Highest Possible Crit Range?

[Dairun Cates]

For all those words, you didn't answer my question.

What is 1/4 + 1/8 +1/16 + 1/32 . . and so on, what is it equal to?

You don't even need to explain your answer or show your work, I just want to know what you think the answer is.

Edit: It occurs to me that maybe you did try to answer the question. If that is the case, then there is no need to reply to this post, your answer can stand for itself.

I know I don't need to, but to reiterate, it WILL hit 1, but only at an infinite number of iterations. Since you can quantify infinity as a number, it will never hit that number, just tend towards it. In the meantime, the answer will inevitably terminate. Since it has an infinite amount of time for that possibility to catch up (no matter how small) it will. The amount of iterations is infinite because that's how long it takes to catch up.

1/99999999999999999999999999999999999999999999999999 9999999999999999999999999999999 is not infinitely impossible. It's just extremely unlikely.

[Signmaker]

So yes. The DIFFERENCE between each probability each turn is going to to be smaller and smaller because of the exponential effect, but at no point will the probability GROW, because you never have a 100% chance or more of success. That number keeps going down and down. There's ALWAYS a chance for success on the next go, but the chances of it to actually keep going in the long run gets lower and lower.

I'll address this due to the interaction with series. If the probability does not converge to 1 or 0, there is leeway, rather than absolute success or absolute failure. That's the entire point about infinite sums: there are certain types of sums (those that follow a distinct pattern, like 1/2^N) that will converge to a number. As I showed before, even the simple case of Heads Nothing Tails Two Flips does not converge to 1 or 0. Therefore there is leeway, a probability that flipping will commence unbounded.

If there is a 10% chance of failure before round 2, an 11% chance before 3, an 11.1% chance before 4, etc, you're left with with a probability approaching infinity, not a certainty.

[Dairun Cates]

I'll address this due to the interaction with series. If the probability does not converge to 1 or 0, there is leeway, rather than absolute success or absolute failure. That's the entire point about infinite sums: there are certain types of sums (those that follow a distinct pattern, like 1/2^N) that will converge to a number. As I showed before, even the simple case of Heads Nothing Tails Two Flips does not converge to 1 or 0. Therefore there is leeway, a probability.

That's the point though, isn't it? If you flip another heads, so to speak, it continues, if you flip tails, it ends. Infinite attacks requires 100% as the sum of the probabilities. The fact that it will end does not require probability to hit 0 though. It merely requires that the other set does not equal 100%. Since that will never happen because infinity is infinite and that takes infinite iterations. The probability to roll not all 1's HAS to hit absolutely 0% for it to never happen in infinite iterations.

So yes, you'll die before you finish your rolls at that point (or your GM will just punch you), but given enough time, you'd eventually fail. It's an arbitrarily high loop, but it is not infinite.

[ocdscale]

Dairun, it seems like you are sticking to your guns here.

EDIT: Incorrect wording of the game. Please see page 249 for the correct version.

So in my hypothetical game:
Start: Go to Round 1
Round 1: 25% chance of ending (1/4), otherwise go to Round 2
Round 2: 12.5% chance of ending (1/8), otherwise go to Round 3
Round 3: 6.25% chance of ending (1/16), otherwise go to Round 4
... and so on ad infinitum

Am I correct in stating that your stance is that:
There is always some chance of failure. Therefore, as the rounds grow towards infinity, it is inevitable that one round will result in failure and end the game. Therefore the chance of failure is 100% (eventually).

[Dairun Cates]

It seems like you are sticking to your guns here.
So in my hypothetical game:
Start: Go to Round 1
Round 1: 25% chance of ending (1/4), otherwise go to Round 2
Round 2: 12.5% chance of ending (1/8), otherwise go to Round 3
Round 3: 6.25% chance of ending (1/16), otherwise go to Round 4
... and so on ad infinitum

Am I correct in stating that your stance is that:
There is always some chance of failure. Therefore, as the rounds grow towards infinity, it is inevitable that one round will result in failure and end the game.

Yes. It will take an arbitrarily high number of turns in some cases (more than a human will roll in his or her life), but since it takes infinite iterations to reach the end, inevitably one round will screw you.

[Signmaker]

It merely requires that the other set does not equal 100%.

