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Author Topic: Math request: Nomad Camp  (Read 43786 times)

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DStu

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Re: Math request: Nomad Camp
« Reply #125 on: January 23, 2013, 08:15:42 am »
0

I wouldn't. The point is, you never actually care about both questions - you only ever care about one - so it doesn't really make sense to compare them. I mean, show me a situation where the difference is important?

It shows that you have to be carefull with conditional probabilities.  Of course, in a given setting, there is only one correct method, but knowing that you have to be carefull and that you can't apply the method from yesterday just because the problems looks similar is something worth knowing...
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AdamH

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Re: Math request: Nomad Camp
« Reply #126 on: January 23, 2013, 08:44:01 am »
+1

OK you win. The rules of the happy little universe I live in don't seem to apply when you get into this business. I think the same goes for Goons games.
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DStu

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Re: Math request: Nomad Camp
« Reply #127 on: January 23, 2013, 08:48:13 am »
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I don't want to win, I want to understand...
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AdamH

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Re: Math request: Nomad Camp
« Reply #128 on: January 23, 2013, 08:57:06 am »
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When I went back to try and figure it out it didn't make any sense. The numbers are all there and they make sense when you just put them down from what the problem tells you to do. It doesn't make any sense to me that it works out that way, but it's the same thing with Goons: doing what makes sense to me is clearly wrong.
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DStu

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Re: Math request: Nomad Camp
« Reply #129 on: January 23, 2013, 09:00:38 am »
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When I went back to try and figure it out it didn't make any sense. The numbers are all there and they make sense when you just put them down from what the problem tells you to do. It doesn't make any sense to me that it works out that way, but it's the same thing with Goons: doing what makes sense to me is clearly wrong.
And I think exactly that is the point of this puzzle. When I think about how that should be, I also don't really see why it should make a difference. When I think longer, I see a reason, but it is the wrong way, i.e. the wrong case gets the higher probability.

But it's not because the puzzle is poorly worder, I don't think you can make the difference any more clear without allcaps, but because intuition clearly doesn't help much here, you just have to count.
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GendoIkari

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Re: Math request: Nomad Camp
« Reply #130 on: January 23, 2013, 09:05:59 am »
0

I'm very tempted to now take a picture of me wearing my shirt that says '2+2=5 (for extremely large values of 2)' but alas, it's in the wash.

So you're saying it's missed the reshuffle?
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theory

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Re: Math request: Nomad Camp
« Reply #131 on: January 23, 2013, 09:40:46 am »
+1

Maybe here's a better, more intuitive explanation.

In one, we are looking for, # hands with at least two Aces / # hands with at least one Ace

and in two, we are looking for

# hands with at least two Aces (one of which is the Ace of Spades) / # hands with the Ace of Spades

In the second problem, the world in which you are operating is slightly different.  There are fewer ways to have hands with 2 A's if one of them has to be the Spade A, and even though there are fewer hands with the Spade A, it doesn't cancel out perfectly.

It is very similar to that "two children one is a boy" vs "two children elder is a boy" question.
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HiveMindEmulator

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Re: Math request: Nomad Camp
« Reply #132 on: January 23, 2013, 12:12:42 pm »
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The way I understand it is this:

The confusion comes from you thinking of the problem as first taking an Ace (or the Ace of Spades), then drawing 12 other cards. In this formulation, it doesn't matter what the first Ace is.

But the real scenario is more like dealing out all the cards into 4 13-card hands and finding the one with the Ace of Spades vs finding the ones with Aces, and then asking if there are 2 Aces in that chosen hand. Finding the hand with the Ace of Spades is just like the previous formulation. It's the Spade Ace and 12 other random cards. But if you are just trying to find a hand with any Ace, you are more likely to land in a 2-Ace hand, since the 2-Ace hands have more Aces in them.
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eHalcyon

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Re: Math request: Nomad Camp
« Reply #133 on: January 23, 2013, 01:19:47 pm »
+1

Part (a) is pretty straightforward.  Take an ace out, and you have 51 cards, 3 of which are aces, so that's 1/17, and then the question is basically just what's the probability of an ace being in 12 of those , so 1/17 * 12/17.

It's not quite like that.  If we work with unordered hands then there are (52 choose 13) hands in total, of which (48 choose 13) have no aces.  So (52 choose 13) - (48 choose 13) hands contain at least one ace.  All we know is that our hand is one of these.

Of these, 4 x (48 choose 12) hands contain exactly one ace.  So, courtesy Wolfram Alpha, the probability of at least two aces given that we have at least one ace is about 0.5071.

For (b), the number of hands containing the ace of spades is (51 choose 12).  Of these, (48 choose 12) contain no additional aces.  So, again courtesy Wolfram Alpha, the probability of at least two aces given the ace of spades is about 0.5612.

Your analysis for (a) is closer to the correct argument for (b).  You are indeed drawing 12 cards from 51, of which 3 are aces, so the average number of aces you expect to draw is 12/17.  But sometimes you'll get more than one ace, so the probability of at least one ace has to be a bit lower to compensate.

I think you input the equation incorrectly on Wolfram Alpha.  The denominator there should have (52 choose 13), not (52 choose 12), right?  With the change, the probability is 0.3696, not 0.5071.

The way I understand it is this:

The confusion comes from you thinking of the problem as first taking an Ace (or the Ace of Spades), then drawing 12 other cards. In this formulation, it doesn't matter what the first Ace is.

