mctrump · Pepe Collector · 2 points ·
This is the kind of argument used by people who don't understand that glitched speedrunning like this is an order of magnitude harder than playing the game normally

mctrump · Pepe Collector · 1 points ·
Send me a message if you're there

mctrump · Pepe Collector · 4 points ·
I bet you don't even know we have a discord server

mctrump · Pepe Collector · OP · 2 points ·
I'm sitting here legitimately wondering if you're being serious or not

mctrump · Pepe Collector · OP · 1 points ·
There are people who have tried to claim what you are claiming. This whole idea of "there two choice from the start" is beyond wrong.

If the host has three doors and just gives you your initial choice without doing anything else, that's a 33% chance of a win, inarguably. If after making your choice, the host opens an unchosen goat door and then gives you the contents of your chosen door, it's still a 33% chance of a win. That's because when you make your initial choice, you lock in the chances of that door being a winning door at the rate at the time of your choice.

At your initial choice, you have an equal chance of it being the car, goat #1, or goat #2, that's a 33% chance of it being the car. The host opening an unchosen goat door does not magically change what your initial choice is. You can't just disregard one of the goats when making your initial choice because it absolutely is a relevant factor when making it.

The effects of the monty hall problem become more obvious when you have a higher number of doors. The generic monty hall problem can have any number of doors as long as it has at least 3, and the host must open every door except 2.

Suppose instead of goats, doors, and a car, a 52 choice version has played with cards. The goal is to find the ace of spades.

First, the player picks a card out of a full deck at random and lays it face down without looking at it, this is the initial choice. The host then thumbs through the remaining 51 cards and lays 50 cards face up that are not the ace of spades and lays the last card face down without showing it to the player.

Which card is more likely to be the ace of spades, the initial card, or the remaining card?

mctrump · Pepe Collector · 8 points ·
You've found my spirit animal

mctrump · Pepe Collector · 4 points · *
izotable's wallpaper

mctrump · Pepe Collector · 6 points ·
My desires have long since surpassed such entry-level smut

mctrump · Pepe Collector · OP · 1 points ·
So I've given some thought regarding the idea of two people playing monty hall.

For starters, you are changing the rules of the system, so of course you're going to come up with a different result. But that doesn't prove anything in regards to the standard monty hall problem. I'm going to work through every possible situation with two players

1) both players pick the same door and stay
No different from one player doing the same

2) both players pick the same door and switch
Also no different from one player doing the same

3) both players pick the same door, one stays, one switches
Well, the host already revealed one door, so the two players have a combined 100% chance of winning as they have selected all possible solutions at this point

4) the players pick different doors and the unselected door is the car door(1/3 chance)
the monty hall problem cannot proceed as there are no unchosen goat doors. This dead end is probably why I've never seen anyone using this "two players" argument

5) the players pick different doors and the unselected door is a goat door(2/3 chance)
This is where new logic has to be invented, as standard monty hall logic cannot apply. There are only two closed door and the car is behind one of them. Both players had an equal chance of being right before the unchosen door was opened. However, as both closed doors were selected before the other door was opened, the potential 1/3 probability of the opened door has to be evenly distributed among the closed doors. In this case, each of the closed doors has a 50% chance of being the car door.

I must reiterate, under standard rules. You have a 33% chance of winning by staying with your initial choice because the choice was made when you had a 33% chance of being right. The fact that the host opened an unchosen goat door does not change this. Since the initial choice has a 33% chance of winning, the other closed door must have a 67% chance of winning.

Suppose a player playes the game many times and always chooses door #1 and stays. The player can only win if the car is behind door #1. The car has an equal chance of being behind any of the three doors.

If the car is behind door #1, he wins
If the car is behind door #2, he loses
If the car is behind door #3, he loses

He can find himself in three possible significant situations. One leads to a win, the other two lead to a loss. He has a 33% chance of winning be staying.

mctrump · Pepe Collector · 1 points ·
If you want votes, post quality content


mctrump
I am the guy she tells you not to worry about.
382,831 Pepe Collector
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