jarljal

[jarljal @jarljal]

@MagicOnline will an in pod swiss draft queue ever be back?

[jarljal @jarljal]

@ereliuer_eteer after you "fix" it, nothing will bloom again

[jarljal @jarljal]

@MillThat It's both ugly and bad

[jarljal @jarljal]

@hftgod *around 0.4, not exactly 0.4. The idea behind 1/(2+x) = x is only a first order approximation, that uses that n=100 is large enough to we should see some "mixing"

[jarljal @jarljal]

@hftgod Let's start with an approximation: assume after an odd number of tails and a single heads, we end up with around 1$ (I'll be more precise later). How late do we expect to see the last odd substring of tails? Or by symmetry, how many throws do we see before our first odd run of T?

[jarljal @jarljal]

@hftgod Hence the optimal price for the game is 2 + the expected amount of money after the last odd run of tails. Doing some extra crude approximations, we expect the amount of money x we have after the last odd run of tails to satisfy: 1/(2+x) = x, which is 0.4. So I'd pay around 2.4$

[jarljal @jarljal]

@hftgod The answer is 10/3. But to be more precise, what we really want to know is how many heads we expect to see before our first odd run of tails, which is 2.

[jarljal @jarljal]

@ANAIP_es esta es la mejor publicidad que se os ocurre? no queda muy claro si estais a favor o en contra

[jarljal @jarljal]

@batallivol quina persona més petita

[jarljal @jarljal]

@joanjorquera Que deies tu de Palestina?

[jarljal @jarljal]

@ZhangTaisu It's unreasonable to compare an assistant prof to undergrads/early grads

[jarljal @jarljal]

@TylerGlaiel Claim: 1>Q(20)\geq P(20). The first inequality comes from the law of large numbers in the second random walk (see https://t.co/D5ln1mLCcv for an example). The second inequality comes from the fact that the first game dominates the second when we start with 20$.

[jarljal @jarljal]

@TylerGlaiel I haven't seen any correct proof so here is mine. Let P(n) be the chance that we hit 0$, I'll show P(20)<1 and from there it follows P(n)<1 for all n>0. Consider a different game where heads -> we gain 20$, tails -> we lose 10$, and let Q(n) be the chance that we hit 10$.