Timeline for Expected outcome for repeated dice rolls with dice fixing
Current License: CC BY-SA 4.0
41 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 25, 2024 at 20:38 | answer | added | Woett | timeline score: 3 | |
| Aug 4, 2020 at 9:46 | history | edited | Joachim Breitner | CC BY-SA 4.0 | Use $, not ` |
| Sep 27, 2019 at 14:07 | comment | added | Joachim Breitner | Every turn you have to fix at least one die. So the maximum number of remaining turns is the number of remaining dies. If you fix two dies, you are essentially forfeiting a turn. | |
| Sep 27, 2019 at 13:46 | comment | added | Aqqqq | Why would re-roll a 6 give me "one extra roll for all other dice"? | |
| Jul 3, 2016 at 9:54 | comment | added | Joachim Breitner | Thanks, I adjusted my story, as I no longer can claim that I don’t know that :-) | |
| Jul 2, 2016 at 8:42 | comment | added | joriki | Ha, I was about to suggest to include a link to your blog post here, but you just did. "I do not know if the tweet made a difference, but a day later some joriki came along": No, I didn't see the tweet; I saw Byron's bounty and I figured it must be an interesting problem :-) | |
| Jul 2, 2016 at 8:01 | history | edited | Joachim Breitner | CC BY-SA 3.0 | added 174 characters in body |
| Jul 1, 2016 at 17:48 | vote | accept | Joachim Breitner | ||
| S May 18, 2016 at 1:58 | history | bounty ended | CommunityBot | ||
| S May 18, 2016 at 1:58 | history | notice removed | user940 | ||
| May 15, 2016 at 21:12 | answer | added | Mark Hurd | timeline score: 0 | |
| May 15, 2016 at 6:12 | history | tweeted | twitter.com/StackMath/status/731729002474405888 | ||
| May 12, 2016 at 23:50 | answer | added | Milo Brandt | timeline score: 1 | |
| S May 12, 2016 at 21:53 | history | suggested | Guilherme Marthe | CC BY-SA 3.0 | There was a duplicated word in the third bullet point of the questions (the word was "all"). |
| May 12, 2016 at 21:20 | review | Suggested edits | |||
| S May 12, 2016 at 21:53 | |||||
| May 12, 2016 at 19:18 | comment | added | Joachim Breitner | Sorry for leaving that implicit. I’m interested in the strategy that maximizes the expected score. | |
| May 12, 2016 at 19:17 | history | edited | Joachim Breitner | CC BY-SA 3.0 | Add that the score should be maximized. |
| May 12, 2016 at 18:55 | comment | added | user940 | @YvesDaoust Oops, sorry. I will leave my comment up anyway. | |
| May 12, 2016 at 18:54 | comment | added | user65203 | @ByronSchmuland: the comment was intended to the OP. | |
| May 12, 2016 at 18:53 | comment | added | user940 | @YvesDaoust I am interested in solving the problem given in the "bounty box". There is a recursive definition for a concrete sequence $v_n$ of numbers, and I ask for a proof or disproof of $v_n\leq v_{n-1}+6$ for all $n\geq1$. You can ignore the original post that motivates this question, if you like. The more people looking at this problem, the better! | |
| May 12, 2016 at 18:46 | comment | added | wolfies | I agree with @Yves - the question appears meaningless. There is no purpose, no stated goal, or objective function. Are we trying to max the sum, or minimise it, or keep rolling dice for as long as possible, or what? | |
| May 12, 2016 at 18:26 | comment | added | user65203 | How can this question have an answer if you don't specify a fixing strategy ? You don't even state a goal to achieve ! | |
| May 12, 2016 at 18:14 | answer | added | joriki | timeline score: 29 | |
| May 12, 2016 at 16:39 | comment | added | mercio | it would be nice if we could prove that $v_{n-1} + v_{n+1} > 2 v_n$ with some kind of strategy stealing argument. | |
| S May 11, 2016 at 0:25 | history | bounty started | CommunityBot | ||
| S May 11, 2016 at 0:25 | history | notice added | user940 | Canonical answer required | |
| Apr 17, 2016 at 17:38 | comment | added | user940 | @JoachimBreitner Well, that explains what I am doing wrong. Thanks! | |
| Apr 17, 2016 at 15:48 | comment | added | Joachim Breitner | Ah, nevermind. Every time, you need to fix at least one die. So if you roll $1,2$, you need to keep one (likely the two) and reroll only the other. With these rules, you cannot reach 8.5. | |
| Apr 17, 2016 at 15:42 | comment | added | Joachim Breitner | Ok, manually doing the calculation I also get 8.5, so the code below has a bug. Thanks! | |
| Apr 17, 2016 at 12:30 | comment | added | user940 | I looked at all 36 pairs $(x,y)$ where $x,y\in\{1,2,3,4,5,6\}$ and applied my strategy to them. | |
| Apr 16, 2016 at 15:57 | comment | added | Joachim Breitner | How do you get this result? | |
| Apr 16, 2016 at 14:53 | comment | added | user940 | I'm trying to understand the rules of your game. Suppose that $n=2$, and that I adopt the following strategy: Roll both dice, and on the second turn re-roll any die showing "1", "2", or "3". After this, count the sum. The average value of this strategy is $17/2=8.5$ which is greater than your value of $8.23611$ below. What am I doing wrong? | |
| Feb 4, 2015 at 9:58 | history | edited | Joachim Breitner | CC BY-SA 3.0 | Removed example calculation, it was wrong. |
| Feb 3, 2015 at 20:45 | answer | added | Joachim Breitner | timeline score: 4 | |
| Jan 24, 2015 at 23:30 | history | edited | Joachim Breitner | CC BY-SA 3.0 | added 32 characters in body |
| Jan 24, 2015 at 23:19 | history | edited | Joachim Breitner | CC BY-SA 3.0 | typo |
| Jan 24, 2015 at 23:19 | comment | added | Joachim Breitner | Yes, at the end all dices are counted. | |
| Jan 24, 2015 at 23:18 | comment | added | advocateofnone | are you calculating the expectation of sum of values on all die ? | |
| Jan 24, 2015 at 23:17 | comment | added | Joachim Breitner | Some possibly related questions: math.stackexchange.com/questions/1010729 math.stackexchange.com/questions/223238 math.stackexchange.com/questions/660523 | |
| Jan 24, 2015 at 23:14 | review | First posts | |||
| Jan 24, 2015 at 23:15 | |||||
| Jan 24, 2015 at 23:14 | history | asked | Joachim Breitner | CC BY-SA 3.0 |