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09-19-2014 , 03:55 PM

#4226

Aesah

Verified Coach NLHE/PLO

Join Date: Sep 2010 Posts: 6,708

Quote:

Originally Posted by Avaritia

The Predictor says you should choose the second box because it has a goat.

Newcomb's isn't even in the same category (it's closer to "do you believe in God" than a math problem), and Monty Hall is trivial IMO

Quote

09-19-2014 , 03:55 PM

#4227

Dubey

old hand

Join Date: Jun 2006 Posts: 1,844

Theoretically, as long as there was a chance that I was going to pick 2, it is a +EV proposition. The chance that I pick 2 exists. The chance that the box has $800 in it does not exist, if the 2 boxes contain $200 and $400.

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09-19-2014 , 03:58 PM

#4228

Aesah

Verified Coach NLHE/PLO

Join Date: Sep 2010 Posts: 6,708

Dubey: I deleted what you originally responded to because while valid, I don't think it was the best way to present the argument. Let's say this instead:

So if the 2 boxes contain $200 and $400 (again, you opened the first box and saw that it had $400, and you know the other is either double or half) then is switching -EV, or is it neutral EV?

Quote

09-19-2014 , 04:01 PM

#4229

Avaritia

Confirmed 2500 hour haver

Join Date: Feb 2013 Posts: 12,248

Quote:

Originally Posted by Aesah

Newcomb's isn't even in the same category, and Monty Hall is trivial IMO

Haha I know I'm just familiar with the problem so I thought I'd drop a hint that I like paradoxes / math problems

Hint for the crowd:

Spoiler:

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09-19-2014 , 04:03 PM

#4230

runruled

newbie

Join Date: Aug 2010 Posts: 37

Always open the second box as well, it means we get to open both boxes in every situation. The outcome is 50%-50% because sometimes we end up with the larger amount and sometimes the smaller amount. but the larger amount is +400 and we only lose 200 when the other box is the lesser. that's always +EV right?

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09-19-2014 , 04:04 PM

#4231

Koss

Carpal \'Tunnel

Join Date: May 2005 Posts: 7,144

Quote:

Originally Posted by Aesah

Ev only applies when the outcome is random. This isnt random. Its not expected value anymore. Its not EV at all. Its just value?

Quote

09-19-2014 , 04:05 PM

#4232

Dubey

old hand

Join Date: Jun 2006 Posts: 1,844

I don't really know how to articulate what I am trying to say, but I know if you ran 1 million trials, and randomly picked a box every time, and switched every time, your overall "profit" at the end would be exactly the same as if you randomly picked a box and didn't switch every time.

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09-19-2014 , 04:07 PM

#4233

Aesah

Verified Coach NLHE/PLO

Join Date: Sep 2010 Posts: 6,708

Yes Avaritia that is a good hint!

EV can apply when the outcome is not random. If I hand you a $100 bill, your EV of that transaction was +$100 EV.

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09-19-2014 , 04:07 PM

#4234

Aleksei

adept

Join Date: Mar 2013 Posts: 1,116

Quote:

Originally Posted by Dubey

my thoughts:

If you are going to switch every time, then your EV should not be calculated by what might be in the second box. Your EV is an actual figure that is based on what is actually in the 2 boxes.

If the 2 boxes contain $200 and $400, your EV is $300, no matter which box you choose, and no matter whether you switch. Your EV doesn't change just because you picked the $400 first.

Therefore, as long as neither amount is indivisble by two, switching does not change your EV, and you can either switch, or not switch. It does not matter.

The catch is that we don't know if the boxes are supposed to contain $200 and $400. We only know that one box contains X and the other contains 2x, which is the point. If we're told this after we know the contents of one box, we still don't know if x+2x = 600, or x+2x = 1200. We're taking a 50% gamble that the unknown box contains either an amount x or an amount 4x, where x is half the known box's contents.

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09-19-2014 , 04:21 PM

#4235

dufusbrain

journeyman

Join Date: Mar 2013 Posts: 278

I would keep the first box. Notice in the OP that the man gives you the chance to switch only AFTER he sees you choose the first box. If you had chosen the other box, he may not have given a second chance. I do this if my read on him is that he wants me to pick the least amount.

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09-19-2014 , 04:24 PM

#4236

Aesah

Verified Coach NLHE/PLO

Join Date: Sep 2010 Posts: 6,708

^ I did forget to specify he will always ask you. Man you guys have lived in the scam-filled poker world for too long.

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09-19-2014 , 04:40 PM

#4237

kleinstein000

old hand

Join Date: Jun 2010 Posts: 1,339

Seems to me that yes you should always switch and the EV of the entire game is very easy to calculate as 1.5x where x is the value of one box and 2x is the value of the other

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09-19-2014 , 04:48 PM

#4238

Aleksei

adept

Join Date: Mar 2013 Posts: 1,116

Quote:

Originally Posted by Aesah

^ I did forget to specify he will always ask you.

