Not sure it's the correct game: Deep Space Bundle leftover for solvers of region-restricted puzzle.
Celebrating the one in "I'm too lazy properly calculate the probability" (see probability in comments) giveaway link.
Thanks for the gesture & the consolation prize m139! But I'll pass, as I only enter GAs for games I think I'll play.
I'm sure you can gather at least 5 people to enter this.
Ah, 4 z is easy to calculate, but what about 3 same symbols in a row and one differing in case next to them? That's a higher probability (more symbols, more placing options), but I'm too lazy now to calculate that %)
Ha, ok. My calcul was wrong anyway, I forgot to take both capital and non-capital letters into account.
So 1/954 307 to have exactly 4 z in a GA code, no matter their position and capitalization.
But I have no idea how to calculate what you have said, my probability knowledge is basic ^^
I think your calculations are wrong again.
The probability would be 150/28629151
The way to solve it (I think) is using the binomial probability expression with the number of trials of 5 (5 symbols), number of events to occur of 4 (4 z's), the probability of "success" of 2/62 and the probability of failure of 60/62 (26 cap letters, 26 lowercase ones, 10 numbers). Here is a better calculator
ha, I thought it would simply be (2/62).(2/62).(2/62)(2/62)(60/62), but I think this is actually having the first 4 characters being a z and not the fifth (and the code in this do have 4 z in a row, so that wasn't really wrong, just different that what I claimed it represents), is that right ?
But yeah, what this page said seems right, probably learned this in school but I haven't used probabilities since then.
So yeah, 1/190 861 is what I get with this method (why did you write it as 150/28629151 ? Where does that come from ?)
Well, the probability you mention (2/62)*(2/62)*(2/62)*(2/62)*(60/62)
is the probability of having 4 z's and something different. The thing is that it's only a probability of having the different character in some particular position. As far, as we have 5 positions we can put that character into we need to multiply it by 5, thus you get 5*(1/31)*(1/31)*(1/31)*(1/31)*(30/31)
whis equals 150/28629151.
Also if you run the Wolfram Alpha link abone and enter 'number of trials"=5, "success probability"=(2/62) and "endpoint"=4 you'll get the same number at (x=4).
Actually, the original probability (3 symbols +1 in different case next to them) might be a bit a bit easier to calculate.
The probability of a particular set of characters is (1/62)*(1/62)*(1/62)*(1/62)*(1/62).
Then we just count the number of sets, that satisfy our conditions.
Say, we want ?zZZZ. There are 62 of those, as the first character can be anything.
Then, there are placement options, but only 4 of them ?zZZZ, ?ZZZz, zZZZ? and ZZZz?.
Finally, there are 52 characters, that can change their case, so, th total probability would be 52*62*4*(1/62)*(1/62)*(1/62)*(1/62)*(1/62)
which is 13/923521
Thanks for the gesture & the consolation prize m139! But I'll pass, as I only enter GAs for games I think I'll play.
I'm sure you can gather at least 5 people to enter this.
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well...good luck then! )
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No worries. I do the same.
Also with my tendency to overcomplicate puzzles I'm already used to giveaways with 0 to 4 entries %)
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You mean the probability of the GA code having 4 z in it?
That would be 1/1 727 605
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awesome code, right?)
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Ah, 4 z is easy to calculate, but what about 3 same symbols in a row and one differing in case next to them? That's a higher probability (more symbols, more placing options), but I'm too lazy now to calculate that %)
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Ha, ok. My calcul was wrong anyway, I forgot to take both capital and non-capital letters into account.
So 1/954 307 to have exactly 4 z in a GA code, no matter their position and capitalization.
But I have no idea how to calculate what you have said, my probability knowledge is basic ^^
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I don't even remember how to calculate the probability that you did)
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I think your calculations are wrong again.
The probability would be 150/28629151
The way to solve it (I think) is using the binomial probability expression with the number of trials of 5 (5 symbols), number of events to occur of 4 (4 z's), the probability of "success" of 2/62 and the probability of failure of 60/62 (26 cap letters, 26 lowercase ones, 10 numbers).
Here is a better calculator
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ha, I thought it would simply be (2/62).(2/62).(2/62)(2/62)(60/62), but I think this is actually having the first 4 characters being a z and not the fifth (and the code in this do have 4 z in a row, so that wasn't really wrong, just different that what I claimed it represents), is that right ?
But yeah, what this page said seems right, probably learned this in school but I haven't used probabilities since then.
So yeah, 1/190 861 is what I get with this method (why did you write it as 150/28629151 ? Where does that come from ?)
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Well, the probability you mention
(2/62)*(2/62)*(2/62)*(2/62)*(60/62)is the probability of having 4 z's and something different. The thing is that it's only a probability of having the different character in some particular position. As far, as we have 5 positions we can put that character into we need to multiply it by 5, thus you get
5*(1/31)*(1/31)*(1/31)*(1/31)*(30/31)whis equals
150/28629151.Also if you run the Wolfram Alpha link abone and enter 'number of trials"=5, "success probability"=(2/62) and "endpoint"=4 you'll get the same number at (x=4).
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Actually, the original probability (3 symbols +1 in different case next to them) might be a bit a bit easier to calculate.
The probability of a particular set of characters is
(1/62)*(1/62)*(1/62)*(1/62)*(1/62).Then we just count the number of sets, that satisfy our conditions.
Say, we want ?zZZZ. There are 62 of those, as the first character can be anything.
Then, there are placement options, but only 4 of them ?zZZZ, ?ZZZz, zZZZ? and ZZZz?.
Finally, there are 52 characters, that can change their case, so, th total probability would be
52*62*4*(1/62)*(1/62)*(1/62)*(1/62)*(1/62)which is
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Hello there.
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sup
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blacklisted D:. =P. HAHA poor guys...really...
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those raging kids...
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and what was the original gift? :p
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Velocity®Ultra
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Thank you
Sorry for making you do all the work xD
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You are wolcome.
What sort of work?
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Ah nvm
I came from here
http://www.steamgifts.com/discussion/FvnrB/dayz-or-h1z1
But those probability things made my brain exploded xD
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Ha-ha! The trolling thread %)
I also came to Toremato's GA from there...
Ah, they were making my brain uncomfortable too until I found some kind of solution.
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1 entry left :D
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Heyo there!
Thank you for the invitation xD
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I need a new brain
Can you give yours to me? xD
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I have some private uses for that one, sorry.
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Hey! Thanks for the giveaway, mate!
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let that guy get points for his GA, enter it)
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But I don't want the game. :(
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np then)
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Doesn't look like my type of game, but hell, free game? xD
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Thank you! 420/420. Still. Exists in account tho.
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can you finally explane dat 0/420, 420/420 cannabis thing?
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It's just circlejerk. 0/420 Means you scored 0 in the scale from 0 to cannabis. and 420/420 means you scored full cannabis.
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Thank you!/
I won't be entering because I don't want the game as well. But I guess with 5 entries he already gets his points, right?
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Yeah, he will
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Thanks for the GA!
I'm redeeming now (:
Have a good day!
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