Incorrect premise. Tending towards infinity, you will have a success rate and a failure rate. Both are convergent numbers, and the two added together will equal one. Picture it as a tree, as a linear-shaped string of rolls, whatever. The series model the probability that a given rolling engine will terminate or continue unbounded. All those tiny, infinitesmal probabilities you're treating as certainties due to infinity? You're not looking at them the correct way.

[Dairun Cates]

Incorrect premise. Tending towards infinity, you will have a success rate and a failure rate. Both are convergent numbers, and the two added together will equal one. Picture it as a tree, as a linear-shaped string of rolls, whatever. The series model the probability that a given rolling engine will terminate or continue unbounded. All those tiny, infinitesmal probabilities you're treating as certainties due to infinity? You're not looking at them the correct way.

Well, of course the two probabilities add up to 1, that's the very basis of probability, but one side has a continue condition and the other has a stop condition. The low probability needs to only occur once to completely end the loop. At this point, the "loss" side "wins" so to speak.

However, the "win" side can NEVER win because succeeding on one roll just leads to another iteration with a lower probability of losing on this hand specifically. There is no end state. If you win again, you just roll again. THAT'S the problem. In order for you to hit infinity, there has to be no quantifiable way to continue, you won infinite times. The equation tends to infinity, but it will never reach it.

The only way you can ABSOLUTELY ensure that you will never fail is to have an ABSOLUTE 0% fail rate.

[Signmaker]

However, the "win" side can NEVER win because succeeding on one roll just leads to another iteration with a lower probability of losing on this hand specifically. There is no end state. If you win again, you just roll again. THAT'S the problem. In order for you to hit infinity, there has to be no quantifiable way to continue, you won infinite times. The equation tends to infinity, but it will never reach it.

The only way you can ABSOLUTELY ensure that you will never fail is to have an ABSOLUTE 0% fail rate.

Problem two. For probability, there is no 'hitting' infinity, but tending towards it. Rather than never approaching the end of the tunnel that is the natural number line, the relevant concept of infinity is to not stop running towards it. To continue unbounded is what matters. There are few absolutes, and this scenario is not one of them (flipping a coin and stopping the second you hit tails is one, though, as p=1). Perhaps this might help.

[ocdscale]

Yes. It will take an arbitrarily high number of turns in some cases (more than a human will roll in his or her life), but since it takes infinite iterations to reach the end, inevitably one round will screw you.

I apologize, I realize now that I incorrectly worded my game. Please let me know if this new game convinces you:

You press a button.
The button turns on a magic light.
The magic light is designed such that 25% (1/4) of the time, it turns off in the first second.
12.5% (1/8) of the time, it turns off in the second second.
6.25% (1/16) of the time, it turns off in the third second.
3.125% (1/32) of the time, it turns off in the fourth second.
and so on ad infinitum

If you will notice, if you make it to the 100th second, the probability it turns off that second is very low, but there is still a chance. The longer you last, the lower the probability the light will turn off, but there is always a chance.

Is it fair to say that your stance is "it is impossible for the magic light to shine forever because there is always a chance it will turn off, even if that chance is small, because in infinite time, eventually you will hit that chance." ?

[Dairun Cates]

Is it fair to say that your stance is "it is impossible for the magic light to shine forever because there is always a chance it will turn off, even if that chance is small, because in infinite time, eventually you will hit that chance." ?

Exactly what I mean.

[Signmaker]

Therein lies the problem.

[ocdscale]

Exactly what I mean.

Okay, I understand your position. I disagree with it, and I'm going to try to convince you to change your mind.

Here is the game for your convenience:

You press a button.
The button turns on a magic light.
The magic light is designed such that 25% (1/4) of the time, it turns off in the first second.
12.5% (1/8) of the time, it turns off in the second second.
6.25% (1/16) of the time, it turns off in the third second.
3.125% (1/32) of the time, it turns off in the fourth second.
and so on ad infinitum

Do you agree with me that there is a 75% chance the light is still on in the first second?

Do you agree with me that there is a 62.5% chance the light is still on in the second second?
(25% of the time it turns off in the first second, 12.5% of the time it turns off in the second second. So 100% - 25% - 12.5% = 62.5% of the time it is not off yet in the second second.)

I don't want to do the math anymore, but do you agree with me that there is a lower than 62.5% chance that the light is still on in the third second?

Do you agree with me that as time goes on, the chance that the light is still on always gets smaller?

