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[Helgoland]

It rests on the incorrect assumption that any largest integer exists. The logic itself is flawless.

[MaximumZero]

If I may? That's fucking stupid.

[Helgoland]

If I may? That's fucking stupid.

Clearly you're not a mathematician then

[Karlito]

If I may? That's fucking stupid.

The theorem that's proven here isn't really best stated as "1 is the largest integer", rather "If a largest integer exists, then 1 is the largest integer" which is vacuously true. You can of course use the same chain of logic to prove "If a largest integer exists, then 2 is the largest integer" or any other number of true but useless statements.

There are mathematicians who are also uncomfortable with proofs like that. I believe reasoning like that in the proof displayed above is disallowed in the school of Intuitionistic logic (which is not widely used).

[Shadowlord]

If n^2 > n is the only criteria deciding whether a number is the "largest number", then 0, 1, all numbers in between, and infinity are all largest number. Which means that 0=1=∞ and also equals all numbers between 0 and 1. (0<=n<=1) = ∞ ?

If this isn't insane troll logic, I don't know what is.

(If there is no largest number, then the theorem is obviously false, 1 cannot be the largest number because no number is the largest number!)

[ZetaX]

There are mathematicians who are also uncomfortable with proofs like that. I believe reasoning like that in the proof displayed above is disallowed in the school of Intuitionistic logic (which is not widely used).

Intuitionists allow that false statements imply everything ("ex falso quod libet"). They only do not use proof by contradiction, or more precisely that one of A or (not A) is true for every statement A.

[Helgoland]

If n^2 > n is the only criteria deciding whether a number is the "largest number", then 0, 1, all numbers in between, and infinity are all largest number. Which means that 0=1=∞ and also equals all numbers between 0 and 1. (0<=n<=1) = ∞ ?

Nope, \infty is not a number, and for 0 the statement 'n > 0 \Rightarrow nē > n' does not hold, because 1 exists and is bigger than 0.

[Shadowlord]

But it's "any n for which n^2 > n is true isn't the largest number" ...

[Culise]

In fact, anything between 0 and 1 would be false for that criterion, since within that range n^2<n.

Yes, but n is chosen to be an integer greater than 1 precisely because any number less than one is tautologically...err, sorry to belabour this, but numbers less than one are by definition not greater than one.  Also, there are no integers between 0 and 1, unless you include the end points of 0 and 1 themselves. 

[Shadowlord]

Yes, but then they went on to say "use this theorem to prove that 1/2 is the largest non-integral rational number" and my brain went "Ia! Ia! Cthulhu fhtagn!"

[i2amroy]

So I've taken statistics, but I have no idea what equation I could use to determine this. Here's the basic setup.
1) You are trying to hit someone. They have a chance to dodge your punches.
2) Your have an anger problem, so every time you miss you are angry enough that your next 3 punches are guaranteed to hit (whether or not they would have hit already).
3) I'm looking for an equation that will allow me to determine the effect on a person's average ability to dodge with your anger problem vs. without it.

This is of course easy to calculate for some nice numbers. For example If they have a 33% chance of dodging without the anger problem then it becomes a 16.5% chance to dodge with it. The reason why is easy to see, if they have a 33% chance then that means on average they will dodge 1 in 3 strikes. Since your anger guarantees you are going to hit for 3 strikes that means that, on average, one of those three punches would have otherwise missed. Since you had to miss a punch in the first place to trigger the effect, the overall average means that for every miss there will be one punch that will hit when it would otherwise have missed, for a net reduction in average dodging power to half of what it would be normally, i.e. 33%->16.5%. 100% and 0% are also easy numbers to calculate, with 100% dodging ability with the anger resulting in a consistent pattern of 1 miss followed by three hits, resulting in a net dodging power of 100%->25%, and 0% means that you will never miss so we get 0%->0%.

Now at first I tried a couple different distribution methods at first but it quickly became obvious that that wasn't what I was looking for. After a bit of digging it seems like I might be looking for something built on top of an "expected value" equation, but I'm not totally sure how to set that up.

Ideally the equation should depend on the number of "guaranteed" anger hits you get after each miss along with their "base" ability to dodge your punches, but in a worst case scenario just one based on their base ability would work for my purposes as well too.

[Reelya]

There are other ways to look at it. e.g. forget the rhetoric of "unblockable attacks", that's a red herring which makes it sound more complicated than it really is. You can think about it this way: consider a "normal" random sequence of dodge/hits. For every successful hit, the attacker scores 1 point and costs 1 turn. For every successful dodge, the attacker scores 3 points and costs 4 turns. What's the average number of points per turn? This is logically equivalent to the puzzle you presented. Just treat the "free" attacks as occurring between the regular random events.

Call the dodge rate "d" and the free hits "h". The formula for points =

d * h + 1 - d

And the formula for "turns" =

d * (h+1) + (1-d) = d * h + 1

Putting that together you have, points per turn of":

P = (d * h + 1 - d) / (d * h + 1)
P = 1 - d/(d * h + 1)
P = 1 - 1/(h + 1/d)

So from that, assuming the logic is sound, for d=1/3, h=3 the formula gives

1 - 1/(3 + 3) = 1 - 1/6 = 83.33%

for the 50% dodge rate, the formula is:

1 - 1/(3 + 2) = 1 - 1/5 = 80.00%

[TheBiggerFish]

Yes, but then they went on to say "use this theorem to prove that 1/2 is the largest non-integral rational number" and my brain went "Ia! Ia! Cthulhu fhtagn!"

Sig'd.

[i2amroy]

-snip-

I think I'm understanding you properly, but now is changing the points per turn into a "dodge percentage chance" (which is what I'm looking for as a final answer) just as simple as doing 1-ans%?

[RedWarrior0]

Number of points is the number of hits. Turns is hits plus misses.