Hmm.

If he will always ask you, then, implicitly, your EV is the same for either box, because before you open a single box the quantity of the two boxes is x+2x. Therefore your EV for either box is 1.5x, and since you don't know what X is, it is the same for either box.

So it makes no difference, EV wise, because you're calculating the EV of an unknown quantity. You're flipping a coin.

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09-19-2014 , 05:28 PM

#4239

Aesah

Verified Coach NLHE/PLO

Join Date: Sep 2010 Posts: 6,708

To recap the puzzle:

1) A guy shuffles two boxes, one with twice as much money as the other, and we randomly pick one. Switching appears to be +EV because let's say that the box we picked has X dollars. Because 50% of the time we randomly picked the box with less money and 50% of the time we randomly picked the box with more money, half the time we'll get 2X and half the time we'll get 0.5X if we switch for an EV of 1.25X, which is greater than X. So for example, we see our box has $400, half the time switching would give us $800, half the time switching would give us $200 for an EV of $500. So it seems like we should always switch no matter what.

2) However, the paradox is that we can clearly see using common sense that switching every time doesn't actually help us since we could have just picked the other box to begin with.

So how can we resolve the paradox presented by 1) and 2)?

Spoiler:

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09-19-2014 , 05:29 PM

#4240

runruled

newbie

Join Date: Aug 2010 Posts: 37

#2 doesn't make sense in that well yeah, we could have picked the "other box" but with no information of what the prize is we have no way of knowing which box to truly pick. which takes us back to #1, pick then switch.. ? If this is presented only one time then switch and pray. if presented over many occurrences than we switch and reap the rewards..

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09-19-2014 , 05:46 PM

#4241

Avaritia

Confirmed 2500 hour haver

Join Date: Feb 2013 Posts: 12,248

FWIW I think Monty Hall would surprise alot of people because of the quick logic/deducing people would use and then adamantly defend.

Quote

09-19-2014 , 06:29 PM

#4242

Dubey

old hand

Join Date: Jun 2006 Posts: 1,844

I was somewhat familiar with the Monty Hall Paradox, one of my profs in university brought up a similar problem but with more than 3 doors, but this two envelopes problem is quite different in my opinion, as there are no changing variables or "known" info in this problem.

I did google this problem and stumbled upon the two envelopes page on wikipedia after posting last in this thread but before Aesah posted his spoiler. Seems like the mathematics and logic community are about as confused as we were regarding the "solution" to this problem.

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09-19-2014 , 06:34 PM

#4243

Koss

Carpal \'Tunnel

Join Date: May 2005 Posts: 7,144

I too read the wiki and was more confused than when i started.

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09-19-2014 , 06:45 PM

#4244

Aesah

Verified Coach NLHE/PLO

Join Date: Sep 2010 Posts: 6,708

Ya that one is a really tough one to wrap your head around.

Here's an easier puzzle someone posted on CLP recently:

What's the worst 5-card hand that win at showdown 9-ways in PLO? To clarify, in a 2-way showdown of 2222 vs. KQJT on a board of 87654r, the winning 5-card hand would be 87622.

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09-19-2014 , 08:40 PM

#4245

tmckendry

Pooh-Bah

Join Date: Jul 2010 Posts: 4,208

Trips? Somewhere between high trips and low straight Is my guess Too lazy to solve. Actually, f that this question probably only exists because the answer is full house. I revise guess to full hizzle

Last edited by tmckendry; 09-19-2014 at 08:45 PM.

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09-19-2014 , 10:09 PM

#4246

Koss

Carpal \'Tunnel

Join Date: May 2005 Posts: 7,144

im drunk now but when i was sober i thinkni figured out that 22336 is possible. Not sure if it gets any better or not.

Quote

09-20-2014 , 01:42 AM

#4247

Aesah

Verified Coach NLHE/PLO

Join Date: Sep 2010 Posts: 6,708

22336 is not the optimal solution

And wtf Tyler, full house?! What if the board just comes 23456r you fish

Quote

09-20-2014 , 02:49 AM

#4248

Illiterate

old hand

Join Date: Mar 2010 Posts: 1,237

66447?

edit: last second realization

Quote

09-20-2014 , 02:52 AM

#4249

Aesah

Verified Coach NLHE/PLO

Join Date: Sep 2010 Posts: 6,708

77664 is not the optimal solution

Quote

09-20-2014 , 02:59 AM

#4250

dom80e

journeyman

Join Date: Jul 2009 Posts: 344

33K76

Board of 73322
Hand of K654
4 way chop.

Losing hands
QJT9
QJT8
QJ98
QT98
JT98

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