vedantmodi wrote:
Why could this not be a case of 1 - the probability that 4 does not come in any of the remaining three throws?
1- (5/6)x(5/6)x(5/6) = 1 - (125/216) = 81/216
It is a bit different.
Scenario 1 ---> All 3 rolls are different from the first two ---> (5/6)*(5/6)*(5/6) = 125/216
Scenario 2 ---> All 3 rolls are different from the first two BUT same (also win) ---> (5/6)*(1/6)*(1/6) = 5/216
Scenario 3 ---> Two rolls are same as the first two rolls (lose because only EXACTLY same numbers count as a win) ---> (5/6)*(1/6)*(1/6)*3 = 15/216
Scenario 4 --->All 3 rolls are SAME from the first two (lose because only EXACTLY same numbers count as a win) ---> (1/6)*(1/6)*(1/6)*3 = 1/216
1 - 125/216 + 5/216 - 15/216 - 1/216 = 80/216