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%?

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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.

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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%?

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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.

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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.

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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.

Sir, thank you very much for your very clear feedback, although I don't fully understand the negative touch.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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Referring back to my original puzzle, may I please ask the following questions, for my better understanding of which mistakes I am making and for me (and others?) to learn which parts of my questions are non-mathematical?

Q1: From your experience, is it the case that it is not possible to calculate the house edge of the related game from these parameters?

* losing 3 units with a probability of 400/1444

* winning 1 unit otherwise (i.e. with a probability of (1 - 400/1444))

Q2: Is it therefore wrong to state that the house edge of the game having these parameters, has a house edge of 3.6%, as ThatDonGuy calculated and which, by the way, would also be my approach to this game's house edge?

Q3: Is it further wrong to state that 400/1444 is the probability of not hitting two even money bets in succession (e.g. not red - not red) at double zero roulette?

Q4: Is it an invalid approach to generate an event occurring with a probability of 400/1444 by observing a double zero roulette wheel and awaiting two even money events not to occur in succession (e.g. not red - not red)?

Q5: Is it wrong to state that a player, losing 1 unit on a first even money bet, then doubling up to 2 units for a second even money bet, and eventually also losing this second bet, has lost 3 units in total?

Q6: Is it false to claim that a player hitting a single even money bet, will win 1 unit to the original wager of 1 unit?

Q7: Is it furthermore false that a player, losing a first even money bet of 1 unit but winning a second even money bet, now wagering 2 units, will eventually have won 1 unit in total?

Q8: Is it wrong to associate the parameters of the puzzle's game and the questioned properties above, as follows:

* losing 3 units (see Q5) with a probability of 400/1444 (see Q3 and Q4)

* winning 1 unit otherwise (see Q6 and Q7) (i.e. with a probability of (1 - 400/1444))

Q9: Assumed, the answer to Q1 is "It is possible to calculate the house edge." and

assumed, the answer to Q2 is "The house edge is 3.6%.", and also

assumed that the answer to Q8 is "The game's parameters can be generated by the associated properties.",

which mathematical components are missing, that would prevent the house edge from being calculated as 3.6%?

Final question:

Which of my questions are non-mathematical and what exactly are these non-mathematical aspects in my questions?

Respectfully looking forward to your answers.

This is extraordinarily rude and you deserve to be censored by the mods for posting incorrectly into the wrong forum in a manner that is flooding or spamming or both. If you had any sense of decency or respect for this site you would stop posting this nonsense here immediately.Quote:ThomasKSir, thank you very much for your very clear feedback, although I don't fully understand the negative touch.Quote:teliotQuote:ThomasK

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Referring back to my original puzzle, may I please ask the following questions, for my better understanding of which mistakes I am making and for me (and others?) to learn which parts of my questions are non-mathematical?

Q1: From your experience, is it the case that it is not possible to calculate the house edge of the related game from these parameters?

* losing 3 units with a probability of 400/1444

* winning 1 unit otherwise (i.e. with a probability of (1 - 400/1444))

Q2: Is it therefore wrong to state that the house edge of the game having these parameters, has a house edge of 3.6%, as ThatDonGuy calculated and which, by the way, would also be my approach to this game's house edge?

Q3: Is it further wrong to state that 400/1444 is the probability of not hitting two even money bets in succession (e.g. not red - not red) at double zero roulette?

Q4: Is it an invalid approach to generate an event occurring with a probability of 400/1444 by observing a double zero roulette wheel and awaiting two even money events not to occur in succession (e.g. not red - not red)?

Q5: Is it wrong to state that a player, losing 1 unit on a first even money bet, then doubling up to 2 units for a second even money bet, and eventually also losing this second bet, has lost 3 units in total?

Q6: Is it false to claim that a player hitting a single even money bet, will win 1 unit to the original wager of 1 unit?

Q7: Is it furthermore false that a player, losing a first even money bet of 1 unit but winning a second even money bet, now wagering 2 units, will eventually have won 1 unit in total?

