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S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 18, 2019, Steven Keyl wrote: Is it truly a 5050 chance otherwise? 

landmark Inner circle within a triangle 4715 Posts 
May I pose another similar but different question which this one brings up for me?
Two real numbers are chosen at random. What is the probability that a third random number lies between them? I have no idea.
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All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 18, 2019, landmark wrote: Assuming that the choices of numbers are all three governed by the same probability distribution, then the probability is ⅓, and a little thought has convinced me that that's true irrespective of the choice of specific probability distribution. 

landmark Inner circle within a triangle 4715 Posts 
Sounds reasonable, but my probability intuition is often wrong in difficult cases. What would your line of thinking be on this?
Edit: my brother just gave me a headslappingly easy explanation!
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 18, 2019, landmark wrote: It's pretty straightforward for a uniform distribution on (0, 1) (or [0, 1]). For every other probability distribution, you can map that distribution to the uniform (0, 1) distribution via its cumulative density function. So, when you choose any three numbers from any probability distribution, you are, in essence, choosing the corresponding cumulative density numbers from a uniform distribution on [0, 1]. If the probability of the latter is ⅓, then the probability of the former is ⅓ as well. 

landmark Inner circle within a triangle 4715 Posts 
Thanks. Okay, one more question. Let's assumed two fixed real numbers L (low) and H (high). M is picked at random. Does the P(M): L<M<H depend on L and H? Intuition says it should, but I have a feeling it might not.
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 18, 2019, landmark wrote: Of course it does. 

landmark Inner circle within a triangle 4715 Posts 
What would be an expression then for P(M) in terms of L and H? Assume our universe is the set of all real numbers, with an equally likely probability distribution for M.
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 18, 2019, landmark wrote: That one's easy: P(M) = 0. Only because for a uniform distribution on all real numbers, the probability density function is zero everywhere. So the probability of being between any two finite numbers is zero; it's the width of that segment divided by the width of the real line. 

landmark Inner circle within a triangle 4715 Posts 
But if P(M) =0, then it doesn't depend on L or H, which was what I was contending above.
Perhaps I'm not stating the problem mathematically correctly. Assume a random process capable of generating all real numbers with equal likelihood. Is the random number more likely to be between 1 and 1000 than between 1 and 2?
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 19, 2019, landmark wrote: It's zero because you specified that the distribution for M is uniform, something you hadn't specified previously. If the distribution for M is not uniform, then P(M) will depend on H and L. Quote:
On Apr 19, 2019, landmark wrote: Your statement of the problem was fine. You simply added a new condition: the uniform distribution of M. Quote:
On Apr 19, 2019, landmark wrote: In that case the probability of being between 1 and 1,000 is zero, and the probability of being between 1 and 2 is also zero. However, if you have a random process capable of generating all real numbers with a normal distribution having mean 0 and variance 1, then the probability of the number being between 1 and 1,000 is bigger than the probability of it being between 1 and 2. 

Steven Keyl Inner circle Washington, D.C. 2494 Posts 
For everyone so inclined, I've got a mystery that needs solving...
The late Lew Brooks once told me about a method for determining the highest number of a group of numbers, the values of which are unknown at the outset. In other words, someone writes down, let's say, 10 numbers on 10 slips of paper. These can be any real numbers, and they do not need to be consecutive: 1, 10.5, 1,000,000, etc. The papers are mixed up and you start pulling out pieces of paper and reading the numbers on them. Somewhere between the 3rd and 7th piece you are able to stop and state which is the highest number in the series. If I recall, the success rate for this was ~6070%, but I may not be remembering this correctly. Does anyone know what I'm talking about? I've never heard about this anywhere else, but it may be a common problem/paradox in mathematics of which I'm unaware. I also don't remember the solution, but it was simple and straightforward. If anyone can point me in the proper direction, I'd appreciate it.
Steven Keyl  The Human Whisperer!
Come visit Magic Book Report.com! "If you ever find yourself on the side of the majority, it is time to pause, and reflect." Mark Twain 

R.S. Regular user CT one day I'll have 194 Posts 
Quote:
On Apr 19, 2019, Steven Keyl wrote: And this is where we're fortunate to have Bill. I look forward to his response to this. Ron
"It is error only, and not truth, that shrinks from inquiry." Thomas Paine


landmark Inner circle within a triangle 4715 Posts 
Quote:
In that case the probability of being between 1 and 1,000 is zero, and the probability of being between 1 and 2 is also zero. Thanks Bill. That makes it very clear for me. I guess when I say pick a number at random, I assume the model of "Hey, think of a number," assuming we can think only of real numbers, and we think of all of them in an equally likely way. But of course, there are other assumptions that can be made. I'm thinking the same argument could be made for the set of rational numbers as opposed to the reals?but not obviously for only the set of integers.
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

tommy Eternal Order Devil’s Island 15737 Posts 
The only real number is One. One cannot have Two horses in reality – One can one horse and another One. One cannot have half a horse but one can have horse meat.
If there is a single truth about Magic, it is that nothing on earth so efficiently evades it.
Tommy 

landmark Inner circle within a triangle 4715 Posts 
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 19, 2019, Steven Keyl wrote: The procedure is this: discard (roughly) the first 37% of the numbers you see. (Don't forget what they are, because you'll need that; you simply remove them as candidates for being the highest.) As you continue looking at numbers, stop when you find one that's higher than any previous number: that's your choice for being the highest of the group. The percentage that you discard is, as I say, roughly 37%. Exactly, it's 1/e, where e is the base of the natural logarithms: 2.718281828.... I don't recall the success rate of this procedure, but you're correct: it's surprisingly high. In one of John Allen Paulos' books  Innumeracy I believe  the author uses this idea to formulate a plan for maximizing the probability that a person will marry the best person they can. He defines a heartthrob as someone who is better than everyone who has come before him or her. You start by estimating how many people you will meet in your lifetime whom you'd consider marrying. As you go through life you discard roughly the first 37% (i.e., i/e) of that estimated number, then marry the first heartthrob you meet after that group. Voilà! 

landmark Inner circle within a triangle 4715 Posts 
It's sometimes called the Secretary Problem:
https://en.wikipedia.org/wiki/Secretary_problem
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 19, 2019, landmark wrote: I'd never heard that before, but it makes sense. Thanks! 

Steven Keyl Inner circle Washington, D.C. 2494 Posts 
That's it, Bill! You found it. Thanks, also, Jack for the wiki reference. I'm going to pick up Innumeracy. Looks like a great read. Thanks for that, too!
Much appreciated, gents.
Steven Keyl  The Human Whisperer!
Come visit Magic Book Report.com! "If you ever find yourself on the side of the majority, it is time to pause, and reflect." Mark Twain 

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