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I'd say 1 in 1000 minus the number of numbers you have already rejected as not being the highest number, if I understand the question correctly.
cheers,
Since there are no limits to the random number generator you will never be sure whether the number is "high" or "low"
eg. If you get 25 as first number. There are infinity numbers above that and below that. Its as good as 0
If you get -1trillion. There are still infinity numbers above and below that.
The numbers generated will not form a standard distribution so thats down the drain.
So to the logician, each number is effectively 0 till the last number comes out.
As such, a logician will be unable to decide whether the number is the highest till the LAST number comes out. (Because each number revealed has an infinite amount of numbers above and below it)
The logician can only make the choice at the last number. The probability that this number is the highest number is 1 in 1000.
I think thats right
EDIT: Well since you said a 'simple' computer program, integer overflow will be at +-4.29 Billion; So that would probably make decision making easier
Since there are no limits to the random number generator you will never be sure whether the number is "high" or "low"
eg. If you get 25 as first number. There are infinity numbers above that and below that. Its as good as 0
If you get -1trillion. There are still infinity numbers above and below that.
The numbers generated will not form a standard distribution so thats down the drain.
Once the first number is generated, wouldnt that be the highest - as you only have 1 number to choose from? answer 'yes'
therefore generating 999 random numbers without stopping.
The logician can only make the choice at the last number. The probability that this number is the highest number is 1 in 1000.
lol
PS say bellenuit, is it ok to use a range of say +/- 1 million? why does it have to be +/- infinity?
I'd say 1 in 1000 minus the number of numbers you have already rejected as not being the highest number, if I understand the question correctly.
cheers,
I agree with you up until your point there
If you wait until the last number you will have one number that is currently the 'highest' of the all the 999 numbers you have so far generated.
Since there are infinite numbers either side of that number you theoretically have a 50-50 chance of being higher or lower than the highest number...
Therefore, I you wait until the the last number it has a 50% chance of being below the current highest number (in which case you lose) but also a 50% chance of being above the highest number and therefore you say 'YES" and you win the game.
50-50 chance is the best that you would get...
Is that right bellenuit?
Cheers
Here's my (flawed) logic.
thanks for that - I'll think about it - probably for a week or so lol ( got a busy one ahead of me) cheers.2020. No, because you would then know what a large number is without reference to what has been generated so far.
Bloomy. Now you have put me into a quandary. Unlike what I said in my previous post, perhaps I shouldn't have used plus or minus infinity as mathematics becomes fuzzy when dealing with the infinities.
You answer is very logical and though it would appear to be correct to say that there is a 50 - 50 chance of the last being higher than the preceding highest, we know that since it is a perfect random number generator, each number has a 1 in 1000 chance of being the highest and the last number is no different from any other.
Perhaps, to remove the anomalies of dealing with infinities, I should rephrase the first line of the problem as follows:
You have a simple computer program that sequentially generates 1000 perfectly random numbers within a finite, but vast range. You do not know what the range is, other than that it is finite.
That should remove the infinity issues while at the same time giving you no information as to what a large number is.
Just a further thought and based purely on assumptionI think, Stan 101, the chance of 1 in 1000 would remain the same with every attempt. A higher or lower number than the previous number may be higher or lower than the remaining numbers at every attempt. Process of elimination is not possible.
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