Quote:teliotOf all things, I got a very cute little trigonometry puzzle from a friend I volunteer with. She asked,

Quote:

I'm trying to calculate the curve so I can sew a stuffed ball for a child, the surface of which consists of six sections as of an orange..

If (for example) the circumference of the sphere is 24 inches, then each section would be 12" long, and 4 " at its widest point. What equation would give the right width?

So, assume the sphere has circumference 24 inches and she is sewing six sections, each of length 12. What is the equation for the width of the section as a function of the number of inches, call it x, from the top of the section?

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M ... the meridian on the sphere, length from north pole to south pole = 12"

X ... the distance walked along the meridian from the north pole, heading south

phi ... the angle defined by distance X along M

measured in the sphere's center

between the line center-to-north-pole and center-to-point-at-the-end-of-X

X phi

- = --- ... same proportion of length of arcs and of angles

M PI

solve for phi:

X * PI

phi = ------

M

r ... radius of the sphere

solve for r from circumference of the sphere:

2 * M = 2 * PI * r ... circumference of the sphere

2 * M M

r = ------ = --

2 * PI PI

rp ... radius of parallel at angle phi

X * PI

rp = r * sin phi = r * sin ------

M

M X * PI

rp = -- * sin ------

PI M

2 * PI * rp ... circumference on the parallel

1 1

- * 2 * PI * rp = - * PI * rp ... width of the section on the parallel

6 3

substitute rp:

+=====================+

1 M X * PI | 1 X * PI |

- * PI * -- * sin ------ = | - * M * sin ------ |

3 PI M | 3 M |

+=====================+

Quote:ThomasKQuote:teliotOf all things, I got a very cute little trigonometry puzzle from a friend I volunteer with. She asked,

Quote:

I'm trying to calculate the curve so I can sew a stuffed ball for a child, the surface of which consists of six sections as of an orange..

If (for example) the circumference of the sphere is 24 inches, then each section would be 12" long, and 4 " at its widest point. What equation would give the right width?

So, assume the sphere has circumference 24 inches and she is sewing six sections, each of length 12. What is the equation for the width of the section as a function of the number of inches, call it x, from the top of the section?

link to original post

...

+=====================+

1 M X * PI | 1 X * PI |

- * PI * -- * sin ------ = | - * M * sin ------ |

3 PI M | 3 M |

+=====================+

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Quote:teliotQuote:ThomasKQuote:teliotOf all things, I got a very cute little trigonometry puzzle from a friend I volunteer with. She asked,

Quote:

I'm trying to calculate the curve so I can sew a stuffed ball for a child, the surface of which consists of six sections as of an orange..

If (for example) the circumference of the sphere is 24 inches, then each section would be 12" long, and 4 " at its widest point. What equation would give the right width?

So, assume the sphere has circumference 24 inches and she is sewing six sections, each of length 12. What is the equation for the width of the section as a function of the number of inches, call it x, from the top of the section?

link to original post

...

+=====================+

1 M X * PI | 1 X * PI |

- * PI * -- * sin ------ = | - * M * sin ------ |

3 PI M | 3 M |

+=====================+

link to original postYes! If you substitute PI*r for M then, that's the solution I got.

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Thank you for confirming my solution.

M = PI * r is given in the middle of my derivation as r = M / PI in order to state the final term based on the predefined, known parameter M: the 12 inches length of a section.

teliot, may I ask you for your opinion and solution on the following puzzle?

Q1: Is it possible to calculate the house edge of this game, based on the given parameters?

Q2: If "yes", how much would the house edge of this game be?

Please note that 400/1444 is the probability of losing two even money bets in sequence at double zero roulette.

Please also note that the player, using the double zero roulette in order to determine the 27.7% case, would win 1 unit either

by winning the first even money bet or

by wagering additional 2 units on a second even money bet in order to recoup the loss of the first roll plus win the 1 unit, aimed for.

Only in the case of losing both even money bets, the player would lose 3 units.

