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| Natura: | Preprint |
| Pubblicazione: |
2024
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| Accesso online: | https://arxiv.org/abs/2411.09716 |
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| _version_ | 1866910699123376128 |
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| author | Peng, Kenny |
| author_facet | Peng, Kenny |
| contents | We consider equilibrium one-on-one conversations between neighbors on a circular table, with the goal of assessing the likelihood of a (perhaps) familiar situation: sitting at a table where both of your neighbors are talking to someone else. When $n$ people in a circle randomly prefer their left or right neighbor, we show that the probability a given person is unmatched in equilibrium (i.e., in a stable matching) is $$\frac{1}{9} + \left(\frac{1}{2}\right)^n\left(\frac{2n}{3} - \frac{8}{9} + \frac{2}{n}\right)$$ for odd $n$ and $$\frac{1}{9} - \left(\frac{1}{2}\right)^n\left(\frac{2n}{3} - \frac{8}{9}\right)$$ for even $n$. This probability approaches $1/9$ as $n\rightarrow \infty$. We also show that the probability \textit{every} person is matched in equilibrium is $0$ for odd $n$ and $\frac{3^{n/2}-1}{2^{n-1}}$ for even $n$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2411_09716 |
| institution | arXiv |
| publishDate | 2024 |
| record_format | arxiv |
| spellingShingle | Stable Tables Peng, Kenny Probability Combinatorics We consider equilibrium one-on-one conversations between neighbors on a circular table, with the goal of assessing the likelihood of a (perhaps) familiar situation: sitting at a table where both of your neighbors are talking to someone else. When $n$ people in a circle randomly prefer their left or right neighbor, we show that the probability a given person is unmatched in equilibrium (i.e., in a stable matching) is $$\frac{1}{9} + \left(\frac{1}{2}\right)^n\left(\frac{2n}{3} - \frac{8}{9} + \frac{2}{n}\right)$$ for odd $n$ and $$\frac{1}{9} - \left(\frac{1}{2}\right)^n\left(\frac{2n}{3} - \frac{8}{9}\right)$$ for even $n$. This probability approaches $1/9$ as $n\rightarrow \infty$. We also show that the probability \textit{every} person is matched in equilibrium is $0$ for odd $n$ and $\frac{3^{n/2}-1}{2^{n-1}}$ for even $n$. |
| title | Stable Tables |
| topic | Probability Combinatorics |
| url | https://arxiv.org/abs/2411.09716 |