succeed on the FIRST try, you will only even get to the second try 4 out of 16 times. So how are you going to set up a situation where you can succeed 9 of the 4 times you get to the second try?

This is a product of decimals between 0 and 1 (not including them), and thus the limit as the number of iterations goes to infinity the probability she will lose all of the iterations goes to zero, thus she will eventually win.

No, she won't.

She will ONLY "eventually win" if she is given an infinite number of tries to do so. Since she cannot shortcut the "loop", she most certainly does NOT have an infinite number of attempts available.
I'm just a Pigment of your imagination.
2goth4U
Joined Oct 2007
10383 Posts
Ok, i can agree that if it is impossible for the game actions to produce a win, then it is stalling. However, at what point does it become impossible? 2goth4u proposed the point at which there is no longer enough game time in which to do all the actions that are needed to win, but how does one determine that point?

well if I were a judge I'd time the length of time to test X number of tests then divide that time by X to find the average time per test.

Finally divide the time remaining by the time for each test and find out how many tests can be completed in the time remaining.

Once the number of successes required becomes greater than that which is possible then you've met your threshold.

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Just_another_Wizard
Joined Jul 2008
968 Posts
Fine. Now, show me an example where that would be the sequence of probabilities.

You can't because it's an impossible sequence for probabilities.

Your second iteration has probability of 9/16, meaning 9 out of 16 times you run the sequence, you will succeed on EXACTLY the second try. But since 12 of 16 times you try, you'll succeed on the FIRST try, you will only even get to the second try 4 out of 16 times. So how are you going to set up a situation where you can succeed 9 of the 4 times you get to the second try?

No, she won't.

She will ONLY "eventually win" if she is given an infinite number of tries to do so. Since she cannot shortcut the "loop", she most certainly does NOT have an infinite number of attempts available.

OK, first iteration (1) the event A has a probability of 3/4 of occuring, assuming the 1/4 it doesn't occur happens, then the next iteration will occur.
This iteration (2), will have a 9/16 chance she wins and a 7/16 chance she won't, if the 7/16 chance occurs, then iteration (3) will occur
(3) Win: 27/64, Lose:37/64
(4) Win: 81/256, Lose: 175/256
etc.

The iterations are INDEPENDENT events, meaning their probabilities DO NOT affect one another. So you see with each iteration her chance to win is less than 1, and her chance to lose is less than 1. That means either outcome CAN occur, and thus she WON'T win if she loses EVERY iteration, and her chances of losing both the first and the second iteration is (1/4)*(7/16)=7/64
Her chance of lose the first 3 iterations is (1/4)*(7/64)*(37/64)=(7/64)*(37/64)=259/4096
etc. etc.
This Product converges to zero (and it represents the probability she will lose the first n iterations). Therefore she will win eventually, barring this being considered stalling, etc.
Which I think it shouldn't since with her eventually winning she should point it out before "doing" those actions and simply skip them just like any other loop. Though this line of discussion is best handled by a judge.