How to choose best option overall? I do not know proper mathematical (and/or English) words, so bear with me...
Lets say, there are 4 possible outcomes. We can distinguish which one is which - I will call them 1, 2, 3, 4 while 1 is "best" and 4 is "worst" outcome.
I want to choose good outcome rather than bad, but (as it works in reality) I can always only take what I see now or skip it (forever) and look at next, until I run out of options and than I have to choose what is left.
I just watched Mathematical Way to Choose a Toilet on Numberphile where pretty lady in red with lovely voice and cute smile explains how to maximize chance to reach "best" outcome. Her answer is look at first ~37% (I expected 42 to be honest) and then choose first better of what you saw untill now.
Since, I am not that picky, I do not really care about "best". More often good enough is better than best.
So, I tried to invent other strategies where I look for most of "goodness" overall without picking best option:


*

*not looking or not taking anything - not interested

*picking bad/worst option on purpose - not interested

*simply by picking first outcome (or anyhow randomly), I get number "2.5" (I call it 50% of
goodness). Looks like any strategy (other than not trying or picking bad option on purpose) is better than this.

*look at first and take any better than first gave me "2.125" (62.5% of goodness)

*look at first and take any better or next worse (when "N" is first, take "<N" or "N+1") gave me "1.875" (70.9% goodness). This is what I aim for - by accepting worse than best I increased overall "goodness" from 62.5% to 70.9%.


So, what are other simple (=to use in daily life) algorithms/strategies which give largest "goodness"?
Just to visualise all possible outcomes for 4 distinct variations:
1234
1243
1324
1342
1432
1423
2134
2143
2314
2341
2431
2413
3214
3241
3124
3142
3412
3421
4231
4213
4321
4312
4132
4123

UPDATE: What are other strategies when number of choices is lets say 100?
 A: You can look at your choices from the end:
If you choose the fourth you see, your expected rank chosen is $\frac52 = 2.5$
If you choose the third you see 


*

*if it is the best of the first three, your expected rank chosen is $\frac54$

*if it is the middle of the first three, your expected rank chosen is $\frac52$

*if it is the worst of the first three, your expected rank chosen is $\frac{15}{4}$


so you will not choose the third if it is the worst of the first three, you might or might not choose it if it is the middle of the first three (it makes not a difference to the expectation, but choosing in the case does reduce the risk of getting the worst), and you will choose the third if it is the best of the first three.  So if you get to the third stage, your expected rank chosen is $\left(\frac54+\frac52+\frac52\right)/3= \frac{25}{12}.$
If you choose the second you see 


*

*if it is better than the first, your expected rank chosen is $\frac53$

*if it is worse than the first, your expected rank chosen is $\frac{10}3$


so you will not choose the second if it is worse than the first and you will choose the second if it is better than the first.  So if you get to the second stage, your expected rank chosen is $\left(\frac53+\frac{25}{12}\right)/2= \frac{15}{8}.$
If you choose the first you see, your expected rank chosen is $\frac52$ so you will not choose the first you see as waiting for the second choice is better.
So the optimal expected rank chosen is $\frac{15}{8}=1.875.$
