One involves 3 plain cards with faces coloured red/red, red/green & green/green. Now if one card is chosen at random and placed on a table so that you can see one side (red say), what it the probability that the other side is also red?
1/3. The chance of it being green is 1/6. The easiest way to comprehend this is to consider that you are always going to bet that the card is the same colour on the other side. Seeing as two of the three cards are the same colour on the other side, your chances of winning are automatically 2/3.
I know it sounds strange, but it actually isn't. Another way of seeing it ; 3 cards have six sides in total. Chances of pulling out red/red? 1/3. Multiplied by the chance of it showing red? One. One times 1/3 is 1/3. Chances of pulling out red/green? 1/3. Multiplied by the chance of it showing red? 1/2. 1/2 times 1/3 =1/6.
As for the Monty Hall problem, the best way of understanding it is to remember that Monty knows where the car is- so it's kind of like Monty "owning" two of the doors, and you only own one.
Monty's always going to have a goat to reveal (there's only one car) but when you switch you are literally exchanging your 1/3 shot for Monty's 2/3.
If Monty doesnt know, and reveals a goat, your chances remain at 1/3- change or swap. All it means is that Monty's 1/3 shot revealed a goat. It doesn't change your odds.
"When man loses God, he does not believe in nothing. He believes in anything" G.K. Chesterton
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