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[MagmaMcFry]

Okay, let's just elaborate on Bauglir's interpretation: Drawing a single random ball will get you a metal ball 1/4 of the time.

In that case, let's again mark a single metal ball red and a single wooden ball blue. Let w be the likelihood of drawing the blue ball divided by the likelihood of drawing the red ball. In my earlier post I had assumed that the description meant w=3. With Bauglir's interpretation, this w changes. To find out this w, let's let p be the probability of drawing the red ball. Since the probability of drawing any metal ball is 1/4, we know that 13p=1/4, so p=1/52. Now the probability of drawing any red ball is equal to 3/4*1/7, but it is also equal to w*p, so w*p=3/28, so w = w*p/p = (3/28)/(1/52) =39/7.

Let's look at the bag after drawing a metal ball, it has 12 metal balls and 7 wooden balls remaining. Let's color another metal ball green again. Now the probability of drawing the blue ball is w times the probability of drawing the green ball, so if q is the probability of drawing the green ball, then w*q is the probability of drawing the blue ball. The probability of drawing any metal ball is equal to 12*q, and the probability of drawing any wooden ball is equal to 7*w*q, and since we know that 12*q+7*w*q=1, we get q = 1/(12+7*w) = 1/51, and the probability of drawing any metal ball is 12*q=12/51 = 4/17.

The total probability of drawing two metal balls is now (as before) equal to the probability of drawing a metal ball first (1/4), multiplied by the probability of drawing a metal ball second after drawing a metal ball first (4/17), so the result is now exactly 1/17 (not 15/247 as Bauglir said).

[frostshotgg]

If I'm understanding the problem correctly, it's saying if you had 1 of both kinds of balls, then it would be 1/4 chance for a metal ball. So when finding the total number of objects for the probability, treat wood balls as 3 objects. In that case, you'd have 3(3) + 17 balls, so 26, and then its just 17/26 + 16/25.

Edit: I suck at reading. Whatever, got super ninjad.

[Jopax]

And I'm officially giving up, I can't focus properly now, I see the numbers and I just can't connect them properly to other numbers, will have to try this in the morning with a fresh pair of eyes. I have other stuff to go over anyways. Thanks for the help guys.

[Bauglir]

Yeah, glancing at what MagmaMcFry just said it seems more reasonable than what I did. Better than arbitrary voodoo factors, which I feel like are a terrible way to be teaching statistics, by the way. Incidentally, if that coin-flipping problem gave 40, it was a coincidence. You've got a 1/10000 chance of any given sequence of 4 flips being all heads, so then you pretty much have to figure out the expected number of such sequences before that happens. Since you don't have to flip four times and start over (by which I mean, if your last two flips in a sequence of 4 and your first two in the next are heads, that works), I don't know how to do that off the top of my head, and with work in 2 minutes I don't have time to reason it out, but it seems analogous to the birthday problem.

Good luck.

[Jopax]

Oh it wasn't four heads in a row if that's what you meant.

It just meant, how many throws would it take to get four heads total, if the chance of getting heads is 1/10, fairly straightforward.

[frostshotgg]

Now that I'm not posting from a phone and can actually read the posts above mine, I *think* you guys may be looking at it wrong. The problem simply states that the odds of pulling a wood are 3 times that of a metal, it doesn't say in the arrangement of 13 metal to 7 wood. My interpretation is that in a bag of 1 each of the ball, 3/4 of the time you'd get the wood ball. So you treat wood as 3 objects, which (redoing my math from misreading in the last post), gives 3(7) + 13 for the total objects, or 34. Then you take the metal balls (which count as 1) over the total number of balls, and then again with 1 less metal. 13/34 + 12/33, which is 4/11, and I don't want to add fractions, but I think that's how you'd do it.

[MagmaMcFry]

Yeah, Jopax's coin-flipping problem is no coincidence at all. Let e be the expected amount of flips required to get the first heads, and p be the probability of a heads. Now if you throw a heads (with probability p), then your revised expected value is 1, and if you throw a tails (with probability 1-p), then your revised expected value is 1+e (since you're basically starting over again after wasting a turn). This means that e = (p)*1+(1-p)*(1+e), which simplifies to e = 1/p.

