Triple your money?

The infinite board is made up of an infinite amount of gray $2 \times 3$ rectangles with a border of $\frac{1}{2}$ around it. Such a larger rectangle has an area of $(2+1) \times (3+1) = 12$ . If the center of the coin is the white border or in the grey rectangle within $\frac{1}{2}$ of the edge of the rectangle, it will not entirely be in the grey rectangle. What remains is a $(2-1) \times (3-1) = 2 \times 1$ rectangle with area $2$ for the coin to land in in order to win. So, the chance of winning is $\frac{2}{12} = \frac{1}{6}$ , hence the expected number of received cents after a throw is

$E = \frac{1}{6} \cdot 150 + \frac{5}{6} \cdot 0 = \boxed{25}.$

For simplicity, let's consider the arrangement to be a "square," with $n$ rectangles in the rows and columns.

The total area of the big square is therefore $\left(2n+n-1\right)\left(3n+n-1\right) = 12n^2-7n+1$

The total area of where the centre of the circle can land to get $150$ tokens is $\left(2-1\right)n * \left(3-1\right)n = 2n^2$

Therefore, the probability that you will get $150$ tokens is $\frac{2n^2}{12n^2-7n+1}$

The probability that the return is $0$ is $1 - \frac{2n^2}{12n^2-7n+1} = \frac{10n^2-7n+1}{12n^2-7n+1}$

The weighted average of the possible outcomes is therefore $\frac{2n^2*150}{12n^2-7n+1} + \frac{0*\left(10n^2-7n+1\right)}{12n^2-7n+1} = \frac{300n^2}{12n^2-7n+1}$

$\lim_{n \to \infty}$ $\frac{300n^2}{12n^2-7n+1} = \frac{300n^2}{12n^2-7n} = \frac{300n}{12n-7}$

$\lim_{n \to \infty}$ $\frac{300n}{12n-7} = \frac { 300}{12} = 25$

So our answer is $25$

For each $2$ by $3$ rectangle, there is $12$ total area allocated to it including the empty white space. This is easy to derive by looking at the $3$ and $2$ adjacent white spaces on the bottom and right, respectively, as well as the $1$ white space on the bottom right corner.

Then, inside of each rectangle, we want to find the area of the locus of points whose perpendicular distance to any of the sides is at least $\frac {1}{2}$ . This creates a $1$ by $2$ rectangle, for an area of $2$ . Thus, there is a $\frac{1}{6}$ chance to triple your starting 50 cents and a $\frac {5}{6}$ chance to lose it all. Thus, the expected value of money is $\frac{1}{6}(150) = 25$ .

Infinite boards are always difficult to deal with, so first try considering finite cases that reflect the characteristics of the infinite board.

Case n=1: Just 1 gray rectangle. The circle of diameter 1 (radius 0.5) must be completely within the rectangle of dimension 2x3. This can be determined by considering all the locations where the center of the circle is not allowed to be. if the center of the circle is with 0.5 of the sides of the rectangle, then part of that circle will be outside the rectangle, so it will not satisfy the condition we're looking for. So we have a smaller rectangle with dimension [2-(0.5+0.5)] \times [3-(0.5+0.5)]. Note that 0.5 is subtracted twice to account for both the left and right as well as both the top and bottom. So the probability that the center of the circle lands in this smaller rectangle is

$\frac {(2-1)\times(3-1)}{2\times3} = \frac{1}{3}$ . This is the probability that our condition is satisfied in just 1 rectangle. Now let's look at bigger (closer to infinity) cases.

Case n=2: a rectangular (3x3) board with 9 rectangles, including the white space between these 9 rectangles. We apply the same principle as before to calculate the probability that our condition is satisfied. The condition is satisfied when the center of the circle lands in that smaller rectangle in any of the 9 rectangles. The overall area can easily be calculated as the area of the giant rectangle that encloses 9 gray rectangles. This is computed below.