But the real scenario is more like dealing out all the cards into 4 13-card hands and finding the one with the Ace of Spades vs finding the ones with Aces, and then asking if there are 2 Aces in that chosen hand. Finding the hand with the Ace of Spades is just like the previous formulation. It's the Spade Ace and 12 other random cards. But if you are just trying to find a hand with any Ace, you are more likely to land in a 2-Ace hand, since the 2-Ace hands have more Aces in them.

qmech's math says that the probability of having more aces is higher given that you have the Ace of Spades, compared to just having an Ace.  Is there something wrong with his math?  It's been a lot time since I've done combinations, but his work seems correct to me.
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HiveMindEmulator

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Re: Math request: Nomad Camp
« Reply #134 on: January 23, 2013, 03:29:29 pm »
0

The way I understand it is this:

The confusion comes from you thinking of the problem as first taking an Ace (or the Ace of Spades), then drawing 12 other cards. In this formulation, it doesn't matter what the first Ace is.

But the real scenario is more like dealing out all the cards into 4 13-card hands and finding the one with the Ace of Spades vs finding the ones with Aces, and then asking if there are 2 Aces in that chosen hand. Finding the hand with the Ace of Spades is just like the previous formulation. It's the Spade Ace and 12 other random cards. But if you are just trying to find a hand with any Ace, you are more likely to land in a 2-Ace hand, since the 2-Ace hands have more Aces in them.

qmech's math says that the probability of having more aces is higher given that you have the Ace of Spades, compared to just having an Ace.  Is there something wrong with his math?  It's been a lot time since I've done combinations, but his work seems correct to me.

Actually my intuitive explanation is is wrong AND qmechs typing into Wolfram is wrong.

You're more likely to find any specific Ace (including the Ace of Spades) in a 2 Ace hand than you are to find it in a 1 Ace hand, so P(2 Aces | Spade A) should be larger. If you're looking for a hand with at least one Ace, you're not more likely to count the hands with multiple Aces than the ones with only one Ace.

So P(2 Aces | specific A) = 0.56
and P(2 Aces | at least one A) = 0.37 (qmech's expressions are right, but entered into wolfram wrong).
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dnkywin

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Re: Math request: Nomad Camp
« Reply #135 on: January 23, 2013, 03:50:21 pm »
+2

f.ds - turning a simple math question into a 6 page argument on probability

>.<
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HiveMindEmulator

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Re: Math request: Nomad Camp
« Reply #136 on: January 23, 2013, 05:04:48 pm »
+1

You're more likely to find any specific Ace (including the Ace of Spades) in a 2 Ace hand than you are to find it in a 1 Ace hand, so P(2 Aces | Spade A) should be larger. If you're looking for a hand with at least one Ace, you're not more likely to count the hands with multiple Aces than the ones with only one Ace.

So P(2 Aces | specific A) = 0.56
and P(2 Aces | at least one A) = 0.37 (qmech's expressions are right, but entered into wolfram wrong).

I wanted to clarify my intuitive argument by giving the example of a typical deal of cards. Typically, you'll find 1 hand with 2 Aces, 2 with 1 Ace, and 1 with none. Of the hands with Aces, 1/3 have 2 Aces (probability ~ 0.37). But half the Aces are in hands with 2 Aces, so if you look for a specific Ace, you'll find it in a hand with 2 Aces 1/2 the time (probability ~ 0.56).
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jeb56

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Re: Math request: Nomad Camp
« Reply #137 on: January 23, 2013, 05:27:06 pm »
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This is another way to frame the argument:

Take four copies of the distribution for a specific Ace (one each for Spade. heart, Diamond, and Club Aces), and mix them together.  This resulting distribution has one copy of each hand with one Ace, two copies of each hand with two Aces, three copies of each hand with three Aces, and four copies of each hand with four Aces.

Compare that with the distribution that has one copy (only) of each hand that contains an Ace.

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qmech

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Re: Math request: Nomad Camp
« Reply #138 on: January 26, 2013, 05:41:03 pm »
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I think you input the equation incorrectly on Wolfram Alpha.  The denominator there should have (52 choose 13), not (52 choose 12), right?  With the change, the probability is 0.3696, not 0.5071.
And you correctly corrected the Alpha typing, but then reported the wrong change! :P

Conclusion: typing is hard.
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eHalcyon

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Re: Math request: Nomad Camp
« Reply #139 on: January 26, 2013, 06:09:44 pm »
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I think you input the equation incorrectly on Wolfram Alpha.  The denominator there should have (52 choose 13), not (52 choose 12), right?  With the change, the probability is 0.3696, not 0.5071.
And you correctly corrected the Alpha typing, but then reported the wrong change! :P

Conclusion: typing is hard.

Wait, what did I do wrong then? I am lost now!
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Morgrim7

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Re: Math request: Nomad Camp
« Reply #140 on: January 26, 2013, 06:46:01 pm »
0

f.ds: Where people spend six pages of discussion talking about the odds of one card.
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ycz6

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Re: Math request: Nomad Camp
« Reply #141 on: January 26, 2013, 08:42:55 pm »
+1

Ain't you guys ever been on the Internet before? I don't think I've seen a forum discussion or comment thread about a riddle in probability that didn't turn into a long-ass argument about conditional probability where several different people on both sides try to use intuitive arguments to make their point but end up failing, and I've seen a few.

Maybe I hang out around the wrong parts of the Internet.
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qmech

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Re: Math request: Nomad Camp
« Reply #142 on: January 27, 2013, 05:55:22 am »
0

(52 choose 13), not (52 choose 12)

That one was right: I had (48 choose 12) when I should have had (48 choose 13).

Hopefully that's an end of this now!
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eHalcyon

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Re: Math request: Nomad Camp
« Reply #143 on: January 27, 2013, 10:57:24 am »
+1

(52 choose 13), not (52 choose 12)

That one was right: I had (48 choose 12) when I should have had (48 choose 13).

Hopefully that's an end of this now!


And then the answer I reported was still correct... Wow. Typing IS hard! :P
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