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If you know how to calculate the house edge of the fire bet using calculus, then you already have almost the entire solution. It’s just a slightly different approach

Quote:Ace2Rephrase: All fire bet winners will have won at least six points before sevening out. When someone wins the fire bet, what is the expected number of total points won?

First you calculate the probability of winning the fire bet. Though this has already been posted, I'll recap as follows. Out of 3,960 possible outcomes/decisions, there are 165 ways to win a 4 (same to win a 10), 264 ways to win a 5 (same to win a 9), 375 ways to win a 6 (same to win an 8) and 2,352 ways to seven out.

The "indirect" way (counting all winning strings regardless of when they win) to calculate the edge is by integrating the following from zero to infinity:

((1 - e^(-165x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-375x/3960))^2 * e^(-2352x/3960) * 2352/3960 dx

The "direct" way (counting all strings at the exact time they win) to calculate the edge is by integrating the following from zero to infinity:

((1 - e^(-165x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-375x/3960)) * e^(-375x/3960) * 375/3960 + (1 - e^(-165x/3960))^2 * ( 1 - e^(-375x/3960))^2 * ( 1 -e^(-264x/3960)) * e^(-264x/3960) * 264/3960 + (1 - e^(-375x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-165x/3960)) * e^(-165x/3960) * 165/3960) * e^(-2352x/3960) * 2 dx

Both methods give the same probability of winning of ~0.0001624.

Next, take the same direct integral from zero to infinity but also multiply by x. Using the indirect integral will not give the correct answer. So :

((1 - e^(-165x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-375x/3960)) * e^(-375x/3960) * 375/3960 + (1 - e^(-165x/3960))^2 * ( 1 - e^(-375x/3960))^2 * ( 1 -e^(-264x/3960)) * e^(-264x/3960) * 264/3960 + (1 - e^(-375x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-165x/3960)) * e^(-165x/3960) * 165/3960) * e^(-2352x/3960) * 2 * x dx

Since this integral sums the individual probabilities of winning the fire bet at all times(x) and then multiplies by time (x), it gives us the weighed average of ~0.001201 decisions to win the fire bet. That is the average all of all outcomes, including the losing ones. We are only interested in outcomes where the fire bet was won, so take 0.001201 divided by 0.0001624 (the probability of winning) to get ~7.39 average total decisions (points won) when the winning the fire bet. The exact answer is:

51,194,997,861,765,388,097,526,966,955,556,874,130,535,332,020,426,690,698,014,074,625,318,137,025,571,849,255/

6,923,708,358,532,324,948,598,712,995,072,892,018,553,301,246,118,708,475,186,839,062,876,563,686,189,522,216

I will use the "direct method fire bet" integral in the example:

(

(1 - e^(-165x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-375x/3960)) * e^(-375x/3960) * 375/3960

+ (1 - e^(-165x/3960))^2 * ( 1 - e^(-375x/3960))^2 * ( 1 -e^(-264x/3960)) * e^(-264x/3960) * 264/3960

+ (1 - e^(-375x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-165x/3960)) * e^(-165x/3960) * 165/3960

) * e^(-2352x/3960) * 2 dx

Let y = e^(-x/3960): the integral becomes

2 *

(

(1 - y^165)^2 * (1 - y^264)^2 * (1 - y^375) * y^375 * 375/3960

+ (1 - y^165)^2 * (1 - y^375)^2 * (1 - y^264) * y^264 * 264/3960

+ (1 - y^375)^2 * (1 - y^264)^2 * (1 - y^165) * y^165 * 165/3960

)

* y^2352 dx

Expand all of the squares, and take 1/3960 out of the sum:

2 / 3960 *

(

(1 - 2 y^165 + y^330) * (1 - 2 y^264 + y^528) * (1 - y^375) * y^375 * 375

+ (1 - 2 y^165 + y^330) * (1 - 2 y^375 + y^750) * (1 - y^264) * y^264 * 264

+ (1 - 2 y^375 + y^750) * (1 - 2 y^264 + y^528) * (1 - y^165) * y^165 * 165

)

* y^2352 dx

I will not bore you with calculating the expansion (hint: use Excel, and a lot of cutting and pasting), but this is:

1 / 1980 *

(

165 y^2517

+ 264 y^2616

- 165 y^2682

+ 375 y^2727

- 528 y^2781 - 330 y^2781

- 264 y^2880

- 750 y^2892 - 330 y^2892

+ 264 y^2946 + 330 y^2946

- 750 y^2991 - 528 y^2991

+ 528 y^3045 + 165 y^3045

+ 375 y^3057 + 330 y^3057

- 375 y^3102

+ 1500 y^3156 + 1056 y^3156 + 660 y^3156

- 264 y^3210 - 165 y^3210

+ 375 y^3255 + 528 y^3255

+ 750 y^3267 + 165 y^3267

- 750 y^3321 - 528 y^3321 - 660 y^3321

+ 750 y^3366 + 264 y^3366

- 750 y^3420 - 1056 y^3420 - 330 y^3420

- 375 y^3432 - 165 y^3432

- 1500 y^3531 - 528 y^3531 - 330 y^3531

+ 375 y^3585 + 528 y^3585 + 330 y^3585

- 375 y^3630 - 264 y^3630

+ 750 y^3696 + 264 y^3696 + 330 y^3696

+ 750 y^3795 + 528 y^3795 + 165 y^3795

- 375 y^3960 - 264 y^3960 - 165 y^3960

)

"Now what?"

d/dx (e^(nx)) = e^(nx) / n

y^n = e^(x * (-n/3960)), so d/dx (y^n) = e^(x * (-n/3960)) * (-3960/n)

For positive n, at x = +INF, this is 0; at x = 0, this is -3960/n, so the integral over x = 0 to +INF = 3960/n

Replace each "a y^b" with "3960 * b / a"; you can take 3960 out of each numerator and multiply the 1/1980 term by 3960 to get 2.

