Critical Thought Experiments

PeyoteReligion said:

At this point we are assuming that we get to continually choose and get shown an empty box. Op never specified if the leprechaun would be so generous or if at that point we need to make a 50/50 shot between your original pick and the other box not yet revealed. I'm pretty high right now, but how do your odds get any better than that? I bet that fucking leprechaun has all sorts of tricky maneuvers to not give the answer away. Btw, is t he already supposed to fork over the gold once you cantch him anyways?

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Yes he does. You always get shown an empty box regardless of which box you chose. If you pick an empty box he shows you the other empty box. If you pick the gold, he shows you one of the empty boxes. Your choice influences which box he shows you, but it is always empty.

You always switch. the 1/3 of the time you picked the right box originally you get fucked. But the 2/3 of the time you picked the wrong box from the start, you end up switching to the only other possible choice which is gold.

guy incognito said:

Wait a minute, you would only be sending 33,554,432 on week 26. You would be sending 67,108,864 on week 27.

2^(w-1) where w is the week number.

2^(1-1) = 1 for week 1,
2^(2-1) = 2 for week 2,
...
2^(26-1) = 33,554,432 for week 26.

So reverse that on week 1 you send 2^(26-1) = 33,554,432 letters, the next week you send 2^(25-1) and so on down to week 1 where you only send a single letter. So you end up with 2^25 + 2^24 + ... 2^0 which is (2^26)-1 = 67,108,863 total.


The final letter is already included in the series. it is 2E26-1, not 2E26

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The converging pyramid of letters for the first 25 weeks meets that criterion, yes. But the accident (unity chance, not part of the pyramid) adds one more letter for the round answer. cn

cannabineer said:

The converging pyramid of letters for the first 25 weeks meets that criterion, yes. But the accident (unity chance, not part of the pyramid) adds one more letter for the round answer. cn

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No, all 26 weeks count. the 25th week he sends 2 letters out, one is right and one is wrong. The one that is correct gets targeted for an accident filling the final letter in the series.

I just reviewed the OP and the mark did indeed receive 26 letters total. I concede the point. cn

Here is one I heard recently. I don't expect this to stump anyone, but it's fun to figure out.


There is a table with three boxes on it, each with an incorrect label. One box is labeled "Buds", another box is labeled "Bongs", and the last box is labeled "Buds & Bongs". You are allowed to pick one box and pull one item from it. How can you label each box correctly?

The box with buds and bongs in it must be the biggest box, so i would pick either the bud box, or the bongs box, the other smaller box will be the opposite of the box i picked.

Or are all the boxes the same size and weight?

I'm not good at these.

hehe

Nothing about the boxes themselves reveals the contents other than the labels.

Heisenberg said:

Here is one I heard recently. I don't expect this to stump anyone, but it's fun to figure out.


There is a table with three boxes on it, each with an incorrect label. One box is labeled "Buds", another box is labeled "Bongs", and the last box is labeled "Buds & Bongs". You are allowed to pick one box and pull one item from it. How can you label each box correctly?

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If each one is wrong, the one labeled 'bud and bongs' has to either be buds or bongs, not both. So, if you pull the buds out of it, the one labeled buds has to be bongs and the one labeled bongs has to be both buds and bongs.

Likewise, if you lift the 'buds and bongs' and its bongs, the 'bongs' box has to have buds under it, as the 'buds' box can't have buds under it.

Beefbisquit said:

If each one is wrong the one labeled 'bud and bongs' has to either be buds or bongs, not both. So, if you pull the buds out of it, the one labeled buds has to be bongs and the one labeled bongs has to be both buds and bongs.

Likewise, if you lift the 'buds and bongs' and its bongs, the 'bongs' box has to have buds under it, as the 'buds' box can't have buds under it.

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Correct. If the one labeled "Buds & Bongs" is wrong, then it can only have one or the other. Once you find out which it is and switch the correct label to it, you only have one move left you can do, which is to switch labels on the remaining two boxes.

Heisenberg said:

Correct. If the one labeled "Buds & Bongs" is wrong, then it can only have one or the other. Once you find out which it is and switch the correct label to it, you only have one move left you can do, which is to switch labels on the remaining two boxes.