Q8: Is it wrong to associate the parameters of the puzzle's game and the questioned properties above, as follows:

* losing 3 units (see Q5) with a probability of 400/1444 (see Q3 and Q4)

* winning 1 unit otherwise (see Q6 and Q7) (i.e. with a probability of (1 - 400/1444))

Q9: Assumed, the answer to Q1 is "It is possible to calculate the house edge." and

assumed, the answer to Q2 is "The house edge is 3.6%.", and also

assumed that the answer to Q8 is "The game's parameters can be generated by the associated properties.",

which mathematical components are missing, that would prevent the house edge from being calculated as 3.6%?

Final question:

Which of my questions are non-mathematical and what exactly are these non-mathematical aspects in my questions?

Respectfully looking forward to your answers.

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What is the expected amount a player will lose per spin (roulette) / comeout (craps) betting even-money bets with an n-step Martingale?

Never mind the house edge as a percentage of the potential ("black box") bet - we're talking actual units here.

The numbers indicate the expected player loss per comeout/spin - not per Martingale cycle - with a base bet of 1.

Max # Rounds | Craps Pass | Roulette Double 0 | Roulette Single 0 |
---|---|---|---|

1 | 0.014141414141414 | 0.0526315789473686 | 0.027027027027027 |

2 | 0.0188994502667423 | 0.0707803992740474 | 0.0361969111969111 |

3 | 0.0243890821482192 | 0.0922467458970009 | 0.0468668421812687 |

4 | 0.0304960453553553 | 0.11684139663243 | 0.0588565960675786 |

5 | 0.0370969846843513 | 0.144312368164884 | 0.0719619355309165 |

6 | 0.0440804199812489 | 0.174410789831217 | 0.0859928975027539 |

7 | 0.0513572701241295 | 0.206933712668106 | 0.100794985737663 |

8 | 0.0588625893455474 | 0.24174222754772 | 0.116254845012615 |

9 | 0.0665522955045133 | 0.278761820982288 | 0.13229632711992 |

10 | 0.0743981826669646 | 0.317973528435176 | 0.148872641013953 |

Hmmm...seems to me that, the longer your Martingale progression is, the more per comeout/spin you are expected to lose.

Let p = the probability of a win (for an even-money bet on double-zero roulette, this is 9/19) and q = 1 - p = the probability of a loss

The number of spins needed to resolve an n-step Martingale, one way or the other, is:

p (i.e. the probability of winning on the first spin)

+ 2 q p (i.e. the probability of needing 2 spins to win)

+ 3 q^2 p

+ 4 q^3 p

+ ...

+ n q^(n-1) p (i.e. the probability of winning on the nth spin)

+ n q^n (i.e. the probability of n losing spins in a row)

= p (1 + 2 q + 3 q^2 + 4 q^3 + ... + n q^(n-1)) + n q^n

= p (

(1 + q + q^2 + q^3 + ... + q^(n-1))

+ (q + q^2 + q^3 + ... + q^(n-1))

+ (q^2 + q^3 + ... + q^(n-1))

+ ...

+ (q^(n-2) + q^(n-1))

+ q^(n-1)

) + n q^n

= p (

(1 - q^n) / (1 - q)

+ q (1 - q^(n-1)) / (1 - q)

+ q^2 (1 - q^(n-2)) / (1 - q)

+ ...

+ q^(n-2) (1 - q^2) / (1 - q)

+ q^(n-1) (1 - q) / (1 - q)

) + n q^n

= p / (1 - q) *

(

(1 - q^n)

+ q - q^n

+ q^2 - q^n

+ ...

+ q^(n-2) - q^n

+ q^(n-1) - q^n

) + n q^n

= (1 + q + q^2 + ... + q^(n-1) - n q^n) + n q^n (since p / (1 - q) = p / p = 1)

= 1 + q + q^2 + ... + q^(n-1)

= (1 - q^n) / (1 - q)

= (1 - q^n) / p

You will lose 2^n - 1 units if you lose n consecutive spins, which has a probability of q^n,

and win 1 unit (i.e. lose -1 units) in any other result, which has a probability of (1 - q^n)

The expected loss is (2^n - 1) * q^n - 1 * (1 - q^n)

= 2^n q^n - q^n - 1 + q^n

= (2q)^n - 1

The expected loss per spin = ((2q)^n - 1) / ((1 - q^n) / p)

= p ((2q)^n - 1) / (1 - q^n)

Thanks for your analysis.Quote:ThatDonGuySpeaking of "math problems," here's one:

What is the expected amount a player will lose per spin (roulette) / comeout (craps) betting even-money bets with an n-step Martingale?

