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### AuthorTopic: Verifying or refuting eHalcyon's very old math  (Read 1161 times)

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##### Verifying or refuting eHalcyon's very old math
« on: November 04, 2017, 02:16:00 pm »
+11

There's an old and very bad thread on the board which I don't want to Necromancer. In it, eHalcyon makes some claims about probabilities. Here I do the math to see if those claims are accurate.

If you had five 5 coppers and 5 estates, there is just as much chance of you drawing ccccc as there is of you drawing ccece, even if one "looks" more random to the human mind.

If we put little stickers on the cards labeled "1" through "10", there would be 10! different orderings. But since we don't distinguish between the 5 estates nor the 5 coppers, there are only 10! / (5!*5!) = 252 = 2*2*3*3*7 orderings.

How many of those orderings start with "ccccc"? As many as there are (non-distinguished) ways of ordering the remaining cards, which is just 1 (that ordering being "eeeee").

The number of orderings that start with "ccece" equals the number of orderings which end with some reordering of "cceee". There are [5 choose 2] = 10 such orderings.

Hence, you're 10 times as likely to draw "ccece" than you are to draw "ccccc". You are even more likely to draw "3 coppers and 2 estates in any order" as you are to draw "ccece" in that specific order, because there are more orders (and all orderings are equally likely).

[D]rawing a hand of all actions, then a hand of all treasures followed by a hand of all greens seems really non-random to us.  But with perfect randomness, there is just as much chance of drawing these clumps as there is of getting three hands each with mixes of all three.

Restated, if you draw three batches each of five balls from a jar with five green, five white and five yellow balls, not putting the balls back between draws, you're equally likely to draw three monochromatic batches as you are to draw three rainbow batches (all colors represented).

The total number of orderings is 15! / (5!*5!*5!) = 756756.

There are six orderings of monochromatic batches: GWY, GYW, WGY, WYG, YGW, YWG. (This is shorthand for ggggg+wwwww+yyyyy etc.); 6 out of 756756 is a bit less than 1 in 100 000.

A rainbow batch has gwyXZ in some order, where X and Z are both members of {g, w, y}. X and Z can either be equal or different. This means that a rainbow batch either has three of one color and one of each of the rest, or two pairs and one loner.

Suppose the first batch is rainbow, with a color occurring thrice; let's say it's green, so it's gggwy in some order. There are 2g+4w+4y left. If all batches are rainbow, the remaining two batches are either [gwyyy+gwwwy], [gwwyy+gwwyy] or [gwwwy+gwyyy].

Note that gggwy has 5! / (3!*1!*1!) = 20 orderings, and gwwyy has 5! / (1! * 2! * 2!) = 30 orderings.

So there are 20 * (20*20 + 30*30 + 20*20) = 34000 orderings that start with gggwy. But picking green was arbitrary; there are just as many orderings that start with wwwgy or yyygw. (Having picked green, we shouldn't count the swap of white and yellow as separate, because that would double-count the gwyyy+gwwwy and gwwwy+gwyyy follow-ups. It would also double-count gwwyy+gwwyy, because it isn't different from gyyww+gyyww.)

So that's 3*34000 = 102000 orderings that start with a rainbow batch with one color repeated thrice.

If instead it starts with two pairs, let's say gwwyy in some order, the rainbow follow-ups are [gwwyy+gggwy], [ggwwy+ggwyy], [ggwyy+ggwwy] and [gggwy+gwwyy], where all hands can be reordered.

That's 30 * (30*20 + 30*30 + 30*30 + 20*30) = 90000 orderings. But this again is with green arbitrarily picked to be the starting loner, counting those with a white or yellow loner first (and the other two colors paired) gives us 3*90000 = 270000 orderings.

In total, that's 102 000 + 270 000 = 372 000 orderings out of 756 756, or approximately half.

So rainbow batches are about 1 in 2 compared to monochromatic batches which are about 1 in 100 000, so rainbow batches are about 50 000 times as common.

Conclusion

I think eHalcyon's quoted claims are incorrect. Here are some similar claims which I think are correct:
• The ordering ccccc eeeee is just a likely as the ordering ccece eceec.
• The ordering ggggg wwwww yyyyy is just as likely as the ordering wggyg wgwyw ywyyg.
I agree that monochromatic hands feel 'less random', whatever the hell that means, to most people (including me). I think eHalcyon's point is accurate, but slightly misstated, which in turns means that this whole thread is a bunch of pedantic nitpicking.

Am I doing f.ds'ing right?
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#### dedicateddan

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##### Re: Verifying or refuting eHalcyon's very old math
« Reply #1 on: November 04, 2017, 02:42:35 pm »
+17

I believe the claims you make are pedantic and correct. Welcome to f.ds!
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#### Gherald

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##### Re: Verifying or refuting eHalcyon's very old math
« Reply #2 on: November 04, 2017, 09:46:36 pm »
+2

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#### trivialknot

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##### Re: Verifying or refuting eHalcyon's very old math
« Reply #3 on: November 04, 2017, 11:42:05 pm »
+6

Maybe eHalcyon was thinking in Bose-Einstein statistics.
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##### Re: Verifying or refuting eHalcyon's very old math
« Reply #4 on: November 05, 2017, 05:56:03 am »
+1

I believe the claims you make are pedantic and correct. Welcome to f.ds!
Why thank you, I do my best to be pedantic

Maybe eHalcyon was thinking in Bose-Einstein statistics.
Obviously a joke. Is there also some element of truth? Is there some way of thinking in B-E stats such that eHalcyon's claims become true?

