Reply With Quotecan you explain a litle more about RMS.
To get the lowest RMS I need much CPU's or is it luck?
With regards schluppy
Straight in at number 4 with a 2.65 RMS.
can you explain a litle more about RMS.
To get the lowest RMS I need much CPU's or is it luck?
With regards schluppy
it looks pretty random but if you start gen 1 with a reasonable RMS it should get better each generation.
It looks like it depends how the algorithm runs on gen 0 as to whether you get a good RMS value for that set.
Schluppy,
the only thing you need is luck.
But, since luck is "computed" on every machine separatly, the chance of getting better (=lower) RMS is higher if you run the project on more boxen (like: you must be quite lucky to win in the lottery but if you play twice at the same game you severely improved your chances of winning).
Than I have more chanses because, I am running on many CPU's thats great.
And I have seen I have found it in the 152th Generation, so it can be that I fond a smaler one, later.
Right?
This isn't true... Say the odds of winning on a single ticket are 1:50,000,000(like: you must be quite lucky to win in the lottery but if you play twice at the same game you severely improved your chances of winning).
The best change in odds you can make is to go from having zero tickets
odds = 1:infinity
to having one ticket
odds = 1:50,000,000
which is an infinite improvement (you can't win if you don't play at all...)
Unfortunately, buying a second tickets does NOT change your odds to being 1 in 25,000,000, and buying 100 tickets does not change your odds to being 1 in 500,000.
The "universe of numbers is unrelated" and your odds of winning on the second ticket is (essentially) 1:49,999,999, on the third it's 1:49,999,998, and so on.
SImplification - since they are "unrelated trials", buying 100 tickets means that you have more on the order of 1 chance in 49,999,900 to win -> think subtraction(50,000,000-100), rather than division (50,000,000/100). Not much of a change in the odds.....
FreeDC Mercenary
I think I'm going to argue with that.
When you buy that second ticket what happens is that there are now two outcomes in the set of favourable outcomes and 49,999,998 in the set of unfavourable ones. To get the probablity of winning you take the size of the set of winning events and divide by the size of the set of all events and you get 2/50,000,000 or 1/25,000,000.
Actually as I finished writing this I'm considering that there may be some other kind of lottery that you are referring to, this would apply to something like a raffle. Maybe you are referring to something else (I can't really imagine anything that would fit off the top of my head though), but I think the comparison is valid with a raffle ticket -- if you think of it as each computer being assigned a certain RMS, then having more computers increases your chances of being assigned a better RMS in the same way as more raffle tickets increase your odds.
I'm no maths genie, but after thinking on this for then minutes I guess I got the solution: we are talking about different things!
IMHO you talked about DIFFERENT (subsequent) lottery games. So, if you don't win todays lottery, you play again next time.
I talked about the same lottery, but buying two tickets. That's what would better fit DF: you can play as long as you like (try it again and again to win with a different "ticket" (generation) ), at least until a new protein is uploaded (and a new game starts).
So, assume there are 50.000.000 possible answers.
When I pick 1 solution, my chances are 1/50.000.000 to win
When I pick 2 solutions, my chances are 2/50.000.000 to win.
When I pick 25.000.000 solutions, my chances are 25.000.000/50.000.000 = 0.5 to win
Greets,
Wirthi
besides the math: invest more cpu power, and you should get better results; and if they are not better they are at least "more".
More is always better.
No - your second sentence is not correct. That's why people think their chances improve so much when they buy more lottery tickets, and this is INCORRECT. Most people, who can't work out the math actually involved, would assume that buying 1,000 tickets in a 1:50,000,000 lottery would give them a 1 in 50,000 chance of winning. THIS IS WRONG - they would have to buy 49,950,001 tickets to change the odds that much...When you buy that second ticket what happens is that there are now two outcomes in the set of favourable outcomes and 49,999,998 in the set of unfavourable ones. To get the probablity of winning you take the size of the set of winning events and divide by the size of the set of all events and you get 2/50,000,000 or 1/25,000,000
Your first sentence is more correct - buying two tickets changes the odds to 2 (favorable) : 49,999,998 (unfavorable) The selection of lottery numbers are independent, unrelated events. With your first ticket you have a 1 in 50M chance to win. Buying the second ticket, with a different number, adds one more chance to win in the remaining 49,999,999 possibilities. There are still 49,999,998 losers that can be drawn, and you only have 2 winners...
This might be more intuitive if you think about it in terms of "ways to lose". It isn't the number of ways to win that is important in a lottery, but the number of ways that you can LOSE.
1 in 25,000,000 would mean that you have one way to win and that there are 24,999,999 ways to lose.
This is NOT the same thing as buying two lottery tickets where the odds are 1:50,000,000. Buying two tickets there means you have 2 chances to win, conversely meaning that there are 49,999,998 ways to lose.
NB: This is a completely different animal and much worse odds than 1 in 25,000,000. To have them mean the same thing - having the same numer of ways to lose be equivalent between the two, you would have to buy 25,000,001 tickets in the 1:50,000,000 lottery.
FreeDC Mercenary
You are absolutly right. BUT:Originally posted by rsbriggs
This is NOT the same thing as buying two lottery tickets where the odds are 1:50,000,000. Buying two tickets there means you have 2 chances to win, conversely meaning that there are 49,999,998 ways to lose.
(lottery with 100 possible answers, easier to handle)
Buying one ticket you have a chance of 1 of 100 to win (1%).
Buying two tickets you have a chance of 2 of 100 to win (2%).
...
Buying 10 tickets you have a chance of 10 of 100 to win (10%)
...
Buying 50 tickets you have a chance of 50 of 100 to win (50%)
...
Buying 100 tickets (=all) you have a chance of 100 of 100 to win (100%)
Sorry, but in my eyes buying just one more (from 1 to 2 tickets) seems to double my chances to win. Double number of tickets => double chance to win. This is linear, isn't it?
If you buy 25,000,000 lottery tickets you actually have a 1 in 2 chance of winning. Visualize a probablity tree with a lot of branches, colour half the branches blue and the other half red. Now if each branch has an equal chance of being traversed then a blue one is equally likely, and if there are 50,000,000 branches and 25,000,000 of them are red then you have a 1 in 2 chance of traversing a red edge.
Another argument which can be made is that what are the odds of losing in the situations you describe -- if you agree that the probability of losing plus the probablity of winning add to 1 then this sum does not make sense given your odds. You are saying that if I buy 49,950,001 tickets out of 50,000,000 that my odds of winning are 1/50,000 -- what are my odds of losing -- 49,000/50,000? This does not make sense if you visualize the tree, if you bought all those tickets the tree would be almost entirely one colour, the probability of following one of the other colour branches would be quite small.
Here is a simulator, if you agree that it is fair:
it takes one number as a parameter which is the number of lottery tickets you buy (to a max of 32767), launch as ./lottery 2 &, let it run for a while and then type rm lottery.lock and it will give you a summary. If you try lottery 1 and lottery 2 it should demonstrate that your odds do double.
Sample run:Code:#!/bin/bash let threshold="$1-1" iter=0 wins=0 touch lottery.lock while test -f lottery.lock do let iter="$iter+1" if test $RANDOM -le $threshold then let wins="$wins+1" fi done echo $wins in $iter games -- `echo 100*$wins/$iter|bc -l -q` %
Code:qed@enigma tester $ ./lottery 32000 & [1] 1523 qed@enigma tester $ rm lottery.lock qed@enigma tester $ 1183053 in 1211635 games -- 97.64103876167327619291 %
Posting Permissions