Now. Would I be able to change your mind if I could prove to you that even though the probability that the light is on always gets smaller, it never gets smaller than 50%?

Edit: Sorry, going to format this a little better, didn't realize how much it hurts the eyes to look at.

[The Glyphstone]

isn't that called the gambler's fallacy?

[ocdscale]

isn't that called the gambler's fallacy?

You might be thinking of two things.
Gambler's fallacy is that things "even out" over time. When someone thinks: If a fair coin lands heads four times in a row, it's more likely to land on tail's this time.

Gambler's ruin is a little more relevant, but does not apply because the proposed build is weighted towards successes instead of being 50/50.
Or the other version of Gambler's ruin (one that incorporates preferential odds) is not applicable because your wager is not a fixed percentage of your bankroll. Your wager is fixed to 1 attack, while your bankroll (number of attacks) grows.
*terminally sick has read a proof that has convinced him that even preferential weighting will not escape Gambler's ruin. I haven't seen the proof, but I disagree (although I trust judgment enough to worry about it).

[Signmaker]

isn't that called the gambler's fallacy?

Nnnnnnnnnot quite. Gambler's fallacy is thinking that at any step in the process, previous rolls are incorporated in to the probability at that specific stage. Which, being independant, are not. The issue at hand is the mental correlation between approaching infinity (big 'thing') with the related infinitesmally small failure rates, and the belief that no matter how small the infinitesmal, infinity wins out.

[Siosilvar]

isn't that called the gambler's fallacy?

No, it's the finite sum of an infinite series.

Ish. finite limit of an infinite series.

[term1nally s1ck]

Well, of course the two probabilities add up to 1, that's the very basis of probability, but one side has a continue condition and the other has a stop condition. The low probability needs to only occur once to completely end the loop. At this point, the "loss" side "wins" so to speak.

However, the "win" side can NEVER win because succeeding on one roll just leads to another iteration with a lower probability of losing on this hand specifically. There is no end state. If you win again, you just roll again. THAT'S the problem. In order for you to hit infinity, there has to be no quantifiable way to continue, you won infinite times. The equation tends to infinity, but it will never reach it.

The only way you can ABSOLUTELY ensure that you will never fail is to have an ABSOLUTE 0% fail rate.

We're not trying to prove that we will never fail, we're trying to prove that there is a non-0 chance that we can get arbitrarily higher than any given finite number, which is the DEFINITION MATHEMATICALLY OF INFINITY. If you want to disregard a few hundred years of mathematical convention, and use your own definition, you're correct, we don't reach YOUR infinity. Ours, and everyone else's on the planet who's put any serious thought and research into the thing is different, and is reached perfectly by this problem.

However, it simultaneously terminates, thanks to a very very clever proof I've found :/

I've tried, but I can't pick out the most important parts and give a summary. Here's the Propositions and Theorems proved:

Proposition 1. For a symmetric random walk (Sn) the probability of changing sign at time n is
O(n−1/2):

Theorem 1. Let X, Y be independent real random variables with the same distribution. Then for any c >0 we have P(|X + Y|<or= c)< 2P(|X − Y|<or= c)

Proposition 2. Let (Ω,B) be a measurable space and let f :Ω^2 →R be a B ⊗ B measurable bounded symmetric function. Let P be the set of all probability measures on B. Then the following dichotomy holds:

Either the Integral (over Ω) of (f (·, y)P(dy)) <or= 0 for some P ∈ P

or the Integral over Ω×Ω of f (x, y)P(dx)P(dy) >0 for all P ∈ P.

Theorem 2. For any one-dimensional random walk (Sn) the probability of a crossing of level L
at time n is O(n−1/2):
P(sgn(Sn − L) =/= sgn(S(n−1) −L))= O(n^(−1/2))

Which proves it trivially by taking L = 0

This one does NOT assume p <or= 1/2, and is valid for our current argument.

The proof is in the last step, where you find that the proof is of the order n^-1/2. This can then be summed to provide the probability of convergence, which would seem to always go to 1, thanks to the basic sum diverging, and taking a diverging series as your probability at n should always give you a probabity at infinity of 1.

[Signmaker]

I...don't quite believe that you can adequately liken the rolling proposition to a random walk scenario.

