Someone writes a bunch of random numbers down on paper and hands you three of them face down. You flip them over one at a time and have to stop when you think you have identified the largest of the three. Move down for correct answer. Flip over the first two. If the second one is larger, stop. If the second number is smaller, flip over the third one. Performing this way gives you a .66 chance of winning.
or, if you prefer cartoonnnnns Hmmmmm. I think your question is MUCH BETTER than the simple Bayesian presented above. A LOT better. Is it *applicable* to trading??? I don't think so -- "not so much", anyway. Why? Cuz the set-up is constant, and in trading, you might *assume* the situation is constant for a minute or three. But if you're working a market, trying to find/anticipate where in a broader Bid_Ask spread that you might get hit (on the BUY or the SELL)...... while you might 'flip cards' and identify things going up or down (as per your set-up), it so often happens that the other parties bidding/offering switch up, go away, cancel or defer orders..... and the info you *thought* you got a moment ago --- IS NOW STALE! So....... I'm going to think about this some more, but, Hell -- even in real estate, price discovery is subject to mad vibrations -- and the constancy requirement of a Bayesian process is not there .... for very long. Nerds unite!
Let's see if this holds true. Someone writes down 5000 numbers and hands you 3. The first one you flip over is 1. The second you flip over is 10. Do you have a 66% chance 10 is the highest number of the 3?
This is not a Monty Hall problem. Without knowing the distribution from which the numbers have been drawn from (or from a distribution of possible distributions), it’s very difficult to say much in general. For example, if the three numbers are unique integers 1,2,3, your strategy should be to stop when you see 3 and go on until you do (which has a 100% success rate). OP’s proposed strategy has a 50% chance of success, succeeding on triples 213, 132, 231 and failing on 123, 312, 321. The point here is that the strategy is conditional on the absolute values observed in relation to the range of the underlying distribution, not on some unconditional strategy. One approach to such situations, which is somewhat counter intuitive, is to assume your own distribution (with nonzero probability density over any interval) and generate a random number yourself as a representative of the unseen one and act accordingly. See the “two envelope problem” and this approach to (re)solving it. https://en.m.wikipedia.org/wiki/Two_envelopes_problem#Randomized_solutions
My bad, by posting videos and distracting from the OP's written words. The question was -- well, the question itself was screwed up: no distribution, no range. ("No [fixed;constant] universe of outcomes" in statistical terms, so we don't know the sample's possible placement within.) That was TheBigShort's point. (And MoreLeverage, too!) My point was that the events constraining parameters (in this case, the individual buyers and sellers) is non-constant for a trading situation, but not so with the 'quiz'. Thus, regardless of any settled distribution, the cards are reshuffling even as the 'game' is underway! The whole conditional probabilities framework is tricky, and sometimes counter-intuitive, and begs for some mental gymnastics -- that makes it fun. ("Yayyyyy!") But to apply it to trading?? I've given it some thought over the past 12 hours, and I think the non-constantcy kills it -- whether for active trading or for use in fundmentals.... I'm jus' not seein' it. (Fun question, though. )
tommcginnis This is why I need better tools! the cards are reshuffling even as the 'game' is underway! ES
Duuuuuuuude! I did my own motorcycle work for a number of years (and many thousands of miles).... If I needed a too to fit a situation?? I *made* it. [Hint hint, hint-hint-hint. Hint.]