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June 14

T-shirt

At [1], they have a t-shirt that poses the question, "If you consider the set of all sets that have never been considered, will it disappear?" Would it be correct to say that such a collection (for some reasonable definition of "considered") would form a proper class? Black Carrot 18:00, 14 June 2007 (UTC)[reply]

Seems to me the super-set (= the set containing the sets that have never been considered) is not a member of itself, because it has been considered. In the process of choosing the sets that form the super-set, is it possible not to consider them? I kinda doubt it. That suggests the super-set is inherently empty, but still existent. -- JackofOz 00:39, 15 June 2007 (UTC)[reply]

I'm not sure, but perhaps we end up in metamathematics (statements about statements, as in "This sentence is false.") or violating one of the set axioms (i.e., trying to define something that doesn't really qualify as a set). —Bromskloss 08:15, 15 June 2007 (UTC)[reply]

Since it seems reasonable to claim that we can only consider at most countably many sets (can you consider an object without being able to describe it?), the class of all unconsidered sets will presumably be a proper class. If we use a weaker definition of considered, it might even be the empty set (I considered all sets just now, while thinking about which ones have been considered). Algebraist 10:54, 15 June 2007 (UTC)[reply]

Which brings us neatly back to Black Carrot's "reasonable definition of considered". That seems to be the stumbling block. I can assert that I have just now "considered" all the individual atoms in the universe, but does it mean anything? Not to me, at least. -- JackofOz 11:04, 15 June 2007 (UTC)[reply]

I can assure you, Sir, that this, however, means a great deal to all those individual atoms out there. On behalf of the individual atoms in the universe, thank you for your kindness in considering us. --Lambiam ^Talk 13:22, 15 June 2007 (UTC)[reply]

Consider yourselves considered. :) -- JackofOz 13:42, 15 June 2007 (UTC)[reply]

Puuhh! How did you do that so easily, without breaking sweat? It took me forever to get it done. I need a rest. Oh, and another thing. Were they uncountable? I lost count. —Bromskloss 20:04, 15 June 2007 (UTC)[reply]

I'm being told that they aren't even infinite! [2] I guess it didn't take me forever, then, but it sure felt like it! —Bromskloss 20:08, 15 June 2007 (UTC)[reply]

So saying that for each real number

x

, there's the set

\{x\}

doesn't qualify? Perhaps I should really stop by at each set, hat in hand, and say it like I mean it? ;-) Like I did with the atoms (though I suspect JackofOz cheated). —Bromskloss 20:14, 15 June 2007 (UTC)[reply]

There seems to be some disagreement. Call the set (if such it be) of all sets that have never been "considered" A. As far as I can see, there are three options:

1. We've considered (in some sense) all sets already, perhaps by stating the axioms of set theory, perhaps merely by trying to find A, in which case A is empty and breaks no rules, hence is a well-defined set.

2. We have not considered all possible sets, where "considered" perhaps means "wrote down a rule individually defining". Then most sets have not been considered. As the collection of all sets forms a proper class, A would have to as well.

3. We have not considered A, where "considered" perhaps means "wrote down a rule individually defining each element of", and have neither considered nor individually defined some other set (making A nonempty), in which case A contains itself. Is a proper class allowed to contain itself? Black Carrot 17:44, 16 June 2007 (UTC)[reply]

By the way, the person above who complained about my "reasonable definition" had a fair point. As there are apparently several reasonable definitions available, but infinitely many trivial variations on each, let's rephrase it as "for each reasonable definition, up to isomorphism." :) Black Carrot 17:50, 16 June 2007 (UTC)[reply]

Regarding your question: No, classes are collections of sets, not of classes. A proper class is not a set, so it cannot be in a class, thus it cannot be in itself. -- Meni Rosenfeld (talk) 10:15, 17 June 2007 (UTC)[reply]

How hard is it to get a six

PUZZLE: How hard is it to get a 6? You have a single normal unbias 6 sided dice.

You can roll the six sided dice as many times as you want (including zero times) and add up all the results.

You can chose when to stop rolling.

The objective is to get a total result of 6.

You are NOT allowed to exclude the result of any roll.

What is the probability of obtaining a 6?

I tried a simpler game of getting a 1. The probability is 1/6

I tried a simpler game of getting a 2. The probability is 1/6 + 1/6 * 1/6

But six is hard. 202.168.50.40 22:00, 14 June 2007 (UTC)[reply]

Well, let's think about what the optimal strategy is, which it seems to me to be "roll until you are at, or over, 6" since if you've got less than 6 you've still got a chance of reaching it. Once you've decided that, let's see what happens. Either:

You get a 6, and you stop.
You get something less than 6, and you roll again.

If you roll again, then there is exactly one number on the die that will give you a total of 6, and up to 3 (why?) that will keep you under 6, allowing you to reroll.