I don't know B-E stats, but the wikipedia page contained the word "quantum", so it's probably above my pay grade until I study for a promotion, which I won't do because I'd rather play Dominion
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#### trivialknot

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##### Re: Verifying or refuting eHalcyon's very old math
« Reply #5 on: November 05, 2017, 09:56:45 am »
+9

Yes it was a joke, but I'm told that jokes are funnier when explained.

Bose-Einstein statistics are a way of dealing with indistinguishable particles.  In classical statistics, if you have two particles you might label them 1 and 2, but in Bose-Einstein statistics you cannot give them separate labels.

Consider the situation where we have two particles and two boxes.  In classical statistics, there are 4 ways to distribute the particles:
12 | _
1 | 2
2 | 1
_ | 12
And if we assign each of these possibilities equal weight, then there's a 1/4 chance that the left box is empty.

In Bose-Einstein statistics, there are only 3 ways:
11 | _
1 | 1
_ | 11
And only 1/3 chance that the left box is empty.

If hands were boxes, and coppers were bosons, then splitting 5/2 would be just as likely as 3/4.  Or something.  Hands aren't really like boxes though, and reinterpreting Dominion in terms of Bose-Einstein statistics is fraught.  The joke is that not even quantum physics can rescue eHalcyon's claims.
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#### Witherweaver

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##### Re: Verifying or refuting eHalcyon's very old math
« Reply #6 on: November 09, 2017, 10:40:24 am »
+2

The gauntlet has indeed been thrown down.

And then picked up, and thrown down again, because the original throwing may have not conveyed the appropriate sense of challenge. And then adjusted because the placement of the thrown gauntlet on the ground did not appropriately capture the gravity of the situation. And then it was wondered, just how likely was it for the gauntlet to land on the ground in this way? Was this configuration any less likely than any other configuration? And then followed a long journey into gauntlet symmetries...
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#### Witherweaver

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##### Re: Verifying or refuting eHalcyon's very old math
« Reply #7 on: November 09, 2017, 11:22:04 am »
0

Yes it was a joke, but I'm told that jokes are funnier when explained.

Bose-Einstein statistics are a way of dealing with indistinguishable particles.  In classical statistics, if you have two particles you might label them 1 and 2, but in Bose-Einstein statistics you cannot give them separate labels.

Consider the situation where we have two particles and two boxes.  In classical statistics, there are 4 ways to distribute the particles:
12 | _
1 | 2
2 | 1
_ | 12
And if we assign each of these possibilities equal weight, then there's a 1/4 chance that the left box is empty.

In Bose-Einstein statistics, there are only 3 ways:
11 | _
1 | 1
_ | 11
And only 1/3 chance that the left box is empty.

If hands were boxes, and coppers were bosons, then splitting 5/2 would be just as likely as 3/4.  Or something.  Hands aren't really like boxes though, and reinterpreting Dominion in terms of Bose-Einstein statistics is fraught.  The joke is that not even quantum physics can rescue eHalcyon's claims.

But Coppers are already like Bosons, because we don't distinguish between the Coppers. I think you'd need Coppers and Estates to be identical Bosons.
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#### filovirus

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##### Re: Verifying or refuting eHalcyon's very old math
« Reply #8 on: November 09, 2017, 11:56:16 am »
0

"If you had five 5 coppers and 5 estates, there is just as much chance of you drawing ccccc as there is of you drawing ccece, even if one "looks" more random to the human mind."

This would be correct if each "c" and "e" were a specific card.

c1c2c3c4c5
c1c2e1c3e2

But because we really don't care which of the 5 c's is drawn in which order, then it is an irrelevant statistic for Dominion. As stated above, there is a difference between "combination" and "permutation".

FYI: 39.68% chance to draw cccee and 0.40% chance to draw ccccc in the OP's first post.
« Last Edit: November 09, 2017, 12:00:46 pm by filovirus »
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#### pacovf

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##### Re: Verifying or refuting eHalcyon's very old math
« Reply #9 on: November 09, 2017, 02:02:07 pm »
0

Yes it was a joke, but I'm told that jokes are funnier when explained.

Bose-Einstein statistics are a way of dealing with indistinguishable particles.  In classical statistics, if you have two particles you might label them 1 and 2, but in Bose-Einstein statistics you cannot give them separate labels.

Consider the situation where we have two particles and two boxes.  In classical statistics, there are 4 ways to distribute the particles:
12 | _
1 | 2
2 | 1
_ | 12
And if we assign each of these possibilities equal weight, then there's a 1/4 chance that the left box is empty.

In Bose-Einstein statistics, there are only 3 ways:
11 | _
1 | 1
_ | 11
And only 1/3 chance that the left box is empty.

If hands were boxes, and coppers were bosons, then splitting 5/2 would be just as likely as 3/4.  Or something.  Hands aren't really like boxes though, and reinterpreting Dominion in terms of Bose-Einstein statistics is fraught.  The joke is that not even quantum physics can rescue eHalcyon's claims.

But Coppers are already like Bosons, because we don't distinguish between the Coppers. I think you'd need Coppers and Estates to be identical Bosons.

We don't, but we could. With bosons, you can't, so you end up with different results. It's uh not intuitive.

Not like it matters to the current discussion.
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