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Oct 28 at 18:16 comment added Tobias Fritz I'm in the process of writing this up as a paper for an expository journal. In case that you're interested in coauthorship @jlewk, just shoot me an email! (Iif not, I'll of course still attribute this idea to you.)
Jul 13 at 8:11 comment added Tobias Fritz Nice! Applying this method with $Y_{i,j} \sim \frac{1}{2} \delta_0 + \frac{1}{6} \delta_2 + \frac{1}{3} \delta_3$, I actually get a probability of $\approx 0.365$. I will try to push this some further and then write another answer on how I came up with this choice.
Jul 12 at 22:23 history edited Michael Hardy CC BY-SA 4.0
This results in proper spacing to the right of \sim.
Jul 12 at 17:30 history edited jlewk CC BY-SA 4.0
typo/aesthetics
Jul 12 at 16:33 comment added Will Sawin In the cases where Tobias Fritz mentioned already finding the optimal value, this procedure (essentially) gets it: It's asymptotically approached here in the limit as $p$ goes to $0$. So it's possible this could be optimal here as well, although I would like to check local optimality also.
Jul 12 at 16:13 comment added jlewk Thanks. With the explicit definition as a sum, simulations confirm around $3.3-3.4$.
Jul 12 at 16:10 history edited jlewk CC BY-SA 4.0
Add simulations
Jul 12 at 15:58 comment added Will Sawin It might be clearer to definine the distribution all at once as $X_i = Y_{i,1} + t Y_{i,2} + t^2 Y_{i,3} + \dots$ where $Y_{i,j}$ are i.i.d. Bernoulli's with the same probability $p$ and $t$ a very small constant. Then $X_1 + X_2 + X_3 < 2 X_4$ if and only if for some $i$ we have $Y_{1,i} + Y_{2,i} + Y_{3,i} < 2 Y_{4,i}$ and for all $j<i$ we have $Y_{1,j} + Y_{2,j} + Y_{3,j} < 2 Y_{4,j}$. One can derive your $f(p)/ (1-g(p))$ formula from this.
Jul 12 at 15:55 history edited jlewk CC BY-SA 4.0
typo where t is located in the weighted sum
Jul 12 at 15:44 history answered jlewk CC BY-SA 4.0