[flabort]

@ cocroach tea party (pg. 7):
google "Smbz", watch episode 7. when Mario and Basilisk are FLYING UPWARDS from the sheer speed of their attacks, their symutaniosly doing what I immagine this build to do, although Mario is closer due to the fact that he's using hammers, and Basilisk is using claws.

THAT's what this build looks like in action.


@ any one arguing about infinite:

Spoiler

Show

I am asuming we are trying to achive infinite hits.

this build threatens crits on 2-20, and scores another attack with each threatened crit due to lightning maces. with Better Lucky than Good, this counts as 1-20. since the attack is used up, i will consider generating another attack to be the attack persisting. as well, when confirming the crit, due to Roundhouse Kick, it generates another attack.

we will asume it "misses", and there by doesn't confirm the crit, when the attack roll is below 18, and we have no bonuses to hit. If we add Blood in the Water, this adds a bonus to hit each time we succesfully hit our target.

Each unit of hypothetical/folded/hammerspace/whatever time, each attack has a 19/20 chance of continuing on to the next unit of time, or with the feat, it's 20/20 chance, or garenteed (sp?) we will asume we are not using Better Lucky than Good from here on, due to the fact that it garentees infinite hits, and so is succesfull. the odds of confirming the crit are the odds of hitting, squared. which is with a roll of 18, 19, or 20. this results in (3/20)^2. this means that the odds of confirming the crit are 0.0225, which are the odds that another attack is generated each round. with Blood in the Water, this makes the formula ((3+n)/20)^2, were n is the number of succesful hits, which i do NOT have a formula for yet. it would equat (sp?) to something like n=(3+((3^X)/20)/20)X, were X is the number of attacks against the target by that point.

I have not enough math-fu to proceed from this point, but with Better Lucky than Good, we can garentee (SP?) infinite hits, and with Blood in the Water, we add more hits per unit of hammer-time. To proceed from this point, we would have to set up an equation for the odds that all the rolls would be 0, based off of the number of attacks, A, for unit of hammer-time, t, were A is based off of the odds that an attack fails to exist, the odds that it does not fail to exist, and the odds that another attack is produced as well, in each instance of t. it will probably have "^t" in it somewere. by that equation, we can figure out the odds that the chain ends at any instance of hammer-time, and as a by-product that is used in the proccess, an average amount of attacks at any instance of hammer-time. however, as I said, My math-fu is weak, so i cannot finnish this prosses. however, since there are odds that it will fail, it's garenteed (sp?...) that it isn't infinite, since infinite is defined that it has 0 chance of failing.

whew, that's longer than i meant it to be...

[Kalirren]

So I did some homework on this, because I don't quite trust term1nallys1ck's "proof by promising I have a proof."

Going back to consider his old structure of begining with a single attack, and going through the following procedure,

At each attack, you have:
A = 133/200 chance of hit and confirming = 2 more attacks
B = 7/200 chance of threatening but not confirming = 1 more attack
C = 3/20 chance of not threatening = terminate

Of course, A + B + C = 1.

Let's try to figure out the probability that the attack sequence terminates.

Let the probability that the attack sequence that is started by a single attack terminates be denoted by p.

This can happen in one of three ways:

Way 1 is that the attack threatens and confirms, i which case the attack tree terminates if the attack sequences generated by the 2 extra attacks -both- terminate. This happens with probability Ap^2.
Way 2 is that the attack threatens but does not confirm, in which case the attack tree terminates if the attack sequence generated by the subsequent attack terminates. This happens with probability Bp.
Way 3 is that the attack fails to threaten, and terminates immediately. This happens with probability C.

So we have the recursive relation

p = Ap^2 + Bp + C

which by cursory examination has the solution p = 1, since A + B + C = 1.

Well, p = 1 means the thing terminates eventually, right?

The only problem I have with this is: What's the other root? If we factor out the monomial (p-1), we get

(Ap^2 + (B-1)p + C) = (p - 1) (Ap - C)

which means that the other root is p=C/A. I think this is an extraneous root, because I can't imagine what significance it actually has. It's positive, and lies between 0 and 1, so it -might- mean something: I just don't know what it would be. If it were negative, or greater than 1, I'd be willing to say that it terminates, period, but the fact that C/A = 30/133 might actually mean something gives me pause.

...halp?

[term1nally s1ck]

For Signmaker:

I make 1 attack.