Never mind the house edge as a percentage of the potential ("black box") bet - we're talking actual units here.

The numbers indicate the expected player loss per comeout/spin - not per Martingale cycle - with a base bet of 1.

Max #

RoundsCraps

PassRoulette

Double 0Roulette

Single 0 10.014141414141414 0.0526315789473686 0.027027027027027 20.0188994502667423 0.0707803992740474 0.0361969111969111 30.0243890821482192 0.0922467458970009 0.0468668421812687 40.0304960453553553 0.11684139663243 0.0588565960675786 50.0370969846843513 0.144312368164884 0.0719619355309165 60.0440804199812489 0.174410789831217 0.0859928975027539 70.0513572701241295 0.206933712668106 0.100794985737663 80.0588625893455474 0.24174222754772 0.116254845012615 90.0665522955045133 0.278761820982288 0.13229632711992 100.0743981826669646 0.317973528435176 0.148872641013953

Hmmm...seems to me that, the longer your Martingale progression is, the more per comeout/spin you are expected to lose.

Let p = the probability of a win (for an even-money bet on double-zero roulette, this is 9/19) and q = 1 - p = the probability of a loss

The number of spins needed to resolve an n-step Martingale, one way or the other, is:

p (i.e. the probability of winning on the first spin)

+ 2 q p (i.e. the probability of needing 2 spins to win)

+ 3 q^2 p

+ 4 q^3 p

+ ...

+ n q^(n-1) p (i.e. the probability of winning on the nth spin)

+ n q^n (i.e. the probability of n losing spins in a row)

= p (1 + 2 q + 3 q^2 + 4 q^3 + ... + n q^(n-1)) + n q^n

= p (

(1 + q + q^2 + q^3 + ... + q^(n-1))

+ (q + q^2 + q^3 + ... + q^(n-1))

+ (q^2 + q^3 + ... + q^(n-1))

+ ...

+ (q^(n-2) + q^(n-1))

+ q^(n-1)

) + n q^n

= p (

(1 - q^n) / (1 - q)

+ q (1 - q^(n-1)) / (1 - q)

+ q^2 (1 - q^(n-2)) / (1 - q)

+ ...

+ q^(n-2) (1 - q^2) / (1 - q)

+ q^(n-1) (1 - q) / (1 - q)

) + n q^n

= p / (1 - q) *

(

(1 - q^n)

+ q - q^n

+ q^2 - q^n

+ ...

+ q^(n-2) - q^n

+ q^(n-1) - q^n

) + n q^n

= (1 + q + q^2 + ... + q^(n-1) - n q^n) + n q^n (since p / (1 - q) = p / p = 1)

= 1 + q + q^2 + ... + q^(n-1)

= (1 - q^n) / (1 - q)

= (1 - q^n) / p

You will lose 2^n - 1 units if you lose n consecutive spins, which has a probability of q^n,

and win 1 unit (i.e. lose -1 units) in any other result, which has a probability of (1 - q^n)

The expected loss is (2^n - 1) * q^n - 1 * (1 - q^n)

= 2^n q^n - q^n - 1 + q^n

= (2q)^n - 1

The expected loss per spin = ((2q)^n - 1) / ((1 - q^n) / p)

= p ((2q)^n - 1) / (1 - q^n)

link to original post

Replacing "amount lost per units wagered" by "amount lost per spin played" doesn't seem to be all that helpful, when trying to discuss "actual units". Your approach again does not take into account that in case of a win, which is ending the Martingale cycle, the previous losses within that cycle are recouped.

From a "per bet resolved" perspective only the 2^n-1 units lost on the n

^{th}spin are the actual loss the player is suffering from in this game. On any winning cyle, the player will walk away 1 unit richer and without an actual loss.

The equation for the expected loss, as you gave it, supports this statement.

E(M

_{n}) = (2^n - 1)*q^n - 1*(1 - q^n) = 2^n * q^n - 1

(2^n - 1)*q^n - 1*(1 - q^n) says that the player will lose 2^n-1 units with a probability of q^n. So the player is required to bring these 2^n-1 units in order to be able to play that game.

