The St Petersburg phenomenon. The paradoxes of probability theory. - page 7
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Monty Hall's paradox
Imagine that you have become part of a game in which you have to choose one of three doors. Behind one of the doors isa car and behind the other two doors aregoats. You choose one of the doors, for example number 1, then the host, who knows where the car is and where the goats are, opens one of the remaining doors, for example number 3, behind which there is a goat. He then asks you if you would like to change your choice and choose door number 2. Will yourchances of winning the car increase if you accept the presenter's suggestion and change your choice?
intuitively really doesn't catch on :)
I don't think they will.
I don't think they will.
Of course everyone thinks so at first :) that's the paradox.
Of course, everyone thinks so at first :) that's the paradox.
Well the probability of winning increases, initially it was 1/3, then 1/2.
But you either win or you lose.
If you take a skewed one and skew it some more, who knows, maybe it will even out.
The number of random number generator states is 32768, not divisible without a remainder by a huge number of numbers. Not divisible by 3, by 7, 9, 10, 11, 12, 13... etc. So it hardly makes sense to worry about skewness due to an error in the dubs.
You can divide numbers by 3, by 7, 9, 10, 11, 12, 13 by them :-) find the largest to RAND_MAX and its.
it is worth worrying about skews because you can easily avoid them
Monty Hall's paradox
Imagine that you have become part of a game in which you have to choose one of three doors. Behind one of the doors isa car and behind the other two doors aregoats. You choose one of the doors, for example number 1, then the host, who knows where the car is and where the goats are, opens one of the remaining doors, for example number 3, behind which there is a goat. He then asks you if you would like to change your choice and choose door number 2. Will yourchances of winning the car increase if you accept the presenter's suggestion and change your choice?
intuitively really doesn't catch on :)
Great Maxime, thank you.
So, let's do the Monty Hall experiment. One experiment easily fits in one line of Excel spreadsheet: here it is (the file is worth downloading to see the formulas), I will give here a column-by-column description:
A. Experiment number (for convenience)
B. Generate a random integer from 1 to 3. This will be the door behind which the car is hidden
C-E. for clarity: in these cells "goats" and "cars"
F. Now we choose a random door (actually we can choose the same door all the time, as randomness in choosing the car door is already enough for the model - check!)
G. The presenter now chooses a door from the remaining two to open for you
H. And here is the most important thing: he does not open the door with the car behind it, but in case you initially pointed to the door with the goat, he opens the other only possible door with the goat! That's his clue for you.
I. Now let's calculate the odds. Let's not change the door yet - i.e. let's count the cases when column B equals column F. Let it be "1" - won, and "0" - lost. Then the sum of cells (cell I1003) is the number of wins. Should get a number close to 333 (we do 1000 experiments in total). Indeed, finding a car behind each of the three doors is an equally likely event, so by choosing one door, the chance of guessing is one in three.
J. Change our choice.
K. Similarly: "1" is a win, "0" is a loss. So what does the total add up to? And the sum is a number equal to 1000 minus the number from cell I1003, i.e. close to 667. Does that surprise you? Could there be anything else? After all, there are no other closed doors! If the originally chosen door gives you the win in 333 cases out of 1000, then the other door must give the win in all remaining cases!
Who has not understood: This is the paradox - initially it seems that the problem "is the same", as in the case with the 1000 doors that 3, but to understand it (and most importantly to understand why you need to change the choice) - consider the problem with 1000 doors, and not with the probability to win, but with the probability of making a mistake: the first choice the probability to make a very high, after narrowing to 2 doors - the probability to make a lower, but for the same door (if not changing the choice) is very high at the time when you made this choice.
From myself: If we don't change the choice, we are left with the same probability as when we started, and when we change the choice the probability is in our favour.
https://habr.com/post/201788/
https://pikabu.ru/story/naglyadnoe_dokazatelstvo_paradoksa_monti_kholla_5393656
Monty Hall's paradox
Imagine that you are part of a game in which you have to choose one of three doors. Behind one of the doors isa car and behind the other two doors aregoats. You choose one of the doors, for example number 1, then the host, who knows where the car is and where the goats are, opens one of the remaining doors, for example number 3, behind which there is a goat. He then asks you if you would like to change your choice and choose door number 2. Will yourchances of winning the car increase if you accept the presenter's suggestion and change your choice?
intuitively doesn't really catch on :)
For the most part, this is a paradox of game theory, not probability theory as stated in the thread title) The problem is that the game is not formalised definitively, and it can be done in many different ways. Although, there are plenty of paradoxes in game theory even when fully formalised (e.g. the famous prisoner's dilemma).
For the most part, this is a paradox of game theory, not probability theory as the thread title states) The problem is that the game is not formalised definitively, and this can be done in different ways. Although there are plenty of paradoxes in game theory even when fully formalised (e.g. the famous prisoner's dilemma).
A bunch is power)))
In the ability to negotiate and stick to agreements.
Who hasn't understood it yet: Therein lies the paradox - initially it seems that the problems are "the same", both in the case of 1000 doors and 3, but to understand it (and most importantly understand why we should change the choice) - consider the problem with 1000 doors and not with the probability of winning, but with the probability of making a mistake: the first choice has a very high probability of making a mistake, after narrowing down to 2 doors - the probability of making a mistake is lower, but for the same door (if not changing the choice) it is very high at the time when this choice was made.
From myself: If we don't change the choice, we are left with the same probability as when we started, and when we change the choice the probability is in our favour.
https://habr.com/post/201788/
https://pikabu.ru/story/naglyadnoe_dokazatelstvo_paradoksa_monti_kholla_5393656
Hi Alexander_K2))