Question Number 141692 by ArielVyny last updated on 22/May/21
$${Daniel}\:{and}\:{Bruno}\:{are}\:{playing}\:{with}\:{perfect}\:{cube} \\ $$$${Daniel}\:{is}\:{the}\:{first}\:{player}\:{if}\:{he}\:{obtains}\:\mathrm{1}\:{or}\:\mathrm{2} \\ $$$${he}\:{wins}\:{the}\:{game}\:{and}\:{the}\:{party}\:{stopping} \\ $$$${or}\:{else}\:{Bruno}\:{plays}\:{and}\:{if}\:{he}\:{have}\:\left\{\mathrm{3}.\mathrm{4}.\mathrm{6}\right\}\:{Bruno}\:{won}\:{and}\:{the}\:{game}\:{stopping} \\ $$$${Determine}\:{the}\:{probability}\:{that}\:{Daniel}\:{winand}\:{the}\:{probability}\:{that}\:{Bruno}\:{win} \\ $$$$ \\ $$
Answered by MJS_new last updated on 22/May/21
$$\mathrm{in}\:\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{of}\:\mathrm{all}\:\mathrm{cases}\:\mathrm{Daniel}\:\mathrm{wins} \\ $$$$\mathrm{in}\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{of}\:\mathrm{all}\:\mathrm{cases}\:\mathrm{Bruno}\:\mathrm{plays} \\ $$$$\:\:\:\:\mathrm{in}\:\frac{\mathrm{3}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{of}\:\mathrm{these}\:\mathrm{cases}\:\mathrm{Bruno}\:\mathrm{wins} \\ $$$$\:\:\:\:\mathrm{the}\:\mathrm{rest}\:\mathrm{end}\:\mathrm{in}\:\mathrm{a}\:\mathrm{draw} \\ $$$$\Rightarrow \\ $$$$\mathrm{Daniel}\:\mathrm{wins}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{Bruno}\:\mathrm{wins}\:\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{draw}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$