Q3: Which reasons would justify, that the game initially described, should have a house edge of 5.26%?

Quote:ThomasKImagine a game where the player loses 3 units with a probability of 27.7% (exact probability = 400/1444) or wins 1 unit in all other cases.

Q1: Is it possible to calculate the house edge of this game, based on the given parameters?

Q2: If "yes", how much would the house edge of this game be?

Please note that 400/1444 is the probability of losing two even money bets in sequence at double zero roulette.

Please also note that the player, using the double zero roulette in order to determine the 27.7% case, would win 1 unit either

by winning the first even money bet or

by wagering additional 2 units on a second even money bet in order to recoup the loss of the first roll plus win the 1 unit, aimed for.

Only in the case of losing both even money bets, the player would lose 3 units.

Q3: Which reasons would justify, that the game initially described, should have a house edge of 5.26%?

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How many threads do you need to ask about "treat the maximum potential loss in a Martingale as the amount bet when calculating HE?"

Anyway, assuming your bet is 3, the ER is -3 x 0.277 + 1 x (1 - 0.277) = -0.108, so the HE is 0.108 / 3 x 100% = 3.6%

The justification is, you aren't betting 3 to win 1, but making up to 2 separate bets, each of which has a HE of 5.26%.

Your justification that the two-step Martingale has an ER of "only" (about) 3.6% is true if you can guarantee to hold discipline - that is, you won't stop after the first loss under any circumstances.

Just as I said the last time you asked this sort of question.

Quote:ThatDonGuyQuote:ThomasKImagine a game where the player loses 3 units with a probability of 27.7% (exact probability = 400/1444) or wins 1 unit in all other cases.

Q1: Is it possible to calculate the house edge of this game, based on the given parameters?

Q2: If "yes", how much would the house edge of this game be?

Please note that 400/1444 is the probability of losing two even money bets in sequence at double zero roulette.

Please also note that the player, using the double zero roulette in order to determine the 27.7% case, would win 1 unit either

by winning the first even money bet or

by wagering additional 2 units on a second even money bet in order to recoup the loss of the first roll plus win the 1 unit, aimed for.

Only in the case of losing both even money bets, the player would lose 3 units.

Q3: Which reasons would justify, that the game initially described, should have a house edge of 5.26%?

link to original post

How many threads do you need to ask about "treat the maximum potential loss in a Martingale as the amount bet when calculating HE?"

Anyway, assuming your bet is 3, the ER is -3 x 0.277 + 1 x (1 - 0.277) = -0.108, so the HE is 0.108 / 3 x 100% = 3.6%

The justification is, you aren't betting 3 to win 1, but making up to 2 separate bets, each of which has a HE of 5.26%.

Your justification that the two-step Martingale has an ER of "only" (about) 3.6% is true if you can guarantee to hold discipline - that is, you won't stop after the first loss under any circumstances.

Just as I said the last time you asked this sort of question.

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Thanks for answering and confirming the math, although my repeated questioning seems to be somewhat irritating. My excuse is, I'm trying to find the most compact way to convey my point.

I agree, the player makes up to two bets in order to win one unit. But the first of these bets does not have a house edge of 5.26%, because the player never actually loses that first one unit. The player either wins this first bet (not losing money to the house = no house edge) or recoups the temporary first drawback by winning the seond bet (not losing money to the house either = no house edge). Only in the case of losing both bets the player will actually lose money to the house. So, only losing the 3 units contributes to the house's take and therefore creates house edge.Quote:The justification is, you aren't betting 3 to win 1, but making up to 2 separate bets, each of which has a HE of 5.26%.

On one hand, discipline shouldn't be an issue when gambling. System players are as serious about their procedures as APs are about their mathematically based strategies for +EV games. On the other hand, the player is not forced to double up on the exact follwing roll. The outcomes of the rolls are independent, which again provides the player with the freedom to play any of the 2-unit-bets at any arbitrary point in time and at any arbitrary roulette wheel, as long as this other wheel produces the same probabilities and as long as the player takes good care to bet exactly as many 2-units as the player has to recoup 1-units.Quote:Your justification that the two-step Martingale has an ER of "only" (about) 3.6% is true if you can guarantee to hold discipline - that is, you won't stop after the first loss under any circumstances.

