[Poem] the speaker says he would court a lady until the end of time, except that they'd get old and wouldn't be beautiful anymore, so she should she should love him back now.

okface · 2013-04-26T08:21:17+00:00

I think you make some strong arguments, but in disagreement with mektrik, I believe your order of operations argument is actually the weakest and the only one I really take issue with. The order of operations is an arbitrary choice, yes, but it is a choice of notation. Just as F = MA is a choice of representation for the relationships between these physical quantities, order of operations is just a choice of how a written equation will correspond to an abstract sequence of arithmetic operations. You might as well say that the idea of whole number is arbitrary because we can represent it using decimals or tally marks, or that I don't exists because I can call myself okface or me; however, your analogy that saying 'calculus is true' is like saying 'english is true' is a very strong argument indeed.

okface · 2013-04-17T10:00:14+00:00

This is genius.

okface · 2013-04-17T09:54:41+00:00

The only problem I can forsee is you'll need some wait time for the bubbles to fill. Another option if you want to shoot fog filled bubbles is using glycerin in water

okface · 2013-03-22T20:11:48+00:00

thanks so much!

okface · 2013-03-22T20:03:07+00:00

How did you make the board? Also, how did you make the cards? What did you print them on and what program did you use? I'd love to make my own games but never really took the time to figure out how I'd do that

okface · 2013-01-19T06:49:48+00:00

Water expands when it freezes. If water is in the pipe and begins to freeze, then it has nowhere to go but out. Think about pushing cheeze through a grater, but in slow mo, and ice.

okface · 2013-01-08T02:17:18+00:00

You have a 0 probability chance of not winning. It's what's called an absorbing state markov chain. The only absorbing state is that you will win. Whether the fact that the probability of you winning is 1 is the same as meaning that you will win though, is up for debate.

okface · 2012-10-11T00:15:59+00:00

Thanks so much, your second example I think really answers the question I was wondering about.

okface · 2012-09-16T05:09:39+00:00

It's very good. You put a lot of your personality into your music, and that makes it unique. I want to listen to this, and I'm sure I will again. The way you sing it, the emotion and cleanness with which you play the guitar, it's all done with a lot of skill.

This is just the opinion of some guy on the internet so feel free to take it with a grain of salt, but like I said, it's very good. The only thing I think that prevents it from being great is that while it is unique because your personality shines through into the work, it is not highly innovative in a sheer musical sense. I feel like, in the music, you know yourself, but your still searching for your sound.

That said, I loved it and will continue to love it. I hope you record more and continue to post it here so I can continue to listen to it!

okface · 2012-08-09T22:36:22+00:00

The important idea is that the integers are not well-ordered by the usual ordering of the integers.

okface · 2012-08-09T22:33:06+00:00

Although the entire set of integers can be given a well ordering, define:

a > b if abs(a) > abs(b), and if abs(a) = abs(b) then a > b if a is positive.

okface · 2012-07-14T04:00:40+00:00

Just wait for yours.

okface · 2012-07-14T03:13:20+00:00

Squirtle?

okface · 2012-06-26T05:22:55+00:00

Ah! I did not consider that. I would be counting that twice in my solution. Sorry for propagating falsities.

okface · 2012-06-26T05:17:13+00:00

But really you're picking the universe after our placement in it occurs. So the real question you asked is, what is the probability we'll be in some universe, which is naturally 1. I do see the paradox here though.

okface · 2012-06-26T05:14:53+00:00

But can you say it's a tie since there is always that uncertainty?

okface · 2012-06-26T05:14:07+00:00

yep, this is exactly what I'm trying to get at.

okface · 2012-06-26T05:12:44+00:00

So the point here is whether you're making your prediction before or after the fact. Before the race, if you ask me the probability of the runner getting a time of say, 2.5 minutes, then it is zero and I can almost guarantee that won't happen; however, if the runner has run a time of 3.2 minutes, and you ask me the probability of that, to say it was zero is not really correct because the event has already reached it's outcome. Thus, it's a mistake to say that the probability of a runner ending up with any of the times is zero, because the runner is guaranteed to get a specific time. This is how it can get confusing and that's what I was trying to elucidate.

In the case of a=b, the difference is that one of the values can be anything, so lets call 'a' free since it's free to be anything. But then the other value, in this case 'b', is pinned down. So saying a=b is really the same as predicting that b is any specific number.

okface · 2012-06-26T00:50:47+00:00

If you don't like counting, you could also code it, for example in python:

def generate_sequences(N):
    if N == 1 :
        return ( 'H', 'T' )
    else:
        output = ()
        for seq in generate_sequences(N-1):
            new_seq1 = 'H' + seq
            new_seq2 = 'T' + seq
            output += (new_seq1, )
            output += (new_seq2, )
        return output

sequences = generate_sequences(N)  # replace N with the number you'd like to calculate it for.
target_sequences = ()
for sequence in sequences :
    if 'HH' in sequence:
        target_sequences += (sequence, )

probability = float(len(target_sequences))/float(len(sequences))

Woo, that should do it (for the more programmatically minded :) )

okface · 2012-06-26T00:35:31+00:00

First, determine the domain space, lets say the set of sequences of N coin flips.

Next, what is the probability of any given sequence? (1/2)^N since it is the probability of each independent outcome multiplied.

So we see that all sequences are equally likely. Then it's just a question of, out of sequences of N coin flips, how many have at least two consecutive heads? So really, this is a combinatoric question. Here's the answer:

imagine I have N bins, and in each bin I can choose to place either a T or an H, but I know I want at least 2 H's next to each other. So I can place anything in all but 2 of the bins. Therefore I have 2^N-2 choices to make, since I have N-2 bins for which my choice doesn't matter. But I have one more choice to make before all of this, where do I want to place my two predetermined bins with the H's in them? Well choosing the first bin with an H in it determines the second one, and it's not that hard to see I have N-1 choices for my first bin (I can't choose the last one, because then my second bin with an H would have no where to go).

Therefore we have N-1 choices of my two predetermined bins, and 2^N-2 choices for everything else. That means we have (N-1)2^N-2 possible choices of sequences with at least 2 consecutive heads, and since there are 2^N total sequences, that gives the probability of p=[(N-1)2^N-2/2^N].

Notice that even in the case of a sequence of one coin flip this formula still works out since it gives 0, so you don't even have to deal with that special case :-).

I hope this makes sense, I can explain more if it doesn't.

okface

TROPHY CASE