Timeline for Hunting an invisible target
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Apr 28 at 7:00 | history | bounty ended | CommunityBot | ||
| S Apr 28 at 7:00 | history | notice removed | CommunityBot | ||
| Apr 24 at 14:50 | comment | added | Nate River | @mathworker21 Yes it is a pretty open ended statement by intention. In this case, the specific result you mention sounds interesting though. | |
| Apr 24 at 14:47 | comment | added | mathworker21 | @NateRiver What do you consider a "good" strategy? Like I said, you can get $\Omega(N)$ correct guesses. | |
| Apr 24 at 14:33 | comment | added | Nate River | @mathworker21 The exact optimal strategy is of interest, but also anything that can be said about a “good” strategy is also of interest, since optimisation problems like these are hard to solve exactly. | |
| Apr 23 at 16:12 | comment | added | mathworker21 | Yuval Peres told me a proof for a $c\, N$ lower bound. Do you care about the optimal constant? Do you care about the exactly optimal strategy? | |
| S Apr 20 at 5:19 | history | bounty started | Nate River | ||
| S Apr 20 at 5:19 | history | notice added | Nate River | Draw attention | |
| Feb 19 at 23:13 | comment | added | user479223 | Updated observation: if you know the optimal strategy for $N-k$ and you guess correctly at $k$ then you are fine. Maybe some kind of Bellman… | |
| Feb 19 at 22:54 | comment | added | Nate River | @user479223 It may be that the two end up equivalent but it is at least not a priori immediate. | |
| Feb 19 at 22:53 | comment | added | Nate River | @user479223 That was my initial thought too, but there is actually quite a subtle but large difference between minimising the time taken to guess the correct location, and maximising the expected number of correct guesses. This is because when the game resets, you now have $N-k$ turns to go, which affects the expected value, and even the optimal strategy of the new game. In other words the expected value is a weighted sum of the time taken to guess correctly, so it’s not immediate that minimising the expected time (an unweighted sum) maximises the expected number of guesses. | |
| Feb 19 at 22:47 | comment | added | user479223 | An observation: when you correctly guess the game effectively restarts. So you want to minimize the number of incorrect guesses until a correct guess. | |
| Feb 19 at 20:31 | history | edited | Nate River | CC BY-SA 4.0 |
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| Feb 19 at 20:25 | history | edited | Nate River | CC BY-SA 4.0 |
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| Feb 19 at 20:25 | history | undeleted | Nate River | ||
| Feb 19 at 20:22 | history | deleted | Nate River | via Vote | |
| Feb 19 at 20:19 | history | asked | Nate River | CC BY-SA 4.0 |