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Reminded me of Sudoku strategy I use... lol

I'm confused, what if the 'buds and bong's box has bongs like you said. How come that makes the 'bongs' box, have to have buds, and the buds box have both buds and bongs? What if the bongs box has buds and bongs, and the buds box has bongs?

I just did it irl. you have the three [buds and bongs] [buds] [bongs] you looks in buds and bongs, there are buds. You remove the false label now you have -buds and bongs- [ ] [buds] [bongs] The problem is, you don't know which one to switch, who is to say [buds] has both buds and bongs to correctly switch the labels, thus making [bongs] labeled correct. Oh wait, if you know for a fact that they are ALL labeled incorrectly, then that means [bongs] must be false, and because you know the previous [buds and bongs] is actually [buds], you switch labels, making
[buds and bongs] = [buds]
[buds]= [bongs] and
[bongs]= [buds and bongs]
...right?

I was over thinking lol, I do that way too much I especially hated when they made test in school that you're too smart for and you start over analyzing what they say thinking "it can't be that simple" when it really was, would just have to deduce that every other answer was wrong. Still got the correct answer but it took a lot longer than needed -.-

Someone posted a similar riddle in T&T I think, but the boxes were labeled "apples", "oranges", "apples&oranges".
The rules were you could pull one article out of one box.
I could not figure it out. cn

Glad you guys found it stimulating. Reminded me a bit of the Monty Hall problem which is my favorite.

Heisenberg said:

Glad you guys found it stimulating. Reminded me a bit of the Monty Hall problem which is my favorite.

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Goddamn Monty Hall problem. It still fucks with me - even though I get the answer intellectually, it doesn't sit right with my gut...

tyler.durden said:

Goddamn Monty Hall problem. It still fucks with me - even though I get the answer intellectually, it doesn't sit right with my gut...

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I follow it.

It still seems strange that in the second set of decisions, that choosing the same door again wouldn't still constitute a 1/2 choice.

sworth said:

"...when you have eliminated the impossible, whatever remains, however improbable, must be the truth..."

Being psychic is impossible

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And the evidence to support said fact?
You can disprove false psychics. But not being psychic. Sorry.
Nice try though

SnakeByte said:

And the evidence to support said fact?
You can disprove false psychics. But not being psychic. Sorry.
Nice try though

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Yes, Psychics currently enjoy the same plausibility and level of evidence as gremlins, unicorns and The Kraken, which also can not be disproved.

Beefbisquit said:

I follow it.

It still seems strange that in the second set of decisions, that choosing the same door again wouldn't still constitute a 1/2 choice.

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It wouldn't. There has to be a switch in order to take advantage of the new odds. Imagine the original odds are 1/100. Monty reveals all doors but the one you chose, and one other. You see all doors that were opened had a goat. The one you chose has the same odds as before he revealed anything because your choice was arbitrary while his choice of which box to leave is not arbitrary. The one he left is either a goat or the real deal. He would only leave a goat IF you picked the car to begin with, so the chances of him leaving a goat are 1/100. If you chose a goat (99/100) he will leave the car. Now, imagine he doesn't give you a chance to change your pick, and just reveals the door he left, forcing you to stay with your original odds. The second thing that would force you to stay with your original odds is if he gives you a choice and you choose the same door; its no different than not having the choice. The door you picked is 1 of 100. The door he offers is 1 of 2. There has to be the action of switching before the odds change.

Lets say it's not doors, but instead he shows you a barrel containing 3 balloons and says one of them has glitter inside. He asks you to pull one out and place it on a table. He then pulls a magic balloon from his pocket and says IF the one you picked has glitter, this one will not. If the one you picked is empty, this one will have glitter. He puts it on the table and tells you to choose one. The magic balloon has the reverse odds of you. The odds you picked glitter are 1/3, making the odds of the magic balloon containing glitter, 2/3. When Monty shows you three doors and by process of elimination shows you that the remaining door's odds are chiral to yours, you have to switch to that door to take advantage of the magic.

^^ Still not sure I get it, Heis. Would you post an analogy using strippers?

Beefbisquit said:

I follow it.

It still seems strange that in the second set of decisions, that choosing the same door again wouldn't still constitute a 1/2 choice.

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The reason the probability changes is based on one important fact. The person revealing the other door has perfect information, which then gives you new information. IOW, Monty KNOWS which of the other door has the goat or the prize. By revealing the goat, he effectively created a new game.