## Wizard Recommends

On average, how many rolls of a fair die are needed to roll every face at least twice?

Ace2

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The answer is 1,172,906,043 / 48,600,000 = apx. 24.13387

Here is my solution. (PDF)

This question is asked and discussed in my forum at Wizard of Vegas.

There is a square dartboard of dimensions 1 by 1. A dart is thrown at it such that it can land anywhere with equal probability. Let the coordinates of where is lands be (x,y), where both x and y are uniformly and independently distributed from 0 to 1.

Let z = round(x/y). In other words, z = x/y, rounded to the nearest integer. What is the probability that z is even?

"Anonymous" .

It will be very useful to know the infinite series in the following hint.

The Leibniz formula for π states:

1/1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ... = π/4

For the answer only, click the following button.

(5 - π)/4 = apx. 0.464601836602552.

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If x/y < 0.5, then the ratio will round dow n to 0, and even number. Any point on the dartboard on the left of the line formed by (0,0) and (0.5) will round down to 0. That area is a right triangle with side of 1 and 1/2. Remember the area of a triangle is (1/2)*base*height. Thus the area of those points rounding down 0 to is (1/2)*(1/2) = 1/4.

The next area on the graph that will round to the next even number, 2, is when 1.5 < x/y < 2.5. This area will be a triangle with base 2/3 - 2/5 and height 1. Note these are the inverses of the bounds of x/y, because x equals 1, so we need to invert y. So, the area that rounds to 2 is (1/2)*(2/3 - 2/5).

The next area on the graph that will round to the next even number, 4, is when 3.5 < x/y < 4.5. This area will be a triangle with base 2/7 - 2/9 and height 1. So, the area that rounds to 2 is (1/2)*(2/7 - 2/9).

The next area on the graph that will round to the next even number, 6, is when 5.5 < x/y < 6.5. This area will be a triangle with base 2/11 - 2/13 and height 1. So, the area that rounds to 2 is (1/2)*(2/11 - 2/13).

Starting to see a pattern? It goes:

1/4 + 1/2*(2/3 - 2/5 + 2/7 - 2/9 + 2/11 - 2/13 + ... ) =

1/4 + (1/3 - 1/5 + 1/7 - 1/9 + 1/11 - 1/13 + ... ) =

Let's move a -1 inside those parenthesis.

5/4 + (-1 + 1/3 - 1/5 + 1/7 - 1/9 + 1/11 - 1/13 + ... ) =

5/4 - (1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 + ... ) =

Next, recall our hint above:

1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11

Getting back to the question at hand ...

5/4 - π/4 =

(5 - π) / 4 = apx. 0.464601836602552.

Interesting how π and e keep popping up all over the place in math.

This question is asked and discussed in my forum at Wizard of Vegas.

Let 9x + 12x = 16x

What is x?

"Anonymous" .

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x = [ln(1 + sqrt(5)) - ln(2)] / [ln(4) - ln(3) ] = apx. 1.67272093446233.

Click the button below for the solution.

9x + 12x = 16x =

Divide both sides by 9x

1 + (12/9)x = (16/9)x

1 + (4/3)x = ((4/3)x)2

(1) Let u = (4/3)x

1 + u = u2

u = (1+sqrt(5)) / 2 (The Golden Ratio)

Putting that back in equation (1):

(4/3)x = (1+sqrt(5)) / 2

Take the log of both sides:

x ln(4/3) = ln[(1+sqrt(5)) / 2]

x = ln[(1+sqrt(5)) / 2] / ln(4/3)

x = [ln(1+sqrt(5) - ln(2)] / [ln(4) - ln(3)] = apx. 1.67272093446233.

This question is raised and discussed in my forum at Wizard of Vegas.

Acknowledgement: I got a variation of this problem from Presh Talwalkar of Mind Your Decisions.

Suppose a fair six-sided die is rolled until a 1, 2, 3, or 6 appears. If a 1, 2, or 3 is the first of these game-ending numbers to appear, then you win nothing. If a 6 is the first of these game-ending numbers to appear, then you win \$1 for every roll of the die. What is the average win of this game?

Klopp

Click the button below for a couple infinite series formulas that you may find helpful.

Hint 1: Sum for i = 0 to ∞ of ni = 1 / (1-n)

Hint 2: Sum for i = 0 to ∞ of i × ni = n / (1-n)2

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Click the button below for the solution.

Suppose a fair six-sided die is rolled until a 1, 2, 3, or 6 appears. If a 1, 2, or 3 is the first of these game-ending numbers to appear, then you win nothing. If a 6 is the first of these game-ending numbers to appear, then you win \$1 for every roll of the die. What is the average win of this game?

Hint 1: Sum for i = 0 to ∞ of ni = 1 / (1-n)

Hint 2: Sum for i = 0 to ∞ of i × ni = n / (1-n)2

The expected win can be expressed as the sum for i = 0 to ∞ of (1 + i) * (1/3)i * (1/6). =

(1/6) * sum for i = 0 to ∞ of (1/3)i + (1/6) * sum for i = 0 to ∞ of (i * (1/3)i).

Let's evaluate these one at a time.

sum for i = 0 to ∞ of (1/3)i =

1 / (1 - (1/3)) =

1 / (2/3) =

3/2

Sum for i = 0 to ∞ of (i * (1/3)i) =

(1/3) / (1 - (1/3))2 =

(1/3) / (4/9) =

(1/3) * (9/4) =

3/4

Putting it all together, the answer is

(1/6) * (3/2) + (1/6)*(3/4) =

(1/4) + (1/8) =

3/8

This question is asked and discussed in my forum at Wizard of Vegas.