There is a probability I gain an attack (confirmed crit), and a probability I lose an attack (no crit)

The value is the number of attacks, the steps are single attacks. Perfect (biased) random walk.

For kalirren: C = 6/20, so 60/133 is the probability, provided there's a chance of it going infinite. That's the same as you get if you merely sum the probabilities of it ending in finite time, and subtract the answer from 1. The other root is also valid however...due to the nature of infinities, the thing is basically able to both reach infinity with positive probability, and terminate guaranteed (which I gess means it terminates at infinity? Strange...)

I've shown as much of the proof as I can, the rest requires notation that I cannot write on this site, so I can't show you the full proof. I've put the main steps in, though, so if you feel courageous, there's a challenge for you

[Signmaker]

For Signmaker:

I make 1 attack.

There is a probability I gain an attack (confirmed crit), and a probability I lose an attack (no crit)

If you only gain one attack per crit, that's analogous to flipping a coin until you hit tails. Did you mean getting a free attack, in addition to another roll?

[ocdscale]

If you only gain one attack per crit, that's analogous to flipping a coin until you hit tails. Did you mean getting a free attack, in addition to another roll?

I understand terminally's analogy, so I hope he doesn't mind if I explain it.
I still disagree with his proof however, but I have no means to verify it. My main disagreement being the proof I can see is better than the proof I cannot.

Anyway, the analogy
Imagine the number you stand on is always equal to the number of attacks you posses.
You start on the number corresponding to your full attack string.
Roll one attack. If it crits, move one step in the positive direction (you gain two attacks to roll, but you used up one). If it misses or otherwise fails to crit, move one step in the negative direction (you lose one attack).

It's a random walk. The proofs I've seen suggest a biased random walk does not have a 100% probability of crossing each value (in this case 0), analogizing from a biased gambler's ruin proof.

To respond to Kalirren:
Edit: Cleaned up my post,
First: The Ap^2 term is clever, but I'm not sure the Bp term is right. Shouldn't it just be B (failing to confirm is just as bad as missing outright)?

Second: I've convinced myself that if anything, the p=1 must be the extraneous root.
Simple proof: Set A=1, B and C to 0.
This simulates the build that never misses, always crits, and always generates double attacks. I think everyone in the thread would agree this hypothetical build is capable of infinite hits.
p = Ap^2 + B + C
Still has p=1 as a root, which of course makes no sense.

The reason p=1 exists as a root is because the roots of the equation aren't the "answer" to what p is in the system modelled, they only represent values of p that are consistent with the model. In other words, the 'right' value of p must be consistent with the model, but not all values that are consistent are 'right'.

The reason p=1 is consistent with a model that never misses is because the model is calculating p in a circular fashion. It's a very clever method of course (I wish I had thought of it first), but the circularity is the flaw.
I'm fairly certain the non p=1 root does represent the "answer" simply because the quadratic can only have two roots, and if one is logically unsound, it other must be right.

Edit 2:
Sigh. Threads like this make me wish I continued to pursue mathematics instead of going into law. Math is fun.

[Kalirren]

Terminally sick:

Thanks for the catch on the 60/133. I should have realized that

3/20 + 7/200 + 133/200 != 1.

Incidentally, this is pretty much the coolest build on the face of this planet. Could you slide me a copy? I have a friend who would greatly enjoy it.

Edit: So if the probability of terminating is 60/133, where did this whole 5/9 business come from? Did I just misread something and end up making it up? I'm sorry if I did...

ocdscale:

On Bp vs. B:
Terminallysick knows the details, but the build apparently generates one attack for threatening and one attack for confirming. So if you threaten but don't confirm (with probability B), you spawn one attack tree, and the probability of that tree terminating is p, so multiply them and you get Bp.

Set A = 1, and B = C = 0. p = 1 is obviously the extraneous root.

Of course, why didn't I think of that? You're right. p=1 is the extraneous root, so p = C/A = 60/133 is the chance that the whole process terminates at all. So this -is- in fact non-deterministically infinite, and the chance that we don't come back to 0 attacks, ever, is 73/133. (That's actually pretty good on a practical level, and holds as long as we hit on a 7.)

Which means that terminallysick's "other proof" must not apply, somehow. I'll look at it later.

[Heliomance]

I'm afraid my maths isn't good enough to begin to understand the propositions that term1nallys1ck provided, but I've seen this build come up before on Gleemax, and I seem to remember they eventually decided that a random walk scenario wasn't an acceptable model.

