**Probability**

The following problems deal with situations and games involving chance. Probability lends itself to "thought experiments" that don't require any material investment except a few sheets of paper, a pen, and a calculator.

**Platonic Dice**

A *Platonic solid* is a geometric solid with flat faces, all of which have identical, regular shape and size. A *regular tetrahedron* has 4 equilateral-triangle faces. A *regular hexahedron* (more often called a cube) has 6 identical square faces. A *regular octahedron* has 8 identical faces, all of which are equilateral triangles.

The following problems involve hypothetical dice that are regular octahedrons, with sides numbered from 1 through 8. Let's call them *octahedral dice*.

**Practice 1**

Suppose you have an octahedral die that is not biased. That is, for any single toss, the probability of the die coming up on any particular face is the same as the probability of its coming up on any other face. (For our purposes, the term "come up" means that a face is oriented facing straight up, so the plane containing it is parallel to the table on which the octahedral die is tossed.) What is the probability that it will come up showing 5 twice in a row? How about 3 times in a row, or 4 times in a row, or *n* times in a row, where *n* is a whole number?

**Solution 1**

The probability *P*_{1} that the octahedral die will come up on any particular face (including 5), on any particular toss, is equal to 1/8. That's 0.125 or 12.5%. The probability *P*_{2} that it will come up showing 5 twice in a row is equal to (1/8)^{2}, or 1/64. That's 0.015625 or 1.5625%. The probability *P*_{3} that it will come up showing 5 exactly 3 times in a row is (1/8)^{3}, or 1/512. The probability *P*_{4} that it will come up showing 5 exactly 4 times in a row is (1/8)^{4}, or 1/4096. In general, the probability *P _{n}* that the octahedral die will come up showing 5 exactly

*n*times in a row is equal to (1/8)

^{n}. Each time we toss the octahedral die, the chance of adding to our consecutive string of 5's is exactly 1 in 8.

**Practice 2**

Imagine a set of octahedral dice, none of which are biased. Suppose you throw a pair of them simultaneously. What's the probability that both octahedral dice will come up showing 3? Now suppose you throw 3 unbiased octahedral dice at the same time. What's the probability that all 3 of them will come up showing 3? How about 4 simultaneously tossed octahedral dice all coming up 3? How about n simultaneously tossed octahedral dice all coming up 3?

**Solution 2**

The progression of probabilities here is the same as that for a single octahedral die tossed over and over. The probability *P*_{1} that the octahedral die will come up on any particular face (including 5), on a single toss, is equal to 1/8. The probability *P*_{2} that a pair of octahedral dice will both come up on any particular face, such as 3, is (1/8)^{2}, or 1/64. The probability *P*_{3} that 3 octahedral dice will come up showing 3 is (1/8)^{3}. The probability *P*_{4} that 4 of them will come up showing 3 is (1/8)^{4}. In general, the probability *P _{n}* that

*n*octahedral dice, tossed simultaneously, will all come up showing 3 is equal to (1/8)

^{n}.

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