Let's discuss Mathematics

[Kyriakos]

@#1305: I didn't expect anyone to have the audacity to ask about :/

 

[dragonprobably]

 

[Mise]

Here's something I couldn't get my head around while lying awake in bed last night...

Scenario 1: 3 people, named A, B and C, each generate a random number between 0 and 1. The probability that C will generate a number higher than both A and B is 1/3. The average number that C will generate is 0.5.

Scenario 2: Knowing that the average number C will generate is 0.5, they elect to not bother with the random number generation step for person C, and instead they just say that C's number is 0.5. However, this changes the probabilities of C beating A and B: the probability is now only 1/4.

Now, I understand the maths; I get how this works, mathematically. But I'm still lost at the intuition step. What I'm about to say smells of pseudoscience to me, but this is the best I can come up with: A's average number will also be 0.5; therefore, if C beats A, then it is likely that C has a higher number than average. Thus, C is more likely to also beat B, if C has already beaten A.

Does this make sense???

 

[Kyriakos]

Here's something I couldn't get my head around while lying awake in bed last night...

Scenario 1: 3 people, named A, B and C, each generate a random number between 0 and 1. The probability that C will generate a number higher than both A and B is 1/3. The average number that C will generate is 0.5.

Scenario 2: Knowing that the average number C will generate is 0.5, they elect to not bother with the random number generation step for person C, and instead they just say that C's number is 0.5. However, this changes the probabilities of C beating A and B: the probability is now only 1/4.

Now, I understand the maths; I get how this works, mathematically. But I'm still lost at the intuition step. What I'm about to say smells of pseudoscience to me, but this is the best I can come up with: A's average number will also be 0.5; therefore, if C beats A, then it is likely that C has a higher number than average. Thus, C is more likely to also beat B, if C has already beaten A.

Does this make sense???

You have two combined issues there, one is generating a purely random positive number between 0 and 1. The other is that one would "win" by having a higher value number than the others. This creates many problems which in my view make it into less of a math issue, cause assuming those people knew they were trying to win something, and moreso knew how to win, the scenario falls apart cause elementary arithmetic would make them all try to write some number which is nearly 1. So we have to assume that they either do not knowingly compete, or that they do not know how one wins (ie they do not have reason to think that the highest number wins).
If you formulate your problem in this manner then maybe probability theory, number sequences, and limits, can be put to use

(ps: why would C generate "an average number of 0.5"? Is this set as part of the scenario itself, or is it itself an assumption? Cause the former would make it far less 'random' regardless of what "an average number of 0.5" means).

 

[azzaman333]

Here's something I couldn't get my head around while lying awake in bed last night...

Scenario 1: 3 people, named A, B and C, each generate a random number between 0 and 1. The probability that C will generate a number higher than both A and B is 1/3. The average number that C will generate is 0.5.

Scenario 2: Knowing that the average number C will generate is 0.5, they elect to not bother with the random number generation step for person C, and instead they just say that C's number is 0.5. However, this changes the probabilities of C beating A and B: the probability is now only 1/4.

Now, I understand the maths; I get how this works, mathematically. But I'm still lost at the intuition step. What I'm about to say smells of pseudoscience to me, but this is the best I can come up with: A's average number will also be 0.5; therefore, if C beats A, then it is likely that C has a higher number than average. Thus, C is more likely to also beat B, if C has already beaten A.

Does this make sense???

You're explaining scenario 1 there, right? If so, that seems a reasonable explanation.

 

[Mise]

Yeah, scenario 1. Thanks, needed a sanity check....

 

[dutchfire]

If A,B and C both generate a random number, if you take all cases where C has the highest number, then the average over those cases is larger than 0.5.

 

[Mise]

Right, yeah, that makes sense. So you'd have to assign C with the number sqrt(1/3) in order to make it equivalent to a scenario 1...

Also, just FYI, Excel tells me that the average of C's random numbers where C had the highest number would be 0.75; it appears to be n/(n+1), where n is the number of players, which makes sense also.

