The Monty Hall problem is easy if you simply write out the possibility tree and count the final nodes. If I get fancier than counting, it gets tricky. I got this one wrong too before I checked using the dumb counting method.
The options above the line are 50% each. All I can do is change the weightings of the possibilities below the line, which must also sum to 50%.
The brain gets confused because there's more Ts in the lower half. However, since they must sum to 50%, they're worth less. More on this at the bottom.
If you run the experiment once, then it is kind of hard to think about. Instead, run it a thousand times.
If I think the probability of tails is 1/2, I will report it half the time, and my probability will match the number of times I'm right.
If I report tails all the time, I will be right half the time. If I report heads all the time, I'll be right half the time.
If I think the probability of tails 2/3rds, I will report it more often in the bottom, but less often than in the top, and be right half the time. (1/9 + 1/9 + 4/9 = 6/9 = 2/3. 2/6 + 1/6 = 1/2.)
I dunno. Doesn't seem like I need a whole big book of calculus to solve the problem. I just have to put aside my pride long enough to do the dumb counting thing.
Self-location information? Conditionalization? Principle of indifference? Imagine rule? Impressive, clever-sounding things I don't need to know or care about. I just make a stupid picture and look at it.
The trick being my choice of perspective. If I want to be right, I can pick whatever I want, as I'll be right half the time. If instead we're talking about the number of reports the experimenter sees, then it's different. If the observer is writing down each report separately, and I want to generate as many 'matching' entries as possible, it favours reporting tails. Tails causes double the records, effectively doing the experiment twice on that branch.
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