Hopefully this should not be too bad a calculation, and I'll give it a shot now just to see. Confusing Manifestation 22:56, 14 June 2007 (UTC)[reply]

OK, I've tried it, and while my method may not have been the most efficient I brought it down to 18 separate probabilities in a form like "1 1 1-3 3-1 = 3/6^4", meaning that there's a 3/6^4 chance that you roll a 1, then a 1, then a 1 2 or 3 followed by a 3 2 or 1 respectively. I get a total probability of roughly 0.365, if anyone wants to check me. (It might have been easier to work with complementary probabilities - i.e., the probability that on any particular roll you have more than 6, especially since you could then probably write them all in terms of rolling six dice but only requiring the first n to add to 6). Confusing Manifestation 23:09, 14 June 2007 (UTC)[reply]

The way I'd approach this would be to find the partitions of 6:

6

5+1

4+2

4+1+1

…

1+1+1+1+1+1

The probability of getting any of these would be

{\frac {c}{6^{n}}}

where

c

is the number of ways to arrange the numbers (ignoring duplicates) and

n

is the number of rolls in that particular sum.

For example, for the sum 2+2+1+1, we have

{\frac {4!}{2!2!}}=6

so that particular probability would be 1/216. We add up all of these and get our overall probability. Donald Hosek 00:04, 15 June 2007 (UTC)[reply]

Oops, I just realized I did a different problem. I showed how to get exactly 6. Donald Hosek 00:05, 15 June 2007 (UTC)[reply]

Isn't that the problem? To get not exactly 6 is rather easy. --Lambiam ^Talk 00:15, 15 June 2007 (UTC)[reply]

So it is, the problem also, though is that the question isn't well-defined. You can find each of the individual probabilities, but you can't use them to get an overall probability since the total number of rolls is not defined. In fact, as the problem is currently stated, the probability of getting a 6 is (asymptotically) 1. (I say asymptotically in that there's no guarantee that you won't roll, e.g., 4-3-4-3 for an infinite number of rolls, but as the number of rolls approaches infinity, the probability gets closer and closer to 1. But you can use the method that I outlined above to at least see the different ways to roll a 6 and get the probabilities with

n

die rolls. Donald Hosek 00:29, 15 June 2007 (UTC)[reply]

The way I interpret it: keep rolling until you feel like stopping. If then you have a total of exactly six, you score. What is the probability, under optimal play, of scoring? I conjecture that the probability of succeeding in getting (exactly) n, where 1 ≤ n ≤ 6, is equal to 7ⁿ⁻¹/6ⁿ. For n = 6, this is 16807/46656 = 0.36023... . I haven't proved this. --Lambiam ^Talk 00:33, 15 June 2007 (UTC)[reply]

I have now also proved this; there is an easy recurrence relation:

6p_n = 1 + Σ_{i < n} p_i.

--Lambiam ^Talk 00:49, 15 June 2007 (UTC)[reply]

The way I resolve the problem is to turn the problem into a similar problem.

Instead of getting to the score 6, I change the problem to starting with the number 6 (what I call the pot) and trying to get to the number 0 by subtracting the result of the dice roll(s) from the pot.

For example:

6 - "roll a 6" = 0 (with probability 1/6)

6 - "roll a 5" = 1 (with probability 1/6)

6 - "roll a 4" = 2 (with probability 1/6)

6 - "roll a 3" = 3 (with probability 1/6)

6 - "roll a 2" = 4 (with probability 1/6)

6 - "roll a 1" = 5 (with probability 1/6)

Let pr[n] be the probability of getting to a pot of 0 when currently you have a pot of n

Thus

pr[0]=1

pr[-1]=0 Once you go negative, you can never get to ZERO

pr[-2]=0

pr[negative number]=0

So

pr[6] = 1/6 * pr[0] + 1/6 * pr[1] + 1/6 * pr[2] + 1/6 * pr[3] + 1/6 * pr[4] + 1/6 * pr[5]

pr[1] = 1/6 * pr[0] + 1/6 * pr[-1] + 1/6 * pr[-2] + 1/6 * pr[-3] + 1/6 * pr[-4] + 1/6 * pr[-5]

pr[1] = 1/6 * pr[0]

pr[2] = 1/6 * pr[1] + 1/6 * pr[0] + ( 1/6 * pr[-1] + 1/6 * pr[-2] + 1/6 * pr[-3] + 1/6 * pr[-4] )

pr[2] = 1/6 * pr[1] + 1/6 * pr[0]

pr[3] = 1/6 * pr[2] + 1/6 * pr[1] + 1/6 * pr[0] + ( 1/6 * pr[-1] + 1/6 * pr[-2] + 1/6 * pr[-3] )

pr[3] = 1/6 * pr[2] + 1/6 * pr[1] + 1/6 * pr[0]

And so on

pr[6] = 16807/46656

er? I see a pattern here.

pr[0] = (1/6)^0

pr[1] = (1/6)^1

pr[2] = (1/6)^2 + (1/6)^1

pr[3] = (1/6)^3 + (1/6)^2 + (1/6)^1

pr[4] = (1/6)^4 + (1/6)^3 + (1/6)^2 + (1/6)^1

pr[5] = (1/6)^5 + (1/6)^4 + (1/6)^3 + (1/6)^2 + (1/6)^1

pr[6] = (1/6)^6 + (1/6)^5 + (1/6)^4 + (1/6)^3 + (1/6)^2 + (1/6)^1

opps! I think my pattern is wrong.

