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I LOVE puzzeling, but this one question always bothered me. And for that I am not good at maths, I couldn't really work it out.

Here is the problem:

Say I have a 3000 pieces puzzle and I blindfold myself. Now I pick out two pieces at random from the box and try to connect them to one another. Bear in mind that every piece consists of 4 "connectors" and even if those two pieces actually were to belong together, I would still have to connect the two correct "connectors" out of the 8 possible ones.

Now, those two pieces don't fit and so I put them back into the box. I only get one chance each time to connect the right pieces at the right connector, if I ever fail I have to disassemble all the pieces I have found matching yet and again start from scratch. Also, once I found two pieces that fit, if I pick another piece from the box, chances become more and more likely that this piece actually goes with those that already lie on the floor since there will always be open connectors UNLESS of course this piece is a corner pieces. Also, once two pieces are connected, the remaining connectors are now 6 out of 8. While connecting another piece to the already connected once, the remaining connectors become 3 out of 4 ( since the start is pretty much an exception already...)

So what are my chances of randomly picking the right pieces with the right connectors and connecting them in the right order one by one without ever messing it up even once until all 3000 pieces sit correclty?

I hope I made this clear, feel free to ask if I didn't.

Good luck! :-)

Little Edit:

Also, I am not requiered to get to pieces from the box. I can get two pieces from the box and try go connect them, or I can get one piece from the box and try to connect it to the ones on the floor. But at some point, I will have many different 2x2 pieces on the floor and those as well have to be connected to one another. So at some point I will have to pick up or rather move to sets of pieces on the floor around and make those connect....

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    $\begingroup$ How many different connectors there are? Is it just two types (one going in and one going out)? Or are there several types? $\endgroup$ – Pedro A Nov 7 '18 at 11:03
  • $\begingroup$ Now, just in and out :) $\endgroup$ – innomotion media Nov 7 '18 at 12:04
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    $\begingroup$ So that's how the protagonist of Greg Egan's Quarantine passes his spare time. :) $\endgroup$ – Calum Gilhooley Nov 7 '18 at 12:09
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assuming each pieces has only one correct place to be, then there are 3000 places, the total possible ways to put the puzzle together is 3000! so $\frac {1}{3000}$ to get first piece right, for the second piece $\frac {1}{2999}$, ..., multiplying the probabilities that is $\frac {1}{3000!}$ chance of getting the puzzle by just luck.

Update : if each piece has 4 connectors then the answer is $\frac {1}{12000!}$

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    $\begingroup$ I think this is not taking all parameters into consideration ;:) $\endgroup$ – innomotion media Nov 7 '18 at 12:09
  • $\begingroup$ @innomotionmedia : Try it with a puzzle of 6 pieces and show me what it is missing, it is better to give an example than to speculate for free. $\endgroup$ – Arjang Nov 8 '18 at 4:53
  • $\begingroup$ Well, I believe that when you find two pieces and put them on the floor you have two different sets of pieces to chose from. But not both sets are equally likely since the two on the floor are just two pieces, while in the box there are 997 pieces remaining... $\endgroup$ – innomotion media Nov 8 '18 at 10:00
  • $\begingroup$ @innomotionmedia just try it wit 9 pieces and then show what you mean $\endgroup$ – Arjang Nov 8 '18 at 10:17
  • $\begingroup$ If there are $N$ tab-slot pairs, treated as independent (in reality, they aren't), then after mating $m$ pairs, there is a one in $N - m$ chance of successful blind mating of the next chosen tab or slot; so $\frac{1}{N!}$ seems a reasonable approximation to the probability. It's pessimistic, but it doesn't look easy to make it more realistic; and a lower bound is still worth having. $\endgroup$ – Calum Gilhooley Nov 8 '18 at 15:55

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