1. Instead of using two standard cubical dice whileplaying a board game, three d

Question

1. Instead of using two standard cubical dice whileplaying a board game, three dice are used so that the game goesmore quickly. In the regular game, doubles (i.e., the samenumber rolled on both dice) are needed to get out of the pit. Inthe revised game, doubles or triples (i.e., the same number rolledon at least two of the three dice) are needed to get out of thepit. How many times as likely is it for a player to get out of thepit on one toss under the new rules as compared to the oldrules? 1. Instead of using two standard cubical dice whileplaying a board game, three dice are used so that the game goesmore quickly. In the regular game, doubles (i.e., the samenumber rolled on both dice) are needed to get out of the pit. Inthe revised game, doubles or triples (i.e., the same number rolledon at least two of the three dice) are needed to get out of thepit. How many times as likely is it for a player to get out of thepit on one toss under the new rules as compared to the oldrules?

Explanation / Answer

Each dice has 6 possibilities, so 36 combinations (ordered, so that(1,2) and (2,1) are different) are possible in total. Six of themare doubles, so for two die the odds are 1/6 of getting out of thepit. With three die, there are 6*6*6 = 216 possibilities in total. Wecan get a double 36 ways with die 1 and 2 (six possible commonvalues of die 1 and 2 and for each of those, six possible values ofdice 3), similarly 36 ways with die 2 and 3 and also 36 ways with 1and 3. Note that these doubles also include the triples. So intotal there are 108 ways to get out of the pit on one toss, makingthe probability 108/216 = 1/2. 1/2 is 3 times as likely as 1/6.

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