Timeline for Minimizing the expected time to get a string of heads with biased coins
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 24, 2023 at 16:26 | comment | added | Lorenzo Pompili | @Henry yes, after some corrections, I see that what you wrote agrees with what I wrote. Thank you! | |
| Jan 24, 2023 at 16:25 | comment | added | Lorenzo Pompili | It's not that simple to do, but it looks straightforward. You could try to reproduce the argument of Henry in his question, which will work at least when $\alpha$ is small enough. For instance, for $n=3$, you first write $p_3=\alpha-p_1-p_2$ in the expression for $E_3$; then you fix $p_2$ and minimize $E_n$ with respect to $p_1$; then you minimize what you get with respect to $p_2$... | |
| Jan 24, 2023 at 16:19 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 | added 7 characters in body; added 46 characters in body |
| Jan 24, 2023 at 16:07 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 | added 1 character in body |
| Jan 24, 2023 at 15:49 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 | deleted 278 characters in body |
| Jan 24, 2023 at 10:01 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 | (Corrected some typos) |
| Jan 24, 2023 at 4:28 | comment | added | S. Green | Okay, I agree with that expression for $E_n$. Using Lagrange multipliers it is possible to derive a set of nonlinear equations that minimize $E_n$. Solving these in closed form does not seem so easy for $n > 2$. Numerically it looks like $p_1$ and $p_n$ converge (but to what?) as $n \to \infty$ but it's hard to even get numerical results for $n > 10$. | |
| Jan 24, 2023 at 2:36 | comment | added | Henry | My calculations suggest $E_n=\dfrac{1+p_1(1+p_2(1+p_3(1+\cdots (1+p_{n-1}(1+p_n))\cdots)))}{p_1p_2p_3\cdots p_{n-1}p_n}$ which I think may be equivalent to what you have written | |
| Jan 24, 2023 at 2:00 | comment | added | S. Green | Your expression for $E_n$ isn't equivalent to mine, I don't think. Consider $n=2$ and $p_1 + p_2 = 1$, then if I'm not mistaken you get $E_2 = \frac{1}{p_1} + 2$, but this possibly doesn't account for the fact that when you get heads and then tails, you have to add 2 to your running total of flips. | |
| Jan 23, 2023 at 23:32 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 | deleted 58 characters in body |
| Jan 23, 2023 at 23:20 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 | deleted 58 characters in body |
| Jan 23, 2023 at 23:10 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 | added 17 characters in body |
| Jan 23, 2023 at 22:58 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 | added 17 characters in body |
| Jan 23, 2023 at 22:53 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 | added 91 characters in body |
| Jan 23, 2023 at 22:42 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 | added 91 characters in body |
| Jan 23, 2023 at 22:36 | history | answered | Lorenzo Pompili | CC BY-SA 4.0 |