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Though I haven't seen a rigorous proof of this, it seems intuitive. (I'm not 100% sure that wasted outs, like when you're QQ vs KK and hit quads on one runout, don't affect EV, but I think they are equivalent to burn cards.) Agreed?
It is indeed 0 EV. Surely someone in your line of work could construct the necessary symmetry argument (plus an appeal to the linearity of expectation). :) The main consideration is a metagame one: do you want your opponents to think they'll be able to run it twice?
Haha, thanks, @NateMeyvis. Unfortunately my discrete math is a bit rusty. 
I was thinking that one should be able to show this via a symmetry argument but didn't think hard enough how to write it down formally.
I agree that the main consideration is meta-game, as expressing willingness to run it multiple times may encourage opponents to get it in incorrectly with draws.
Does this mean you’re always doing business in order to flatten variance and also to try to encourage people to chase without odds in hope of getting multiple run outs?
I certainly am
Ended up picking this up again 11 months later and now realize it trivially follows from linearity of expectation; didn’t previously appreciate that it applied to non-independent random variables. Given your CS background, your discrete math skills are likely stronger than mine
I’m much more about solving PDEs and studies analysis, differential geometry and other more physics-relevant subjects.
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