http://gothamist.com/2014/09/23/rock_paper_scissors_hustler.php
rock paper scissors…lose
That is awesome
I think the AF’ers in NYC should seek him out and see if he can overcome the superior performance guaranteed by the charter.
There used to be a professional rock, paper, scissors league. www.usarps.com. I know they used to have sanctioned events in Vegas. Lol
This guy quit a job at Apple to make shapes with his hands in dive bars.
Yeah, but he was probably a “genius” working in an Apple Store. Rock, Paper, Scissors is more prestigious.
He brings in as much as $300/night. Only Uber drivers do better than that.
$300/night, 4 days a week is a little more than $60k a year without taxes. Not shabby, but how much of it does he spend on alcohol.
So more like an all time best rather than a consistent figure. More power to him, though.
But why doesn’t he just go sell cars or something?
The dominant strategy for each player that results in a Nash equilibrium for RPS is to randomly decide if you’ll go for a rock, paper or scissors on each turn. Analytically, two things are left to be exploited in a real-world scenario:
To find a pattern in the opposing player’s game with some confidence level that will make it reasonable for you to exploit it, you need a rather long sequence of consecutive games with the same player. For instance, a simple pattern could be that you suspect he is more likely to play rock than any of the other two options. You can start wih a prior assumption of beta(30,60) for the probability of playing rock - peaked distribution centered around 1/3 with stadard deviation of +/- 5%. After G games, you observe he played R rocks, so your posterior becomes beta(30+R, 60+G-R). The expected value is no longer 1/3, but it is (30+R)/(90+G). Is this significantly different from my prior assumption that the guy plays rock with probability 1/3? That depends on what “significantly different” means to you, but let’s say you want to see a result of 43% (1/3+10%) to be sure. To get there, if you’ve only played 30 games (G=30), you’ll need to see at least 22 rocks (R=22) - seems pretty unreasonable. Your opponent must be drunk or have a strong bias for rocks to engage in such predictable pattern. But if you played 300 games, you only need to see at least 139 rocks - more subtle. You’d likely get bored after 50 games though.
For example, how likely is it to see a consecutive stretch of “no rock” if you have played 10,000 games? The probability of no rock is 1-1/3=2/3 in a single game that is decided in a truly random fashion. The expected value for the “longest run” for an event with probability p for a sequence of n trials is LN(n*(1-p))/LN§. So the expected stretch of “no rock” in 10,000 games would be 20 (±6), between 14 and 26 with 95% confidence. Most people only have a short memory and would perceive a stretch of 20 consecutive games with no rock as non-random, so they’ll switch more often. You can exploit that by anticipating shorter stretches from a human player than what would be normal in a truly random sequence, but you need to find a really patient player for an opponent.
I call BS on this guy.
^whoa
^^ Search him out in a NYC bar and prove him wrong.
Mob Strip, your new sig;
8=============================D
thanks homie, I ain’t getting played by some halfway RPS crook… he’s just a shook one, as quantitative analysis suggests.
^ Respect.
Awesome