Kumdlalo omncinci weBowser's Big Blast evela kuMario Party 2 kunye neMario Party Superstars, abadlali bakhetha iswitshi enye ngexesha. Ngokuchanekileyo iswitshi enye ibangela ibhombu, ezinye zikhuselekile. Xa uqala nge \(4\) abadlali kukho \(5\) iiswitshi. Nabani na ocofa kuqala ulahlekelwe ngu \(20\,\%\) ; Umntu wokugqibela ome phambi kokutshintsha kuphela \(2\) une \(50\,\%\) . Nangona kunjalo, indawo yokugqibela ayinayo amathuba aphezulu elahleko. Ukuphikisana okubonakalayo kuvela ngenxa yokuba sithelekisa izinto ezinokwenzeka kwiindawo ezahlukeneyo ngexesha.
Sicinga ukuba i-detonator isasazwa ngokulinganayo phakathi kokutshintsha kwamahhala kwaye ayibonakali. Umdlali wokugqibela ufumana ithuba lakhe kuphela ukuba abadlali bokuqala \(3\) bacinezele ngokukhuselekileyo, oko kukuthi ibhombu ayizange iqhume. Ithuba lale nto \(\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{2}{3}=\frac{2}{5}=40\,\%\) . Kwintsalela \(60\,\%\) umjikelo uphela kwangaphambili. Ukuba uphinda-phinda ithuba lokufumana ithuba ngomngcipheko olandelayo, ufumana okufanayo \(20\,\%\) njengowomdlali wokuqala. Oku kubangela ukuba kwindawo yokugqibela:
$$\frac45\cdot\frac34\cdot\frac23\cdot\frac12=\frac15=20\,\%.$$
Nge \(q\) sibonisa ukuba kunokwenzeka ukuba umdlali onikiweyo ngolandelayo ukuba apheliswe, nasemva kwalo naliphi na inani lokuphindaphinda. Kumjikelo wokuqala kukho iimeko ezintathu: Nge \(1/5\) ibetha lo mdlali. Nge \(3/5\) ibetha enye kwezinye; umdlali wethu akayi kususwa ngokulandelayo. Nge \(1/5\) wonke umntu uyasinda kwaye iiswitshi zisetwa kwakhona. Emva koko imeko ihlala ingatshintshi, ngoko ke ithuba elifunwayo kwakhona \(q\) . Yiyo loo nto isebenzayo:
$$\begin{aligned}q&=\frac15+\frac15q,\\\frac45q&=\frac15\quad\Longrightarrow\quad q=\frac14.\end{aligned}$$
Emva kokupheliswa okulandelayo, abadlali abaseleyo baqala kwakhona ngokwemiqathango elinganayo. Ithuba lokuphila kude kube ngoko ngu \(3/4\) xa uqala nge \(4\) abadlali, emva koko nabadlali \(3\) \(2/3\) kwaye ekugqibeleni nabadlali \(2\) \(1/2\) . Ukuba uphinda-phinda la mathuba anemiqathango okusinda, ufumana \(25\,\%\) ithuba lokuwina kwizithuba zokuqala nezokugqibela. Ngokubanzi, nge \(n\) abadlali, wonke umntu uphumelela nge \(1/n\) . Imibala ayincedi kwimodeli efanelekileyo. Kuba uloyiso jikelele lulandela:
$$P(\text{Sieg})=\frac34\cdot\frac23\cdot\frac12=\frac14=25\,\%.$$
Bowser's Big Blast – Spiel und Wahrscheinlichkeiten von David Vielhuber auf CodePen.