(1) A Paradox A few people gather together to play a game in which one needs to

Question

(1) A Paradox A few people gather together to play a game in which one needs to roll 3 six-sided dice. One person notices that the sum of the number of spots on three dice comes up as 11 (the event E1) more often than it does 12 (the event E2) even though it looks like it should be even. This person argues as follows: E1 occurs in just six ways: {(1,4,6),(2,3,6),(1,5,5),(2,4,5),(3,3,5),(3,4,4)}, and E2 occurs also in just six ways: {(1,5,6),(2,4,6),(3,3,6),(2,5,5),(4,4,4),(3,4,5)}. Therefore E1 and E2 have the same probability P(E1) = P(E2). Why isn’t it?

Explanation / Answer

This paradox comes into picture because we have not taken order of outcomes for a roll into consideration here for event counting ; like event (1,4,6) should be represented in 6 ways and counted 6 times which are

{(1,4,6),(1,6,4),(4,1,6),(4,6,1),(6,1,4),(6,4,1)}

number of ways E1 occurs ={(1,4,6),(1,5,5),(1,6,4),(2,3,6),(2,4,5),(2.5.4).(2.6.3).(3,2,6).(3,3,5),(3,4,4),(3,5,3),(3,6,2),(4,1,6),(4,2,5),(4,3,4),(4,4,3),(4,5,3),(4,6,1),(5,1,5),(5,2,4),(5,3,3),(5,4,2),(5,5,1),(6,1,4),(6,2,3),(6,3,2),(6,4,1)}

=27 ways

number of ways E2 occurs ={(1,5,6),(1,6,5),(2,4,6),(2,5,5),(2,6,4),(3,3,6),(3,4,5),(3,5,4),(3,6,3),(4,2,6),(4,3,5),(4,4,4),(4,5,3),(4,6,2),(5,1,6),(5,2,5),(5,3,4),(5,4,3),(5,5,2),(5,6,1),(6,1,5),(6,2,4),(6,3,3),(6,4,2),(6,5,1)} =25 ways

therefore from above E1 occurs more times then E2 if we consider order of outcomes ; which is required to reach individual probability,

please revert for any clarification required

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.