Dice and Coins

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PUTNAM PREPARATION PROBLEMS: PROBABILITY

1. The envelope game

Bob is presented with two distinct real numbers, each concealed in its ownenvelope. Bob chooses one of the envelopes according to a flip of his trustyfair coin. He reveals the number contained in the envelope and guesseswhether the number in the other envelope is larger or smaller than the num-ber he just revealed. Does Bob have a strategy that gives him a better thaneven chance of guessing correctly?

Hints:

(1) Think about where chance lies in this game.(2) What if you assume that one number is positive and the other neg-

ative?(3) What if for a known fixed number b, you assume that one number

is greater than b and the other is smaller than b?(4) Can you probablistically generate such a number b?

2. Summing random numbers

An experiment consists of summing random numbers, distributed uni-formly in (0, 1), until the sum exceeds 1. What is the expected number of summands?

Hints:

(1) First, find the volume of the region in Rn bound by the coordinate

hyperplanes and the hyperplanen

i=1xi = 1. It might be easier to

find a recursive formula first.(2) Find the probability that the number of summands is at least n.(3) Find the probability that the number of summands is exactly n.

3. Loading dice

A die is called fair if each of the six possible outcomes of a die toss are

equally likely. A die that is not fair is called loaded .• Can a pair of standard dice be loaded so that each of the possible

sums occur with the same probability?• Can a pair of standard dice be loaded so that each possible sum

occurs with the same probability as it would with a fair pair of stan-dard dice?

1



PUTNAM PREPARATION PROBLEMS: PROBABILITY 2

Here are some observations that may help with these problems:

Let X , Y and Z be random variables taking values in {x0, . . . , xl}, {y0, . . . , ym},{z0, . . . , zn} with probabilities p0, . . . , pl, q 0, . . . , q m, and r0, . . . , rn, respec-tively. Associate to X, Y and Z , polynomials:

P X (t) :=l

i=0

pitxi P Y (t) :=

m

j=0

q jtyj P Z (t) :=n

k=0

rktzk

Observe that if Z = X + Y then

rk =

xi+yj=zk

piq j , k = 0, . . . , n

It follows that

(1) P Z = P X P Y




Solution to the first dice problem:

Take X and Y to be the random variables associated with two loaded andtake Z to be their sum. Then;

{x0, . . . , xl} = {y0, . . . , ym} = {1, 2, 3, 4, 5, 6}

{z0, . . . , zk} = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

P X (t) = t( p0 + p1t + p2t2 + p3t3 + p4t4 + p5t5)

P Y (t) = t(q 0 + q 1t + q 2t2 + q 3t3 + q 4t4 + q 5t5)

P Z (t) = t2(r0 + r1t + · · · + r10t10)

Finally, from (1), we have

(2) ( p0 + p1t + . . . + p5t5)(q 0 + q 1t + . . . + q 5t5) = r0 + r1t + · · · + r10t10

The first dice problem above asks if there exist pi’s and q j’s so that r0 =r1 = · · · = r10 = 1

11. If such pi’s and q j’s do exist then, by equation (2),

( p0 + p1t + . . . + p5t5)(q 0 + q 1t + . . . + q 5t5) =1

11(1 + t + · · · + t10)

The roots of the polynomial on the right are the ten unreal 11 th roots of unity. (Why?). On the other hand, the polynomials on the left are of odddegree and hence each have at least one real root. Contradiction!

Thus, the sum of two standard dice never behaves like a single fair (non-standard) die.




Solution to the second dice problem:

The problem asks if there exist pi’s and q j’s so that

rk =1

36|{i, j | i + j = k}|, k = 0, . . . , 10

Equivalently,

P Z (t) =1

36t2(1 + 2t + 3t2 + 4t3 + 5t4 + 6t5 + 5t6 + 4t7 + 3t8 + 2t9 + t10)

Observing that

1+ 2t+3t2 +4t3 +5t4 +6t5 +5t6 +4t7 +3t8 +2t9 +t10 = (1+t+t2 +t3 +t4 +t5)2

(Write down a multiplication table for the powers of t and then add upthe entries of the table along lines of slope −1) we see that the roots of P Z = P X P Y are the five nontrivial 6th roots of unity each with multiplicity

two. Letting ω denote a primitive sixth root of unity, the distinct roots of are given by:

−1, ω , ω , ω2, ω2

Since the polynomials on the left side of (2) are real polynomials of degree5, they factor into a product of a real linear factor and two pairs of conjugatecomplex factors. The only possibilities for p0 + p1t + . . . + p5t5 are:

p0 + p1t + . . . + p5t5 = (t + 1)(t − ω)(t − ω)(t − ω)(t − ω)

p0 + p1t + . . . + p5t5 = (t + 1)(t − ω)(t − ω)(t − ω2)(t − ω2)

Expanding the first polynomial (writing −ω2 for ω and −ω for ω2) gives apolynomial with unreal coefficients - clearly invalid. Expanding the second

polynomial gives p0 + p1t + . . . + p5t5 = 1 + t + t2 + t3 + t4 + t5

and hence

P X (t) = P Y (t) =1

6t +

1

6t2 +

1

6t3 +

1

6t4 +

1

6t5 +

1

6t6

Thus, if the sum of a pair of dice behaves as if it is a sum of fair dice thenit is the sum of a pair of fair dice.

Dice and Coins

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