$\frac{9\times(2\times1)}{(2+1+2+1+2)\times(3+1+3+1+3))}$ . Note that to determine the denominator (the area of the big rectangle containing the 9 gray rectangles), we determined the height of this rectangle by adding 3 heights of gray rectangles and 2 (=3-1) spaces between gray rectangles, and similarly, we determined the length by adding 3 lengths of gray rectangles and 2(=3-1) spaces between gray rectangles and multiplied height and length to get area. It's easy enough to simply compute the result here, so we won't explicitly do it and it's not necessary to do so. Let's look at 1 bigger example and then we'll be ready to generalize to infinity.

Case n = 3. A rectangle (5x5) with 25 rectangles, including the white space between these rectangles. The probability that our condition is satisfied here is computed in exactly the same way as before, so let's just do it.

\(frac{25\times(2\times1)}{(2\times5+1\times4)\times(3\times5+1\times4)}. Now again, we don't need to compute it, we just need to look at what we did here and before and we are ready to generalize to infinity.

So in general, for any n, we're creating rectangles that are m by m, where m is the nth odd number (1 is the 1st odd number, 3 is the 2nd odd number, 5 is the 3rd odd number). m = 2n-1. We're calculating the numerator of the probability we want by taking the area that satisfies our condition in each little rectangle \(2\times1)\) and multiplying it by $m^2$ . Also, we are calculating the denominator of our probability by taking $2\times m + 1\times (m-1))\times(3\times m + 1\times (m-1))$ . So we have that our probability that our condition is satisfied is:

$\frac{(2 \times 1) \times m^2}{(2 \times m + 1 \times (m-1))\times(3 \times m + 1 \times (m-1))} = \frac{2 \times m^2}{(3m-1) \times (4m-1)} = \frac{2 \times m^2}{12 \times m^2 - 7 \times m + 1}$ .

If we take the limit as m approaches infinity (the limit as the board becomes infinite and not just these small cases we're looking at), we ignore the smaller m and constant terms and just look at the quadratic term $\frac{2m^2}{12m^2} = \frac{1}{6}$ . Hence the probability that our condition is satisfied in an infinite board is $\frac{1}{6}$ so we can easily calculate expected value.

$E(x) = P(conditionSatisfied)\times payoutIfSatisfied + P(conditionNotSatisfied)\times payoutIfNotSatisfied = \frac{1}{6} \times 150 + \frac{5}{6} \times 0 = 25$

Due to the symmetry of the situation, we can concentrate completely on a single rectangle. Now , area b\w any 2 rectangles will be divided into 2 equal parts(Or the sides of the grey rectangle rectangle are to be extended by $\frac{1}{2}$ from both sides). Thus sides of the bigger rectangle about the grey rectangle = $(\frac{1}{2} +\frac{1}{2} +2)(\frac{1}{2} +\frac{1}{2} + 3) = 12.$ To find favourable area , we have to subtract $\frac{1}{2}$ twice, because the coin has to land COMPLETELY on the grey rectangle. $\\$ P = $\frac{favourable \ area }{total \ area } = \frac{2}{12} = \frac{1}{6} \\$ . Hence expected cents = $(\frac{1}{6})250 =\fbox{25}$

The infinite board can be divided into infinitely many $3 \times 4$ rectangles, with a $2 \times 3$ gray rectangle in one corner.

Since the coin has radius $\frac{1}{2}$ , it must land in a $(2 - 2\cdot\frac{1}{2}) \times (3 - 2\cdot\frac{1}{2}) = 1 \times 2$ rectangle inside the gray rectangle.

Hence, there is a $\frac{3 \cdot 4}{1 \cdot 2} = \frac{1}{6}$ chance of winning. This means one can expect to receive an average of $\frac{150}{6} = \boxed{25}$ cents after each throw.

Since this game costs $50$ cents, I'd suggest staying away from it.

We can solve the problem by considering the area of the game where the center of the coin can land and score. for each rectangle, this is an area 1x2, as you lose .5 (the radius of the coin) from each edge. Then note that the total area is:

$(4*columns -1)*(3*rows-1)$

(Subtracting one because the last row and column don't have any white space after them.)

Then we can calculate the probability of the coin landing in a scoring zone by the ratio of areas, considering WLOG of generality the pattern having an equal number of rows and columns of rectangles. (It's actually not much more difficult to do the problem in general, but regardless of the ratio between rows and columns as we expand to infinity, the probability at infinity should be the same.)