Now if p = 1/10, then e = 10. Let e₄ be the expected amount of flips required to get four heads, which is the expected amount of flips to get one head, then another, then another, then another. Since expected values are additive, the expected amount of flips to get four heads is equal to the expected amount of flips to get one head, plus the expected amount of flips to get another head, plus the expected amount of flips to get another head, plus the expected amount of flips to get another head. In short, e₄ = e+e+e+e = 10+10+10+10 = 40. More generally, if you have a coin that shows heads with a probability of p, then the expected amount of flips required to get k heads is exactly k/p.

As an encore, the expected amount E of flips to get four heads in a row is equal to 0.0001*4+0.0009*(4+E)+0.009*(3+E)+0.09*(2+E)+0.9*(1+E), which simplifies to E=11110. So you'll need on average 11110 flips to get four heads in a row.

Now that I'm not posting from a phone and can actually read the posts above mine, I *think* you guys may be looking at it wrong. The problem simply states that the odds of pulling a wood are 3 times that of a metal, it doesn't say in the arrangement of 13 metal to 7 wood. My interpretation is that in a bag of 1 each of the ball, 3/4 of the time you'd get the wood ball. So you treat wood as 3 objects, which (redoing my math from misreading in the last post), gives 3(7) + 13 for the total objects, or 34. Then you take the metal balls (which count as 1) over the total number of balls, and then again with 1 less metal. 13/34 + 12/33, which is 4/11, and I don't want to add fractions, but I think that's how you'd do it.

This was the interpretation I used in my first reply to the problem.

[da_nang]

For example, we had a coin flipping one, where the chance of heads was 1/10 and we were supposed to figure out how many the expected number of throws it took to get four heads. It was fairly straightforward and gave the expected answer of forty. I think the same should apply here, because if I've learned anything, it's that math shouldn't make real-world sense in most cases but instead works within any ruleset you give it as a sort of mental exercise.

FTFY.

[GavJ]

(however little sense that makes)

I honestly like your approach better Bauglir, since MMF's is just doubly confusing me. We just haven't had any problems so far that had you use both the real odds of something happening and the odds they gave you.

"Metal balls are 3x more likely" (or the reverse? I don't remember) is NOT necessarily "arbitrary."

Metal and wood have different properties. So depending on the method of drawing a ball, they could have significantly different weightings than their numbers alone imply. For example, if you shake up the group of them, and then release one ball from a hopper opening at the bottom.  Since metal is more dense, far more metal balls are going to settle near the bottom than the proportions alone imply.

Alternatively, if you shake them up, then winnow one off the top, it's disproportionately to their ratios going to be wooden.

You would only EXPECT it to have odds equal to the number of balls if the balls are all physically identical except for some superficial quality like paint color.

[Bauglir]

Welp. I suck. >________________>

brb learning statistics forever

[frostshotgg]

No amount of learning statistics can save you from ambiguously written questions.

[Vector]

.

[frostshotgg]

No amount of learning statistics can save you from ambiguously written questions.

GOOD LORD I am learning this right now. A bag has a course, textbook, and everything. Some things may be fucked.

ftfy

[da_nang]

Speaking of statistics... (well probability)

A challenge:

One of the formulations of the negative binomial distribution is the probability of the nth Bernoulli trial being a success given k-1 successes prior. If p∈(0,1) is the independent probability for a single success, then the PMF is
  (cf. coin tossing earlier)
with n = k, k+1, k+2, ... and k > 0. This can also be understood as the probability of getting k-1 successes in n-1 i.i.d. Bernoulli trials (Binomial distribution) followed by a success.

Now using this latter definition, let's extend it by assuming each trial is not in fact identically distributed but still independent. This changes it to the probability of getting k-1 successes of n-1 independent Bernoulli trials (Poisson-binomial distribution) followed by a success. The PMF would most likely be horrible, given the PMF of the Poisson-binomial distribution. In any case, what would the expected value and the variance be for the sequence p_i ∈ (0,1) representing the probability for success in the ith trial? Bonus points if you can do it for the closed interval p_i ∈ [0,1].

[Skyrunner]

The formula to convert angular velocity and linear velocity doesn't seem to match dimensionally.

angular velocity = linear velocity / r

But angvel is a scalar, isn't it? While linear velocity is a vector?