The solution becomes

2 *

(

165 / 2517

+ 264 / 2616

- 165 / 2682

+ 375 / 2727

- 858 / 2781

- 264 / 2880

- 1080 / 2892

+ 594 / 2946

- 1278 / 2991

+ 693 / 3045

+ 705 / 3057

- 375 / 3102

+ 3216 / 3156

- 429 / 3210

+ 903 / 3255

+ 915 / 3267

- 1938 / 3321

+ 1014 / 3366

- 2136 / 3420

- 540 / 3432

- 2358 / 3531

+ 1233 / 3585

- 639 / 3630

+ 1344 / 3696

+ 1443 / 3795

- 804 / 3960

)

Easy, right?

The sum is done one term at a time (i.e. start with the first two, then add the third, then add the fourth, and so on), using:

a / b + c / d = (ad + bc) / bd

No, you can't do this in Excel, as you will soon run out of digits of precision.

Questions:

1. What is the area of the red rectangle*?

2. What is the least possible area of the entire figure?

In case it's not clear, the height of the yellow triangle is 4 and the base of the blue triangle is 5.

* Corrected from square

Quote:WizardMy apologies if business is not concluded on Don's problem. Here is something easier.

Questions:

1. What is the area of the red square?

2. What is the least possible area of the entire figure?

In case it's not clear, the height of the yellow triangle is 4 and the base of the blue triangle is 5.

Clarify: the red portion is supposed to be a square, and not just a rectangle. The way it appears, it is not a square.

Or perhaps question 1 requires that it be a square, but question 2 requires only that it be a rectangle?

Also, my "problem" was not a problem, but an explanation as to how to calculate the exact rational number solution of a Poisson-based probability problem, rather than depending on, e.g., Integral Calculator to do it for you.

Quote:ThatDonGuyClarify: the red portion is supposed to be a square, and not just a rectangle.

Yes, I should have said red rectangle.

Let b be the "missing" length of a leg of the blue triangle, and y the "missing" length of a leg of the yellow one.

Both problems depend on the fact that the blue triangle, the yellow triangle, and the large triangle are all similar; 4 / y = b / 5 = (4 + b) / (5 + y).

For #1, y = b, so 4 / b = b / 5, which means b = 2 sqrt(5), and the area of the large triangle = 1/2 (4 + 2 sqrt(5)) (5 + 2 sqrt(5)) = 20 + 9 sqrt(5)

For #2, 4 / y = b / 5, so y = 20 / b. The area of the triangle is 1/2 (4 + b) (5 + y) = 1/2 (4 + b) (5 + 20/b) = 20 + 5/2 b + 40/b.

Let A denote the area; the first derivative with respect to b, dA / db = 5/2 - 40/b^2. This equals zero when b^2 = 16; since b > 0, b = 4.

Note that the second derivative, d^2A / db^2 = 80/b^3, which is positive when b = 4, so A is a minimum at b = 4.

b = 4, so y = 5, and the minimum area = 1/2 (4 + 4) (5 + 5) = 40.

Note that there is no maximum area, as 20 + 5/2 b + 40/b approaches infinity as b approaches infinity.

Quote:ThatDonGuyWell, the area of the red rectangle could be anything

I disagree. It has a specific answer.

If the area of the rectangle were a billion, you couldn't inscribe it a right triangle with the 4 and 5 distances given. If you feel otherwise, please give me the dimensions of a the rectangle with area one billion.

Quote:WizardI disagree. It has a specific answer.

If the area of the rectangle were a billion, you couldn't inscribe it a right triangle with the 4 and 5 distances given. If you feel otherwise, please give me the dimensions of a the rectangle with area one billion.

1) x= width of rectangle, y = height of rectangle, 4/x=y/5, xy = 20

2) Entire figure area, A = (4+y) * (5 +x)/2 = (20 + 4x + 5y + xy)/2, and xy = 20

So, A = 20 + 2x + 50/x

dA/dx = 2 - 50/x^2 = 0

x= 5

Minimum A = 20 + 2(5) + 50/5 = 40

What if x=0, y= 0 ? still applicable to your question ?

Quote:WizardI disagree. It has a specific answer.

If the area of the rectangle were a billion, you couldn't inscribe it a right triangle with the 4 and 5 distances given. If you feel otherwise, please give me the dimensions of a the rectangle with area one billion.

Oh, you're looking for the area of the rectangle. I misread it - I thought you were looking for the area of the large triangle.

In that case...

As stated in my earlier solution, if b is the length of the undetermined leg of the blue triangle and y the length of the undetermined length of the yellow triangle, then 4 / y = b / 5, so by = 20. Since the rectangle has height b and width y, its area is yb, or 20.