I'm sure there's a way I can use this to my advantage somehow. I once thought of a thing that exploited people's inability to do probabilities in their heads. The odds of rolling a 6 with 3 dice is about 42%, but most people would think it was 50%. So you offer people odds at 50% (i.e. double your stake if you roll a 6 with 3 dice), and collect the 8% margin. I'm sure you could do something similar here, where instead of agreeing that C (the mark) has a fixed number of 0.5, you simply do a random number between, say, 0.3 and 0.7, which to the average punter looks like a reasonably random number between 0 and 1, but is in fact stacking the odds in the house's favour. Reason being, C only wins ~25% of the time (or a number that approaches 25% the narrower the random range is), rather than the 33% of the time the punters assume. So you can even give them "favourable" odds and still collect a margin...

EDIT: Of course, you don't tell them that the random number is between 0.3 and 0.7, cos they'd smell a rat. You tell them it's a random number between 0 and 1. And it looks to any outsider as if it's random btwn 0 and 1, since it's still distributed around 0.5. I'm not sure how you'd work this into a game though. The dice game is easy; this? No idea. Pick a card maybe?

 

[Kyriakos]

^Should be noted that probability theory does not always calculate any practical probability, ie how often something would happen given a finite number of attempts/other. It pressuposes an infinite number of attempts, unless it is about groups formed in very specific manner which does not take into account any factor of reforming the group/state by starting anew for a finite number of times.

So while they are both part of Probability theory, you cannot know practically how often one will roll a 6 with some dice, but you can entirely know how many different groups of (let's say) triads will be formed by the placement of X numbers of different objects in those groups. Eg there are exactly 8 differing triads of 2 different objects with repetition of those and taking position into account.

Spoiler :

AAA, AAB, ABA, ABB, BAA, BBA, BAB, BBB

 

[dutchfire]

Right, yeah, that makes sense. So you'd have to assign C with the number sqrt(1/3) in order to make it equivalent to a scenario 1...

Also, just FYI, Excel tells me that the average of C's random numbers where C had the highest number would be 0.75; it appears to be n/(n+1), where n is the number of players, which makes sense also.

I'm sure there's a way I can use this to my advantage somehow. I once thought of a thing that exploited people's inability to do probabilities in their heads. The odds of rolling a 6 with 3 dice is about 42%, but most people would think it was 50%. So you offer people odds at 50% (i.e. double your stake if you roll a 6 with 3 dice), and collect the 8% margin. I'm sure you could do something similar here, where instead of agreeing that C (the mark) has a fixed number of 0.5, you simply do a random number between, say, 0.3 and 0.7, which to the average punter looks like a reasonably random number between 0 and 1, but is in fact stacking the odds in the house's favour. Reason being, C only wins ~25% of the time (or a number that approaches 25% the narrower the random range is), rather than the 33% of the time the punters assume. So you can even give them "favourable" odds and still collect a margin...

EDIT: Of course, you don't tell them that the random number is between 0.3 and 0.7, cos they'd smell a rat. You tell them it's a random number between 0 and 1. And it looks to any outsider as if it's random btwn 0 and 1, since it's still distributed around 0.5. I'm not sure how you'd work this into a game though. The dice game is easy; this? No idea. Pick a card maybe?

Or, to reformulate it once again, A<C and B<C are not independent 'events'.

 

[MCdread]

Another way of seeing it, although not sure it's the clearest intuition is:

At each round, the chance C wins given it's current value is C^2.

If C is fixed at 0.5 (scenario 2), then the probability is 1/4 as you say. Imagine that you now allow C to fluctuate a bit (but uniformly !) around 0.5, e.g, between 0.45 and 0.55. The scenario changes because the "reward function" y = C^2 is not a linear function.
In other words, even though C still gets an average score of 0.5, C gets a greater reward by landing on 0.55 than he otherwise looses by landing on 0.45 (i.e., compared to what he gets by being stuck on 0.5). So, as you give C greater and greater freedom, his chances slowly grow from 1/4 to 1/3.