Quote:ThomasKQuote:ThatDonGuyQuote:ThomasKImagine a game where the player loses 3 units with a probability of 27.7% (exact probability = 400/1444) or wins 1 unit in all other cases.

Q1: Is it possible to calculate the house edge of this game, based on the given parameters?

Q2: If "yes", how much would the house edge of this game be?

Please note that 400/1444 is the probability of losing two even money bets in sequence at double zero roulette.

Please also note that the player, using the double zero roulette in order to determine the 27.7% case, would win 1 unit either

by winning the first even money bet or

by wagering additional 2 units on a second even money bet in order to recoup the loss of the first roll plus win the 1 unit, aimed for.

Only in the case of losing both even money bets, the player would lose 3 units.

Q3: Which reasons would justify, that the game initially described, should have a house edge of 5.26%?

link to original post

How many threads do you need to ask about "treat the maximum potential loss in a Martingale as the amount bet when calculating HE?"

Anyway, assuming your bet is 3, the ER is -3 x 0.277 + 1 x (1 - 0.277) = -0.108, so the HE is 0.108 / 3 x 100% = 3.6%

The justification is, you aren't betting 3 to win 1, but making up to 2 separate bets, each of which has a HE of 5.26%.

Your justification that the two-step Martingale has an ER of "only" (about) 3.6% is true if you can guarantee to hold discipline - that is, you won't stop after the first loss under any circumstances.

Just as I said the last time you asked this sort of question.

link to original post

Thanks for answering and confirming the math, although my repeated questioning seems to be somewhat irritating. My excuse is, I'm trying to find the most compact way to convey my point.I agree, the player makes up to two bets in order to win one unit. But the first of these bets does not have a house edge of 5.26%, because the player never actually loses that first one unit. The player either wins this first bet (not losing money to the house = no house edge) or recoups the temporary first drawback by winning the seond bet (not losing money to the house either = no house edge). Only in the case of losing both bets the player will actually lose money to the house. So, only losing the 3 units contributes to the house's take and therefore creates house edge.Quote:The justification is, you aren't betting 3 to win 1, but making up to 2 separate bets, each of which has a HE of 5.26%.

On one hand, discipline shouldn't be an issue when gambling. System players are as serious about their procedures as APs are about their mathematically based strategies for +EV games. On the other hand, the player is not forced to double up on the exact follwing roll. The outcomes of the rolls are independent, which again provides the player with the freedom to play any of the 2-unit-bets at any arbitrary point in time and at any arbitrary roulette wheel, as long as this other wheel produces the same probabilities and as long as the player takes good care to bet exactly as many 2-units as the player has to recoup 1-units.Quote:Your justification that the two-step Martingale has an ER of "only" (about) 3.6% is true if you can guarantee to hold discipline - that is, you won't stop after the first loss under any circumstances.

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ThomasK, have you looked at craps bets that quote both a house edge per roll and house edge to bet resolution? Such as place bets or hard ways. You might find some analogy there.

unJon, thank you for having me read the Wizard's analysis again.Quote:unJonQuote:ThomasKQuote:ThatDonGuyQuote:ThomasK

Q1: Is it possible to calculate the house edge of this game, based on the given parameters?

Q2: If "yes", how much would the house edge of this game be?

Please note that 400/1444 is the probability of losing two even money bets in sequence at double zero roulette.

Please also note that the player, using the double zero roulette in order to determine the 27.7% case, would win 1 unit either

by winning the first even money bet or

by wagering additional 2 units on a second even money bet in order to recoup the loss of the first roll plus win the 1 unit, aimed for.

Only in the case of losing both even money bets, the player would lose 3 units.

