## Poll

8 votes (47.05%) | |||

6 votes (35.29%) | |||

3 votes (17.64%) | |||

2 votes (11.76%) | |||

6 votes (35.29%) | |||

2 votes (11.76%) | |||

3 votes (17.64%) | |||

2 votes (11.76%) | |||

8 votes (47.05%) | |||

6 votes (35.29%) |

**17 members have voted**

August 3rd, 2020 at 8:35:51 PM
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Quote:ssho88Quote:ChesterDogI cannot solve it, but at www.WolframAlpha.com, you can type in y''=1/(1-y) to get the general solution.

Can I solve it this way ?

(1-y) d2y = dx2

(y - y^2/2 + c1)dy = (x + c2)dx

y^2/2 - y^3/6 + c1(y) + c3 = (x^2/2) + c2(x) + c4

Correct ?

I have seen separation of variables for first degree differential equations but I haven't seen that trick for second degree equations.

Your original equation is: d

^{2}y/dx

^{2}= 1 / (1-y).

The meaning of this is: d[dy/dx] / dx = 1 / (1-y), which can be rearranged to: (1-y)d[dy/dx] = dx. I can integrate the right to get x + c

_{1}, but I can't integrate the left because I don't know an expression for y in terms of dy/dx.

August 3rd, 2020 at 8:44:07 PM
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Quote:ChesterDogQuote:ssho88Quote:ChesterDogI cannot solve it, but at www.WolframAlpha.com, you can type in y''=1/(1-y) to get the general solution.

Can I solve it this way ?

(1-y) d2y = dx2

(y - y^2/2 + c1)dy = (x + c2)dx

y^2/2 - y^3/6 + c1(y) + c3 = (x^2/2) + c2(x) + c4

Correct ?

I have seen separation of variables for first degree differential equations but I haven't seen that trick for second degree equations.

Your original equation is: d^{2}y/dx^{2}= 1 / (1-y).

The meaning of this is: d[dy/dx] / dx = 1 / (1-y), which can be rearranged to: (1-y)d[dy/dx] = dx. I can integrate the right to get x + c_{1}, but I can't integrate the left because I don't know an expression for y in terms of dy/dx.

Agree, I don't think it can be solved that way. Thanks

August 4th, 2020 at 8:00:24 AM
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The Babylonians used a base 60 number system.

What shape inspired them to decide that a circle has 360 degrees?

Have you tried 22 tonight? I said 22.

August 4th, 2020 at 8:35:31 AM
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Yikes, that means I would not only need to take my shoes off, but also to find 2 other barefoot friends just to count to 10! ;)Quote:Gialmere

The Babylonians used a base 60 number system.

Quote:What shape inspired them to decide that a circle has 360 degrees?

A hexagon, perhaps?

"Dealer has 'rock'... Pay 'paper!'"

August 4th, 2020 at 1:42:41 PM
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Quote:JoemanA hexagon, perhaps?

Correct!

Since the perimeter of a hexagon is exactly equal to six times the radius of the circumscribed circle, they chose to divide the circle into 6 x 60 = 360 degrees.

I should note there are other reasons for the 360 degrees. What they are and why the number 360 is special makes for some interesting reading if you're bored sometime.

------------------------------------

Things I learned in organic chemistry...

1. How to draw hexagons.

Have you tried 22 tonight? I said 22.

August 4th, 2020 at 1:46:24 PM
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Quote:ssho88Quote:ChesterDogI cannot solve it, but at www.WolframAlpha.com, you can type in y''=1/(1-y) to get the general solution.

Can I solve it this way ?

(1-y) d2y = dx2

(y - y^2/2 + c1)dy = (x + c2)dx

y^2/2 - y^3/6 + c1(y) + c3 = (x^2/2) + c2(x) + c4

Correct ?

edit: Chesterdog appears to have beat me to this.

I am rusty on my calculus. But I question whether you can integrate one side of an equation over x and integrate the other side of the equation over y and still have an equality. Does anyone know for sure?

Basically this is y(x) such that d

^{2}y(x)/dx

^{2}= 1/(1-y(x)). Can you expand 1/(1-y(x)) into an infinite series and then integrate it with respect to x?

So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.

August 5th, 2020 at 6:05:42 PM
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Two craps players make a side bet on a side bet.

Player A always plays "All Small" and "All Tall". Player B always plays "Make em All".

Starting with a new shooter, A tells B he will win both of his wagers at least twice before B wins his once. In this casino, All Small, All Tall, and Make em All bets can only be made before a new shooter's first roll.

This is an even money bet. They will play until their wager is resolved. What is A's edge (or disadvantage)?

Player A always plays "All Small" and "All Tall". Player B always plays "Make em All".

Starting with a new shooter, A tells B he will win both of his wagers at least twice before B wins his once. In this casino, All Small, All Tall, and Make em All bets can only be made before a new shooter's first roll.

This is an even money bet. They will play until their wager is resolved. What is A's edge (or disadvantage)?

It’s all about making that GTA

August 9th, 2020 at 7:30:41 AM
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Quote:Ace2Two craps players make a side bet on a side bet.

Player A always plays "All Small" and "All Tall". Player B always plays "Make em All".

Starting with a new shooter, A tells B he will win both of his wagers at least twice before B wins his once. In this casino, All Small, All Tall, and Make em All bets can only be made before a new shooter's first roll.

This is an even money bet. They will play until their wager is resolved. What is A's edge (or disadvantage)?

Is there an "easy" solution to this?

I can see this as a Markov chain, but the number of states makes it really hard to solve.

The way I see it, each state consists of a Small Bet state, a Tall state, and an All state.

There are 63 Small states, ranging from 0 to 62: (31 x the number of Small bets won, up to 2) + (16 if 2 has been rolled in the current Small bet) + (8 if 3 has been rolled) + (4 if 4 has been rolled) + (2 if 5 has been rolled) + (1 if 6 has been rolled)

There are also 63 Tall states, replacing 2-6 with 8-12

There are 1024 All states: (512 if 2 has been rolled) + (256 if 3 has been rolled) + ... + (2 if 11 has been rolled) + (1 if 12 has been rolled)

This is over 4 million states, although some are not possible (for example, if there have not been any Small or Tall bets won yet, then each All state has exactly one Small/Tall state pair.)

There is the further complication that, when the Small or Tall bet is won for the first time, it does not "reset" until there is a new shooter.

August 9th, 2020 at 10:07:01 AM
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There are just three relevant possibilities for every shooter: wins small, tall or all. Then there are only four states in terms of small and tall: 1 win, 2 distinct wins, 2 similar wins and 3 wins. You should be able to solve with linear equations

I solved it using calculus which, in my opinion, is much easier than linear equations.

Or my logic/math is wrong.

I solved it using calculus which, in my opinion, is much easier than linear equations.

Or my logic/math is wrong.

Last edited by: Ace2 on Aug 9, 2020

It’s all about making that GTA

August 9th, 2020 at 3:42:58 PM
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A surveyor wants to approximate the width of a lake. He places 5 markers near the lake and measures the distances shown in the diagram above.

What is the width of the lake in meters from point P to point Q?

Have you tried 22 tonight? I said 22.