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Hi,

Im not really a math guy here but more of a logical thinker and this question really confused me because of the way people explain it.

Namely i do agree with the fact that taking a wrong door away after the contestant has chosen initially, rises the chance that the contestant will pick the right door by switching.

However: people say there is a one third chance initially and a two third chance after. Maybe I am digging in to deep but let me explain with my next example.

Me and two of my friends John and Mike used to play a lot of cards.

We would play a simple game which only took 3 cards from the deck. The 1 clubs. the 2 clubs and the 3 clubs.

We played as followed: John would shuffle the 3 cards and then would deal each of us a random card from the top of the deck. The one who got the highest card would win (It was a really exciting game).

Now let me ask this: what are the odds each game Mike will get the highest card?

Well if you ask me it isnt 1 in 3.

Which one will be getting the highest card is really a matter of how the deck is getting shuffled. If John shuffles in a way which makes the 3 Clubs go on the top alot time and he deals to Mike first then it wouldn't be 1 in 3 would it?

What im saying is that it really depends on the coincidence of the way the card is getting shuffled and therefore you can never predict (no matter the amount of games) how often (not even approximately) you will get the highest card.

Now what if there would be 2 “3 clubs” and 1 “1 clubs” does that mean there now is a 2/3 of a chance of getting the high card? The way I look at it the chance has indeed increased but the ratio still isnt 2/3.

The only way of saying what the ratio is of the highcard falling is is by testing each case uniquely. And then afterwards you could say what the ratio is of the amount of times you try d in that case. and not before hand where coincidence and random human decision making has way to much influence on the outcome.

Its the same thing with coinflip.

The ratio the coin is falling heads or tails depends on the way the person throws the coin.

If the person throws the coin in a similar way every time, it will also fall on the same side everytime.

In case of the Monty Hall scenario the the chance of knowing the ratio to the winning door is impossible to know because it depends on the random choice of the contestant and the random doors the gameshow put the goats at.

Still i agree that if you switch you will win more often, but that is only because randomness will always try and go toward switching between 3 possible options(from which 2 in this case are the right ones).

However there is no way to know for certain where the 2 goats will switch and which door the contestant will choose.

Makes sense right?

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# I Monty Hall Problem -- Questions

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