Safecracker Tokens?

Baron Rubik

New member
Mar 21, 2013
1,852
1
Played about 10 games without getting a token, then got 2 in one game.
Nice how you can view your collection and get duplicates too.

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Kolchak357

Senior Pigeon
May 31, 2012
8,102
2
Looks very good. I like how they did this.
It will be cool to see the coins up close. My local safecracker doesn't have coins in it and it's the only Safe Cracker I've ever played.
 

kinggo

Active member
Feb 9, 2014
1,024
0
To tell the truth, that's exactly where I'm at. The token thing is neat, but the table itself, meh.
I like AotV, those 90s when I actually play pinball. But keeping the ball alive just to play something ludo-like really isn't much fun.
 

eckless

New member
Sep 11, 2012
76
0
Collecting all 26 tokens is a wizard goal. That just seems tedious. Not difficult, but it might take a little while.
So do any mathletes here want to figure out the expected number of tokens you will need to win to achieve this wizard goal? (Assuming the tokens are equally likely and there is an infinite supply of all 26?)
 
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Kaoru

New member
Mar 29, 2012
230
0
I quite like this table, especially because it offers something a little different. A round might not take too long, but the rules are still deep enough.

I only wish Farsight would stop doing these luck based wizard goals. Collecting the full token set? I'm sorry, that has nothing to do with skills. More with praying for the moment in which the right stars perfectly align. Reminds me bit of that "Collect 5 aces" or whatever it was in "Lights Camera Action". When does something like that ever happen?
 

Zaphod77

Active member
Feb 14, 2013
1,316
2
Yeah, that wizard goal is simply tedious. no skill involved. just play enough and eventually the game will gift you the tokens. it's the only one i got left. it's cool how the y roll don the game though. :)
 

Kaoru

New member
Mar 29, 2012
230
0
I got it.:D And then I stopped playing that table forever.
Hah! I gave up on that table, let alone even that one particular wizard goal, a long time ago. ;) And I don't think I'll ever look back.

Anyway, so far I've got 9 tokens and a couple of dublicates. Feels a bit like grinding, really. They probably should have went for something like activiating the Vault Jackpot via the board game instead or something. Because THAT is actually really hard, haha. The furthest I ever got was collecting three letters - and that was it.

BTW, what was the price for starting a game back then? Was it the same as for any other table? If it was I can see why it wasn't a hit back in the day, considering the shortened time of play.
 

invitro

New member
May 4, 2012
2,337
0
PCs are meant to do work; game consoles are meant to play games, and tablets for casual web surfing.
Game consoles are stripped-down PCs. So are tablets. So are mobiles. So are microwave ovens and most other electronic devices. PCs are meant to do anything and everything (except being put in your pocket).

touch screens are a poor substitute for gamepads.
And gamepads are poor substitutes for keyboards.
 

invitro

New member
May 4, 2012
2,337
0
I would like the option to view each token individually more up close in the future if possible. Maybe with the new UI

Sparkles!

$(KGrHqNHJDkFBjY4PYeJBQo-MsS2Wg~~60_57.JPG


image-19.jpg



I'm curious to find out the cause of the discrepancy between this number of 20 tokens, versus the number of 26 in the TPA version... EDIT: looks like the TPA version has six more silver tokens, all versions of the brass ones.
 
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invitro

New member
May 4, 2012
2,337
0
Coupon Collector's Problem

So do any mathletes here want to figure out the expected number of tokens you will need to win to achieve this wizard goal? (Assuming the tokens are equally likely and there is an infinite supply of all 26?)
I thought someone said somewhere that you didn't get duplicates, but a couple of you say you can... I've got I think 7 of them with no dupes. Anyway, if you can get duplicates, the expected number of tokens needed to have all 26 is:

the expected number needed to get the 1st token, which is 26/26 = 1, PLUS
the expected number needed to get the 2nd different token, once you have one, which is (26 possibilities)/(25 successes) = 26/25, PLUS
the expected number needed to get the 3rd different token, once you have two different ones, which is 26/24, PLUS
the expected number needed to get the 4th different token, once you have three different ones, which is 26/23, PLUS
...
the expected number needed to get the 25th different token, once you have 24 different ones, which is 26/2 = 13, PLUS
the expected number needed to get the 26th different token, once you have 25 different ones, which is 26/1 = 26

= 26 * (1/26 + 1/25 + 1/24 + ... + 1/2 + 1/1)

= 100.2, to one decimal place. I got an average value of 99-101 with a simulation, so I'm pretty sure this is correct.

There's a nice write-up on this problem at https://en.wikipedia.org/wiki/Coupon_collector's_problem.
 

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