A problem by Spock Weakhypercharge

2 sec - 2 hits 5. Teams are 2, 4, 5 vs 3.
3 sec - 3 hits 5. Teams are 2 , 4 vs 3, 5.
4 sec - 2 hits 3, 4 hits 5. Teams are 2, 3, 4, 5. Game ends.

I\quad think\quad the\quad answer\quad is\quad \frac { 3 }{ 8 } \quad and\quad here's\quad my\quad reasoning:\\ \\ \frac { }{ } There\quad is\quad a\quad \frac { 1 }{ 2 } \quad chance\quad the\quad game\quad ends\quad at\quad t\quad =\quad 3\quad s,\\ \\ it\quad happens\quad when\quad the\quad 2nd\quad grader\quad hits\quad the\quad 3rd\quad grader\quad at\quad t\quad =\quad 2\quad s.\\ \\ Therefore\quad the\quad probability\quad that\quad it\quad will\quad not\quad end\quad at\quad t\quad =\quad 3\quad s\quad is\quad \frac { 1 }{ 2 } .\\ \\ \\ At\quad t\quad =\quad 4\quad s,\quad there\quad are\quad three\quad possible\quad team\quad distributions:\\ \\ each\quad is\quad equally\quad probable\quad (\frac { 1 }{ 3 } \quad chance)\\ \\ 2,4\quad vs\quad 3,5;\quad 2,5\quad vs\quad 3,4;\quad and\quad 4,5\quad vs\quad 2,3\\ \\ we\quad evaluate\quad each\quad case\quad separately.\\ \\ \\ 2,4\quad vs\quad 3,5\\ \\ The\quad game\quad lasts\quad more\quad than\quad 5\quad s\quad if\quad 5\quad does\quad not\quad get\quad hit\quad by\quad 2\quad or\quad 4,\\ \\ since\quad game\quad ends\quad at\quad 4\quad s\quad if\quad 2\quad and\quad 4\quad have\quad different\quad targets,\\ \\ and\quad game\quad ends\quad at\quad 5\quad s\quad if\quad both\quad 2\quad and\quad 4\quad target\quad 5.\\ \\ The\quad probability\quad is\quad \frac { 1 }{ 4 } \\ \\ \\ 2,5\quad vs\quad 3,4\\ \\ Since,\quad 2\quad and\quad 4\quad are\quad on\quad opposite\quad teams,\\ \\ at\quad 5\quad s,\quad there\quad will\quad be\quad two\quad people\quad on\quad either\quad team.\\ \\ since\quad only\quad 1\quad person\quad (5th\quad grader)\quad will\quad throw\quad at\quad 5s\\ \\ The\quad game\quad cannot\quad end\quad within\quad 5s,\\ \\ therefore\quad the\quad probability\quad is\quad 1.\\ \\ \\ 4,5\quad vs\quad 2,3\\ \\ this\quad case\quad is\quad the\quad same\quad as\quad the\quad previous\quad case.\\ \\ probability\quad is\quad 1.

we\quad multiply\quad the\quad probabilities\quad of\quad each\quad case\quad with\quad \frac { 1 }{ 3 } \\ \\ which\quad is\quad the\quad probability\quad that\quad each\quad case\quad occurs.\\ \\ then\quad add\quad them,\quad since\quad they\quad are\quad mutually\quad exclusive\\ \\ We\quad get\quad \frac { 1 }{ 3 } (\frac { 1 }{ 4 } )\quad +\quad \frac { 1 }{ 3 } (1)\quad +\quad \frac { 1 }{ 3 } (1)\quad =\quad \frac { 3 }{ 4 } \\ \\ then\quad we\quad multiply\quad the\quad probability\quad that\quad the\quad game\quad will\quad not\quad end\quad at\quad 3s\\ \\ with\quad \frac { 3 }{ 4 } \quad since\quad the\quad two\quad events\quad are\quad not\quad mutually\quad exclusive.\\ \\ \frac { 1 }{ 2 } \times \frac { 3 }{ 4 } =\quad \frac { 3 }{ 8 }