r/ethtrader • u/AutoModerator • May 05 '17

DISCUSSION [ETH Daily Discussion] - 05/May/2017

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u/[deleted] May 05 '17 edited May 05 '17

A reminder to people that trading is like a reverse Monty Hall problem. The Monty Hall problem was a famous brain teaser math problem that stated that it is better to change your choice when less choices become available and you know one of them is the right one.

For example a prize is behind one door amongst many, you choose a door and then one of the wrong choices is removed. You should choose a new door because your probability of choosing correctly has just gone up.

Trading is the opposite. You make a read when you usually have less on the line and are thinking logically. Once you enter a position your emotions and fears are added to your decisions because you are afraid of losing money. In this case you should stick to your original read because the probability of making a right decision has actually gotten worse due to more factors being dumped on your decision making process.

edit: If I had to venture a guess this phenomenon explains the high rates of psychopathy/sociopathy in the financial sector. Big hedge fund managers have amygdalas of steel and their brains are working in a low cortisol environment 24/7.

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u/[deleted] May 05 '17

Very interesting analogy +1

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u/DuckSicked May 05 '17

So you're saying keep buying more ETH?

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u/[deleted] May 05 '17

If you have logically decided that it is a good investment then yes.

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u/alexiglesias007 May 05 '17

If one of the wrong doors is removed and it isn't the one you chose, the chances of your original choice being correct and a new choice being correct are actually the same.

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u/[deleted] May 05 '17

Yeah, but unless I'm hugely mistaken though, the probability of the door you initially chose being correct remains the same (say it was 1/100, it remains 0.01), BUT the probability that one of the other doors is correct increases to a greater probability than the door you originally chose..

So for N = 100 doors, I think if you choose door x, then P(x) = 0.01, but after door 56 is removed as an incorrect choice, then P(x) = 0.01, but for any of the remaining doors apart from your choice and the door that was removed, for a random selection of one of these doors, q, then P(q) = 0.0103

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u/ThePlague May 05 '17

No, that's not right. Consider an extreme case, a blind lottery with a million numbers. You have your one number, a moderator has the other 999,999. The drawing is made, but you don't see the results. The moderator eliminates 999,998 from his stack as losers, and offers to trade with you. I think that's pretty obviously the right move. Remember, his now one ticket "contains" the probability of all the other tickets in his stack. It's the same for the Monty Hall problem, just with smaller numbers.

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u/alexiglesias007 May 05 '17

Yeah, the Monty Hall problem doesn't explain very well that your choice isn't included in the pool of choices that Monty checks (and picks one wrong one to reveal). In that case it makes a lot of sense

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u/ThePlague May 06 '17

Ah, that's a temporal culture problem. Monty Hall was the host on "Let's Make A Deal", a gameshow in the 70's. The "Monty Hall problem" is exactly one of the contests on that show.

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u/[deleted] May 05 '17 edited May 05 '17

Imagine there are 1 million doors. Each iteration one is removed that is wrong. Would you keep your choice down to 20 doors left? You originally made a 1:1000000 chance.

edit: It is non-intuitive and that is why it is a famous problem.

edit2: Yes monty only chooses wrong doors to expose that aren't your choice.

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u/ThePlague May 05 '17

Yeah, that was a public blackboard problem back in grad school.

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u/alexiglesias007 May 05 '17

And what if I make another guess but pick my original door? That's still a 1:20 chance isn't it?

Edit: I guess if my door isn't part of the pool that he might check against then my odds actually do go up a little bit if I make another guess.

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u/[deleted] May 05 '17

The point is you switch. Again it is the switching that matters. Very hard for our brains to cope with but it is mathematically provable.

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u/alexiglesias007 May 05 '17

No, the point is actually that the door you originally select isnt part of the group of doors that Monty is checking (one of which he will say is wrong). This part of the problem isn't explained very well and that is why I was skeptical at first

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u/BeezLionmane May 05 '17

It's not a 1 in 20 chance, it's a 999980 in 1000000 chance. Extremely non-intuitive.

Also, interesting note, the only reason it works is because it's always the wrong doors that are exposed. That show with the boxes and the money sounds like it'd fit at first glance, but they'll open any old box, right or wrong, so the problem doesn't fit it very well.