randomness, statistics and quiz

LucaMs

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I read the thread "Rnd Functionality?" and I would like to ask some questions, but not closely related to the function.

I remember that the function in the various versions of MS basic, was changed, to make it as random as possible (based on system clock, as rightly stated by JohnD).

I would like to talk about repetition, randomness, referring also to what Tom wrote in his post.

Some people claimed that I had wrong about the following.

The concept of delay and "recovery" of a random event.

The probability that a coin falls on the face A is 50%, obviously. But the proof that the previous launches affect the subsequent, is the fact that no one will ever be able to flip the coin 10,000 times always getting A (if he does not cheat!).

Also, since I'm not a statistician, I rely on empirical evidence, I am convinced that an event that has had in the past a very low frequency compared to the odds, is to "recover", repeat frequently in subsequent sorties.

Maybe you do not believe me, but I developed a project of a roulette for fun. Well, playing with some tactics and relying on the "recovery", simulating x number of days at the table (2/3 hours), I was able to "back to home" winner at least 8 times out of 10.
In fact, I will develop this system with B4A and I will enrich :p.

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Little (old) quiz.

There are 3 doors. Behind two of them is hidden a goat; behind the third is a car to win.
You choose a door, for example, the number 3.
I tell you that behind the door number 1 there is a goat and I invite you to change your 3 to 2.
At that point, you should (TO WIN) change (A), confirm your choice (B) or is it indifferent (C)?

(I hope all this is understandable, despite my poor english)
 
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LucaMs

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It is strange: no one is intrigued.

Well, the answer is: you would have greater odds changing your initial choice (66.66% vs. 33.33%: all of us would have answered 50% vs. 50%).


(Will someone ask why? :eek:)
 

LucaMs

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When I choose one of 3 doors, I have about 67% chance of being wrong (choose a goat).
There remain two doors, one hiding a goat, the other hides the car. The door that hides the goat is opened.
The probability that I have chosen a door that hides a goat is still 67%, while the probability that the door still closed hides a goat is now 50%, because one of the two certainly hides the car.
So I have a better chance of winning by changing my choice.

(I have modified the attached app, using some pictures. I'm sorry, but my English does not allow me greater clarity)
 
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