$P=\frac{A_{scoring}}{A_{total}}=\lim_{n \to \infty} \frac{2n^2}{(4n-1)(3n-1)}=\frac{1}{6}$

From this we calculate the expected value, noting that the payout is 3*50=150.

$E=\sum x_iP_i=\frac{150}{6}=\fbox{25}$

http://i1094.photobucket.com/albums/i452/alephvn/Untitled_zpse3e8f0d2.png We just need to consider a rectangle with size $3 \times 4$ (rectangle $ALMN$ in the image). The coin is completely inside the $2 \times 3$ rectangle $ABCD$ if and only if it center is inside the rectangle $HIJK$ .

The area of $HIJK$ is 2 and the area of $ALMN$ is 12. Therefore, the expected number of cents is $\frac{2 \times 150 + 10 \times 0}{12} = 25$ .

The infinite board is a tessellation of a larger rectangle (call this "LR") composed of one gray rectangle ("GR") bordered by a white "L" shape (with dimension 3 x 4). So consider LR instead.

For the token to be completely inside GR, its center must be on or in the interior of a smaller rectangle ("SR"), wherein SR is inside GR, has sides 1/2 unit away from the sides of GR and has dimension 1 x 2.

Hence, the probability of success is (area of SR)/(area of LR) = 2/12 = 1/6, and expected gain = 150 x probability of success - 0 x probability of failure = 150 x 1/6 = 25.

The infinite plane is made up of an infinite number of 2x3 rectangles with a 0.5 border around them, resulting in a 3x4 rectangle. The expected number of cents you will receive if you confine the centre of the token in one of these rectangles is the same as that on an unbounded plane.

In order for you to win this game, the centre of the token must land in the 1x2 area in the very middle of any grey rectangle. The probability of this happening (and winning 150 cents) is equal to the area of the target area divided by the total area, or $\frac{1*2}{3*4}=\frac{2}{12}=\frac{1}{6}$

Landing anywhere else will result in receiving 0 cents, so the expected number of cents that you will receive after a throw is $\frac{1}{6}*150+\frac{5}{6}*0=25$

We extend the sides of each rectangle by $0.5$ on each side to form new rectangles that fill all the white space. As the new rectangles are the only possible areas to land on, and they are all the same, we only need to consider if a coin lands in one. The area of each rectangle is $3\times4=12$ , while the area of each of our original rectangles is $2\times3=6$ . So we have a $\frac{6}{12}=\frac{1}{2}$ chance of getting in one of our target rectangles.

Now we consider when the coin will lie completely inside of a target rectangle. The center of the coin must be at least $\frac{1}{2}$ away from each side, thus the "target zone" for the center of the coin is a rectangle with sides $1$ less than our first rectangles. Because the center of the coin completely determines where the coin lands, and the area of our target zone is $1\times2=2$ , the probability of landing in the target zone if we land in one of the target rectangles is $\frac{2}{6}=\frac{1}{3}$ . Thus, in total there is a $\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}$ chance of tripling our money. As this is the only way to affect our expected value, this value is thus $150\cdot\frac{1}{6}=\boxed{25}$

Since the pattern is infinitely repeating, I took a selection that does not over lap another part of that same section. I took the grey rectangle, and the white boarder up to the edge of the next grey border in the top and right section. This together will form a 3 x 4 section, with a 2 x 3 rectangle in the lower-left most section.

In order for the coin to remain completely in the grey, the center of the coin must be more than half way inside of one unit of the grey section. Since the radius of the coin is 0.5 units, reduce the length on each edge of the grey rectangle by 0.5 units. This results in a 1 x 2 hit box. If the single coin was a point in space, if it landed in the 'hit box' it counts as a win.

The area of the hit box is 2, and the area of the selected patern is 12. $\frac{2}{12}$ = $\frac{1}{6}$

There is a 1 in 6 chance of winning. Point of the story, never play this game. For the 1 in 6 times you do win, you multiply your coin by three. $3 \times ( \frac{1}{6}$ ) = $\frac{1}{2}$

Therefore, you will get an average of half of what you put in, so the expected result of 50 cents is 25 cents.