Q3: Which reasons would justify, that the game initially described, should have a house edge of 5.26%?

link to original post

How many threads do you need to ask about "treat the maximum potential loss in a Martingale as the amount bet when calculating HE?"

Anyway, assuming your bet is 3, the ER is -3 x 0.277 + 1 x (1 - 0.277) = -0.108, so the HE is 0.108 / 3 x 100% = 3.6%

The justification is, you aren't betting 3 to win 1, but making up to 2 separate bets, each of which has a HE of 5.26%.

Your justification that the two-step Martingale has an ER of "only" (about) 3.6% is true if you can guarantee to hold discipline - that is, you won't stop after the first loss under any circumstances.

Just as I said the last time you asked this sort of question.

link to original post

Thanks for answering and confirming the math, although my repeated questioning seems to be somewhat irritating. My excuse is, I'm trying to find the most compact way to convey my point.I agree, the player makes up to two bets in order to win one unit. But the first of these bets does not have a house edge of 5.26%, because the player never actually loses that first one unit. The player either wins this first bet (not losing money to the house = no house edge) or recoups the temporary first drawback by winning the seond bet (not losing money to the house either = no house edge). Only in the case of losing both bets the player will actually lose money to the house. So, only losing the 3 units contributes to the house's take and therefore creates house edge.Quote:The justification is, you aren't betting 3 to win 1, but making up to 2 separate bets, each of which has a HE of 5.26%.

On one hand, discipline shouldn't be an issue when gambling. System players are as serious about their procedures as APs are about their mathematically based strategies for +EV games. On the other hand, the player is not forced to double up on the exact follwing roll. The outcomes of the rolls are independent, which again provides the player with the freedom to play any of the 2-unit-bets at any arbitrary point in time and at any arbitrary roulette wheel, as long as this other wheel produces the same probabilities and as long as the player takes good care to bet exactly as many 2-units as the player has to recoup 1-units.Quote:Your justification that the two-step Martingale has an ER of "only" (about) 3.6% is true if you can guarantee to hold discipline - that is, you won't stop after the first loss under any circumstances.

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ThomasK, have you looked at craps bets that quote both a house edge per roll and house edge to bet resolution? Such as place bets or hard ways. You might find some analogy there.

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I've read it quite a while ago and haven't understood what a house edge per roll could be useful for, when rating a pass line bet, knowing that the initial bet must not be taken down, anyway. The player has to wait for the pass line to resolve.

Even now, reading it again and understanding that the question is about taking down e.g. a place bet before it is resolved, I still don't completely get the usefulness of the concept.

* After how many rolls should the player take down the place bet?

* When playing this strategy repeatedly, will it always be the same number of rolls, before the player takes down the bet?

* If the number of rolls may vary, how does the player decide, when to take down?

(I suspect that a Marcov Chain Analysis is needed to determine the corresponding house edges.)

Following the Wizard, ...

... I'm trying to argue, that a finite progressive betting sequence is a game of its own, with a house edge ("per bet resolved") different from the house edge of its single seperate rolls.Quote:the Wizard of Odds... I prefer to define the house edge in craps on a "per bet resolved" basis ...

Please take this progression system nonsense to the appropriate forum (Betting Systems) and stop polluting this very good thread on Easy Math Puzzles with this off-topic, annoying and non-mathematical voodoo.Quote:ThomasKI'm trying to argue, that a finite progressive betting sequence is a game of its own, with a house edge ("per bet resolved") different from the house edge of its single seperate rolls.

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Quote:teliotPlease take this progression system nonsense to the appropriate forum (Betting Systems) and stop polluting this very good thread on Easy Math Puzzles with this off-topic, annoying and non-mathematical voodoo.Quote:ThomasKI'm trying to argue, that a finite progressive betting sequence is a game of its own, with a house edge ("per bet resolved") different from the house edge of its single seperate rolls.

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Agree this isn’t the thread for it. I don’t think he’s pushing non mathematical voodoo though. It’s actually a math quirk that I think just shows that EV is a better metric than HE.