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A problem from /sci/
I wrote some code that runs through and tries out different cutoffs, keeping the winner each time and giving the loser a new random cutoff (between 0.5 and 0.7 - this seems to be a good range).
Anyways, the answer comes out to around 60% consistently.
How to prove mathematically?
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>>59094725
the code
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>>59094725
bump?
does it involve integrals. do you nerds know about integrals?
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>>59094725
So you've written a program to try and solve the problem in the OP image. I dont mean to be an ass but i dont understand how this is /g/ related. Does the program not work?
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>>59094791
The fact that it specifies an uniform distribution should tell you this is a probabilities problem.
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>>59094725
What you do is, you calculate what is the average number that comes up, and anything above that number is stupid to cutoff, that is all.
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call me retarded but isn't this a regression problem?
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>>59094725
there is no optimal cutoff
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python's RNG is garbage, use an external source like entropypool
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bump, I got it, don't want to 404 while I'm typing
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>>590947251 - e^(-1)
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>>59096588
We are looking for x. If you get a number below x, you go on, if you get one above it, you keep it. The optimal strategy is if you have a 50% chance of getting at least x on two tries.
Chance of getting a number above x on the first try: 1-x
Chance of getting a number below x on the first try: x
Now, if you get a number below x, you keep playing:
Chance of getting a number above x on the second try: 1-x
So, you have an 1-x chance of getting a number above x on the first try, and you have an x*(1-x) chance of getting a number above x on the second (you have to fail the first). so
1-x + x*(1-x) = 0.5
(x+1)*(1-x)=0.5
x1=0.7070707...
x2=-0.7070707...
so x1 is the solution, meaning the cutoff is 0.7070707, or 1/sqrt(2)
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>>59096692
Sorry, goofed up the decimal. The solution is still 1/sqrt(2), however that is not 0.7070707..., but 0.70710678118something
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That sounds like a really awful game show.
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>>59096692
>>59096720
Tested it with a program, you are indeed 50% likely to get a number above 1/sqrt(2) with this strategy (left value is when the number was bigger, right value is when the number was smaller).
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>>59096809
At least you're not going home with a stinky goat :^).
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