202.168.50.40 01:19, 15 June 2007 (UTC)[reply]

For some reason, I'm getting ~~0.171123~~ 0.19876, which doesn't match anyone's. Let me look at it again... --Wirbelwindヴィルヴェルヴィント (talk) 01:41, 15 June 2007 (UTC)[reply]

Ok, assuming the problem isn't a trick question and that the ratio is 0 (cause you roll an infinite times every time, so you never get a 6), then I'm pretty sure that ~~0.171123~~ 0.19876 is right. Have a look at my sandbox, where the Xs represent a roll that adds to 6, and the 0s represent no roll. I'd paste the results here, but that's way too long. --Wirbelwindヴィルヴェルヴィント (talk) 02:02, 15 June 2007 (UTC)[reply]

Oh, and to clarify, on the results I did, once you roll a sum of 6 or greater, you stop. And if you haven't gotten there, you roll again. Thus, I guess that's the optimal probabilities. And oh, fixed the ratios because I had runaway 7s in my rolls the first time. --Wirbelwindヴィルヴェルヴィント (talk) 02:05, 15 June 2007 (UTC)[reply]

The reason this returns the wrong result is that the 161 rolls in your list don't have equal probabilities. You're much more likely to roll 6-0-0-0-0-0 than 1-1-1-1-1-1, for example. -GTBacchus^(talk) 02:36, 15 June 2007 (UTC)[reply]

I'm getting ${\frac {1}{6}}+{\frac {5}{6^{2}}}+{\frac {10}{6^{3}}}+{\frac {10}{6^{4}}}+{\frac {5}{6^{5}}}+{\frac {1}{6^{6}}}={\frac {16801}{46656}}\approx 0.36010374$ .

The first term is the probability of winning in one roll, the second term is the probability of winning in (exactly) two rolls, etc. I'm assuming the optimal strategy is to keep rolling until you either win or lose. I found this by simply drawing the complete tree of possibilities, writing down a probability for each branch, and adding up all the wins. I'm not sure why the binomial coefficients are popping up there.

Someone could program a computer to play the game a million or so times and see whether the empirical frequency agrees with any of our predictions. -GTBacchus^(talk) 02:29, 15 June 2007 (UTC)[reply]

As an example, the 10 ways of winning in exactly three rolls are: 4-1-1, 3-1-2, 3-2-1, 2-1-3, 2-2-2, 2-3-1, 1-1-4, 1-2-3, 1-3-2 and 1-4-1. Each of those occurs with a probability of 1/216. -GTBacchus^(talk) 02:33, 15 June 2007 (UTC)[reply]

Computationally, I'm getting ~0.3602. That's by rolling until the total equals or exceeds six. →Ollie (talk • contribs) 03:44, 15 June 2007 (UTC)[reply]

OK, I concur with the 0.3601 - I'd missed a term in the 1/6^5 lot and double-counted one in the 1/6^3, which gave me the 0.365 I stated above. Confusing Manifestation 04:43, 15 June 2007 (UTC)[reply]

I think that ${\frac {1}{6}}+{\frac {5}{6^{2}}}+{\frac {10}{6^{3}}}+{\frac {10}{6^{4}}}+{\frac {5}{6^{5}}}+{\frac {1}{6^{6}}}={\frac {16807}{46656}}$ . That's 16807/46656 202.168.50.40 04:48, 15 June 2007 (UTC)[reply]

You're right, of course. I somehow missed that 6 in the ones place in the first term. Thanks. -GTBacchus^(talk) 09:27, 15 June 2007 (UTC)[reply]

... which is the answer I gave four hours earlier, together with a more general form and a proof. --Lambiam ^Talk 06:55, 15 June 2007 (UTC)[reply]

\sum _{n=1}^{6}{}^{n}C_{r}/r^{n}

?

So you did. Isn't that interesting, that

\sum _{n=0}^{5}{^{5}C_{n}6^{5-n}}=7^{5}

, when you add it up properly? There must be some more general identity lurking behind that. -GTBacchus^(talk) 09:27, 15 June 2007 (UTC)[reply]

\sum _{n=0}^{5}{^{5}C_{n}6^{5-n}}=\sum _{n=0}^{5}{^{5}C_{n}6^{n}}=(1+6)^{5}=7^{5}

...binomial theorem (or did I miss the joke ?). Gandalf61 09:52, 15 June 2007 (UTC)[reply]

Huh... I guess the joke is that I was fairly drunk for my above post. I guess applying the binomial theorem to 1 and 6 just didn't occur to me. I still had fun working out the probability. -GTBacchus^(talk) 10:32, 15 June 2007 (UTC)[reply]

With a quick simulation, I've got an approximate of 0.360 as the probability. I'm lazy to compute an exact result now. – b_jonas 15:47, 16 June 2007 (UTC)[reply]