Emdlalweni omncane we-Bowser's Big Blast kusukela ku-Mario Party 2 naku -Mario Party Superstars, abadlali bakhetha ukushintsha okukodwa ngesikhathi. Iswishi eyodwa impela icupha ibhomu, amanye aphephile. Uma uqala ngabadlali \(4\) kukhona \(5\) amaswishi. Ocindezela kuqala ulahlekelwa ngokuthi \(20\,\%\) ; Umuntu wokugcina omi phambi kokuthi \(2\) amaswishi kuphela uno- \(50\,\%\) . Noma kunjalo, indawo yokugcina ayinawo amathuba aphezulu okulahlekelwa. Ukungqubuzana okusobala kuvela ngoba siqhathanisa amathuba ezindaweni ezahlukene ngesikhathi.
Sicabanga ukuthi i-detonator isatshalaliswa ngokulinganayo phakathi kwamaswishi amahhala futhi ayibonakali. Umdlali wokugcina uthola ithuba lakhe kuphela uma abadlali \(3\) bokuqala becindezele ngokuphephile, okungukuthi ibhomu aliqhumanga. Ithuba lalokhu ngu- \(\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{2}{3}=\frac{2}{5}=40\,\%\) . Kokusele \(60\,\%\) umzuliswano uphela ngaphambili. Uma uphindaphinda ithuba lokuthola ukujika ngobungozi obulandelayo, uthola \(20\,\%\) ofanayo nowomdlali wokuqala. Lokhu kubangela isikhundla sokugcina:
$$\frac45\cdot\frac34\cdot\frac23\cdot\frac12=\frac15=20\,\%.$$
Ngo \(q\) sibonisa amathuba okuthi umdlali onikeziwe nguye olandelayo okufanele asuswe, ngisho nangemva kwanoma iyiphi inombolo yezimpinda. Emzuliswaneni wokuqala kunezimo ezintathu: Nge \(1/5\) ishaya lo mdlali. Nge \(3/5\) ishaya enye kwezinye; umdlali wethu ngeke akhishwe ngokulandelayo. Nge \(1/5\) wonke umuntu uyasinda futhi amaswishi asethwa kabusha. Ngemva kwalokho isimo sihlala singashintshile, ngakho ithuba elifiswayo liphinde \(q\) . Yingakho kusebenza:
$$\begin{aligned}q&=\frac15+\frac15q,\\\frac45q&=\frac15\quad\Longrightarrow\quad q=\frac14.\end{aligned}$$
Ngemva kokukhipha okulandelayo, abadlali abasele baqala futhi ngemibandela elinganayo. Ithuba lokusinda kuze kube yileso sikhathi lithi \(3/4\) uma kuqala \(4\) abadlali, bese kuba \(3\) abadlali \(2/3\) futhi ekugcineni \(2\) ngabadlali \(1/2\) . Uma uphindaphinda lawa mathuba anemibandela okusinda, uthola \(25\,\%\) ithuba lokuwina esikhundleni sokuqala nesokugcina. Ngokuvamile, ngabadlali \(n\) , wonke umuntu uwina nge \(1/n\) . Imibala ayisizi kumodeli efanelekile. Ukunqoba okuphelele kulandela:
$$P(\text{Sieg})=\frac34\cdot\frac23\cdot\frac12=\frac14=25\,\%.$$
Bowser's Big Blast – Spiel und Wahrscheinlichkeiten von David Vielhuber auf CodePen.