 

[Mise]

Ahh yeah, I like that MCdread - I think that's the clearest intuition for me. It has the added benefit of making it a lot easier to tweak the odds on my casino scam

 

[dutchfire]

An random number between 0 and 1 will give 0.5 on average. A coin flip (1 for heads, 0 for tails) also gives 0.5 on average. So you and MCdread both generate a random number in [0,1] and I will flip a coin, the person with the highest outcome gets a dollar from the other two.

 

[MCdread]

Ahh yeah, I like that MCdread - I think that's the clearest intuition for me. It has the added benefit of making it a lot easier to tweak the odds on my casino scam

Yeah. Mathematically, and for any C_min and C_max between 0 and 1, you just have to calculate what the mean of y = C^2 would be for a large number of trials (so long as you sample from uniform distributions of course).

 

[dutchfire]

Given any distribution p(x), x in [0,1], with unit volume, you just calculate the integral p(x) x^2 dx, x from 0 to 1, right?

edit:
and if players a,b and c are given by distributions p_a, p_b, p_c, then you calculate:

P(C wins) = integral dz dy dx p_a(x) p_b(y) p_c(z)

with boundaries x,y in [0,z] and z in [0,1]

 

[MCdread]

Yeah, that's the integration you need to do for Mise's scenario 1.

For the general case, where you sample C in range [C_min, C_max] (with 0 < C_min < 1; 0 < C_max < 1), you do the same thing and integrate z in [C_min, C_max].
But with p_c(z) corrected by a factor that reflects "player C's loss of freedom" (to use non-rigorous language).
Integrating 1/(C_max - C_min)*p_c(z), where p_c(z) has unit volume, should work, I think.

EDIT: Actually, I'm overcomplicating things. Because the p.d.f. for C equals zero outside the sampling interval anyway, and dutchfire defines it as having an area of 1, his formula will work just fine for any case.

 

[Harv]

Bump for Paradigm Shifter...

From memory, this is called the hailstones problem, or series, or something.

Take any integer x

If x is even, halve it
If x is odd, then do 3x+1

rinse & repeat.

The conjecture is that it will always reach the loop 1-->4-->2-->1, but that's yet to be proven.

Prove that no matter which number you start from, you'll always reach a number divisible by 4.

Bump. You did mean positive integer, correct?

Start with -11
(-11)x3+1 = -32
(-32)/2 = -16
(-16)/2 = -8
(-8)/2 = -4
(-4)/2 = -2
(-2)/2 = -1
(-1)x3+1 = -2
(-2)/2 = -1 and so on

Nobody has proved this? I did not read all the posts here, but I did read the Wikipedia link provided by Paradigm Shifter.

EDIT: The Wikipedia link does specify a positive integer. My mathematical notation is rusty, and typing out a proof in mathematical notation on a forum like this is a bit awkward.

Second EDIT:
It looks like in fact we eventually do get to a number divisible by 4.
If we express an odd number as A*2^N-1, (where A is an odd number and N is an integer greater than zero) then:
a. After (2N-1) iterations, we will arrive at a number divisible by 4.
b. This number divisible by 4 will be 2*(A*3^N-1).

 

[PlutonianEmpire]

How do I use this calculator: http://www.1728.org/ellipse.htm ? I'm entering numbers, but nothing's working.

At all.

Nothing shows up, zilch.

 

[Samson]

How do I use this calculator: http://www.1728.org/ellipse.htm ? I'm entering numbers, but nothing's working.

At all.

Nothing shows up, zilch.

What are you trying to do? I follow the instructions : "click on the "Eccentricity to yx" button, enter .8, click "Calculate"" and it shows " Y/X Ratio =" " 0.6".

 

[PlutonianEmpire]

EDIT: Turns out it refuses to work offline. It's an online-only calculator, apparently. (Those) Bastards.