I was awake all night again. I boofed the very last of my bong water in a single syringe that gave me an all day boost from noon till now. Part of my evening was obsessing over the Monty Hall math problem. Although tired, I worked on that task from 1am till 3am changing my mind a few times. I’m still not 100% the commonly accepted answer is the correct one

Math can be tricky when chances and odds get involved. It’s known as the Monty Hall 3 Door math question because Monty Hall came to fame by hosting the original LETS MAKE A DEAL game show on Tv from 1963 to 1975. The show was revived later staring Wayne Brady. As far as I know, the show still ends the same way.

The highest prize winner plays the last game, consisting of 3 doors. Behind one door is a great prize and the other two doors contain lesser prizes or their branded Zonk donkey or goat.

The math problem is as follows. The player gets to choose one of the three doors and is then shown one of the other two that they didn’t choose at this point, the player is asked whether they wish to change their mind and choose the other door instead.

The accepted answer online is that it makes mathematical sense to always change your mind at this point. The reasoning is that the first choice you make is one of three doors meaning there are two possible losing choice but after one is revealed the odds are in favor of the third door claiming the odds to Be better than the original one out of three.

I struggled with this concept over and over in my mind chiefly because the moment you choose one of the three, a door is revealed and you are set with your choice or the other choice. It seems logical at first glance that it’s 50/50 at this point.

When’s the third door is discarded The choice seems clear. Either the one you chose or the other one. It gets tricky when you realize the one you chose had a one in three chance of being the right one meaning two out of three chances of it being the wrong one and then they claim the third one doesn’t have those odds. It either is or isn’t the winner.

At first, I struggled with the concept that the first term doesn’t make any difference at all no matter which one you choose you are always greeted with two boxes one winner and one loser. If you choose box one box two or box three no matter what your choice, you are greeted with an a or b choice.

The trick is that your first choice really has a very good chance of being a loser. Two out of three and this does not change when you remove the third box. It just seems like it should.

Confusing things more I believe the third door also has a one in three chance of being the winner from the beginning even though you didn’t choose it and therefore two doors to be the loser. After one of the doors is removed you still don’t know where your choice was correct or not and even though your initial choice was one out of three, so is the third box.

Maybe I’ll have to go online when I’m not quite so tired from being up all night and look at some other experts solution as to why that third box has a better chance of being correct than the other three or two. I suppose when I’m choosing it it really is one out of two at this point. Again, the first turn seems less relevant to the whole puzzle.

Of course all of this thinking doesn’t take into account the fact that the boxes are chosen by human and the third box discarded is also chosen by a human and there is probably a lot of math around human preference of choosing the middle box more than the other two or choosing box number one more than three. That’s a separate math problem I suppose.

Anyway it’s 8:00 a.m. and I fed the animals for breakfast so maybe I’ll go back to sleep if this problem still doesn’t nag on my brain and keep me awake..

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