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Pope asked in Science & MathematicsMathematics · 10 years ago

Are these experiments equivalent?

A mathematics teacher, wishing to get caught up on some paperwork, gives his students a tedious activity of questionable educational merit. The students are paired off. Each team is given a fair cubic gaming die. They are instructed to roll the die 500 times and record the number of sixes.

Looking up from his work, the teacher notices that one team is not following his instructions to the letter. One student is rolling the die on a glass-top table and counting the sixes. The other student is sitting under the table and counting the sixes that appear on the bottom. The students insist that their procedure is equivalent to the one that was assigned. The probability of a six on bottom is equal to the probability of a six on top. This way, they argue, they can roll the die only 250 times and still record 500 trials.

Are the students in fact conducting an equivalent experiment? Let X be the number of sixes recorded the usual way in 500 rolls. Let Y be the number of sixes recorded in 250 rolls using the modified procedure. Do X and Y have the same distribution?


There seems to be some confusion. Of course a six is more likely when two faces are checked. That was not the question. The students were told to record the total number of sixes in 500 rolls. The two uncooperative students claim to be simulating that total while rolling their die only 250 times. If X and Y have the same distribution, then they are correct. Are the distributions equivalent?

By the way, this is based on an actual event.

2 Answers

  • 10 years ago
    Favorite Answer

    It is true that X and Y have the same expectation of 6's, and so in that respect, the two deviants are extremely likely to get results close to that of their peers'. However, the two experiments are not equivalent.

    E(X) = 500/6; (1 in 6 chance on 500 trials)

    E(Y) = 250/3; (1 in 3 chance on 250 'trials')

    The difference is that X is a binomial distribution on 500 with p = 1/6, but Y is a binomial distribution on 250 with p = 1/3. It is possible for X to be > 250, but not so for Y, and so it follows necessarily that Y has a greater chance to achieve lower values than does X.

  • 10 years ago

    No. The odds of a 6 coming up on any side of the die is the same each time (1 in 6) so counting twice from two different sides doubles the probability that I 6 will come up (or down). It is essentially changing the odds to 1 in 3 since the team has 2 changes in 6 that the 6 will show up on top or on the bottom.

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