[Krazddndfreek]

I just had a bad bad bad idea. Isn't there an ability that lets you take 10 on an attack? If we can work that into the build, it IS infinite if you hit on a 10.

If no one has mentioned this before, Aura of Perfect Order is a Devoted Spirit stance that allows you to "take 11". Albeit its a 6th level maneuver so you'd have to take 11 levels in crusader first. Or... you could have a 20 level build that includes at least 2 levels in any martial class but that would use up at least one feat slot.

[term1nally s1ck]

Yeah, there's a build a page or so back which is Fighter 2 Swordsage 6 DoD 8, and I'm pretty sure it qualifies with ease. I'll have a look, see what prereqs I need again...

I need Improved Critical (Light Mace), Improved unarmed strike, roundabout kick, combat expertise, power attack, disciple of darkness, two weapon fighting, and weapon focus (light mace), Lightning Maces, along with enough swordsage to get Aura of Perfect Order. That's 9 feats.


Lawful Evil, Human (2 flaws)
Fighter 2 (Any Full BAB) 3 Swordsage 2 Disciple of Despite 8

Take Martial Stance as your Level 12 feat to get Aura of Perfect Order, and you're golden (IL 7 minimum at that point, and you should take at least 2 Devoted Spirit maneuvers in your swordsage levels)

Doable at Level 15...possibly 14 if you don't mind just burning 3 feats just to get the one stance, and taking no swordsage...


You can even do it without Flaws, if you take Fighter 4 and use your L15 feat for that last one you need.

[Krazddndfreek]

*grins sheepishly*

I had only read up to the end of the first page when I wrote that. Probably should have included that.

EDIT: And on another note, it doesn't seem like anyone noticed this at all:

I feel as though I've aided in making a monster.

His name will be Chav'Xaa of the Lost Heresy and he will gleam like a wildfire. Behold my horrific creation!



He's also the sun.

It just seemed so amusing to me I had to put it up there, because it makes you wonder. Was Krimm just checking to see if people would actually notice him trying to derail the thread? The world may never know.

[PhoenixRivers]

If the chance the attack string ends on the first round is 1/4
And the chance the attack string ends on the second round is 1/8
And the chance the attack string ends on the third round is 1/16
And so on in that fashion.

Once started, what is the probability this attack string will end?
In other words, what is 1/4 + 1/8 + 1/16 + 1/32 and so on into infinity?
I say the sum of all the probabilities of failure, as the rounds go into infinity, is 50%. Therefore there is a 50% chance of failure, and a 50% chance the attack string will not fail.

You argue there is a 100% chance this attack string will fail eventually because every round has a non-zero chance it will fail.
I invite you to show me how to add 1/4 + 1/8 + 1/16 ... in order to sum it up to 1 (100%).

This is not how to determine whether it will fail. If it were, then the following would be true:

I flip a coin twice. The odds of getting a heads is 100%. Why? Because 1/2 + 1/2 = 1.

Odds are computed by the odds of it not failing.

Premise 1: For any given step to succeed, all previous steps must succeed.

Premise 2: If a step has a 1/4 chance of failing, then it must have a 3/4 chance of succeeding. The odds of success are (1 - Chance of failure).

Premise 3: A single failure at any step indicates end of the string.

Premise 4: Odds of failure are (1 - Chance of success)
Odds of succeeding are:

1. Iteration | Chance of success at this step | Overall chance to reach and succeed this step
   1 | 3/4 | 3/4
   2 | 7/8 | 21/32
   3 | 15/16 | 315/512

Odds drop slowly, approaching a 0% chance of success as the iteration approaches infinity.

By premise 4, this means a 100% chance of failure as it approaches infinity.

[term1nally s1ck]

^^Actually, the product tends to a finite number.

it's the product to infinity of ((2^n)-1)/(2^n)

which is doable easily by taking logs.

log of the product = SUM(log(((2^n)-1)/(2^n)))

=SUM(log((2^n)-1) - log(2^n))

Which at every n is negative, so the sum must be negative. lf log(X) is negative, then X is between 1 and 0, hence the product to infinity is between 1 and 0.

However, it does NOT diverge to negative infinity, so the product is not 0.

In fact, the product converges to a little over 0.29.