The equivalent expression 2^n * q^n - 1 defines a slot machine with a payout of 2^n units with that same probability of q^n and a coin in of 1 unit. The payout is positive, so this is the house's perspective (player loses to the house). The 1 unit is all the house needs to play against a player's Martingale cycle, as I have shown earlier.

While it is true that with a longer Martingale progression the player loses a higher amount, it is also true that the probability of the player losing becomes much smaller. In total the player loses less relative to the higher amount required to play.

Using geometric progression ('grin') is rather impressive and somewhat intimidating at first glance. Simple multiplication does that job as well:

= p(1+2q+3q^2+4q^3+...+nq^(n-1))+nq^n

=(1-q)*(1+2q+3q^2+4q^3+...+nq^(n-1))+nq^n

= 1+2q+3q^2+4q^3+5q^4+...+(n-1)q^(n-2)+nq^(n-1)

-q*(1+2q+3q^2+4q^3+........+(n-1)q^(n-2)+nq^(n-1))+nq^n

= 1+2q+3q^2+4q^3+5q^4+.......+nq^(n-1)

-q-2q^2-3q^3-4q^4-...-(n-1)q^(n-1)-nq^n +nq^n

=1+q+q^2+q^3+q^4+...+q^(n-1)

=(1-q^n)/(1-q)

A reality check if you will:

A player walks up to the $10 minimum craps table with a bank roll buy-in of $1000. His goal is to win $200, i.e., leave the table with $1200. Using all of the above "math," what should be his wagering approach? How long will it take?

tuttigym

Quote:ChumpChange$10 PL with $10 odds and two $10 come bets with single odds. $15 if it's a $15 table.

link to original post

Great. so how does that relate to all the "math" and charts? Perhaps ThatDonGuy, Teliot (sp?), or even the Wizard can provide the "math" clarity as expressed in those rambling posts.

tuttigym

Quote:tuttigymWithout producing all of the previous posts showing the endless equations and touting the charting of house edge components, how does any of this actually relate to table play?

A reality check if you will:

A player walks up to the $10 minimum craps table with a bank roll buy-in of $1000. His goal is to win $200, i.e., leave the table with $1200. Using all of the above "math," what should be his wagering approach? How long will it take?

tuttigym

link to original post

I get: (corrected}

$10 Martingale: 81.33% (simulated)

$200 Martingale: 82.574% (simulated)

$200 flat (no odds): 82.123% (calculated)

Quote:ThatDonGuyQuote:tuttigymWithout producing all of the previous posts showing the endless equations and touting the charting of house edge components, how does any of this actually relate to table play?

A reality check if you will:

A player walks up to the $10 minimum craps table with a bank roll buy-in of $1000. His goal is to win $200, i.e., leave the table with $1200. Using all of the above "math," what should be his wagering approach? How long will it take?

tuttigym

link to original post

I get:

$10 Martingale: 81.33% (simulated)

$200 Martingale: 81.36% (simulated)

$200 flat (no odds): 82.123% (calculated)

link to original post

Thanks, I knew i could count on you to provide an answer.

I am assuming these bets are all PL only?

On the $10 Marty sim, how many PL wagers, on average, would it take to reach that $200 win?

Same question for the $200 Marty?

The $200 flat would achieve the win on one roll, correct?

So, the realty questions are: Would you play like that? and Do you believe that any "educated" player would play like that?

I also would like to know how your answers above relate to the wild algebraic posts above.

p.s. I personally do not regard computer sims to be reflective of actual play. (Kinda just my thing.)

tuttigym

Quote:ChumpChangeI would plan on losing at least 10 PL bets in a row, maybe even 15 or 20.

link to original post

What percentage of the time would one end up winning?

Again, the answer I am seeking is how does the posted "math"/algebra/HE/HA relate to the reality of actual play as exhibited by my question?

tuttigym

If you're flat betting $10, you'd lose $90 then win $10 back for an $80 loss; or lose $190 then win back $10 for a $180 loss.

Quote:tuttigymQuote:ThatDonGuyQuote:tuttigymWithout producing all of the previous posts showing the endless equations and touting the charting of house edge components, how does any of this actually relate to table play?

A reality check if you will:

A player walks up to the $10 minimum craps table with a bank roll buy-in of $1000. His goal is to win $200, i.e., leave the table with $1200. Using all of the above "math," what should be his wagering approach? How long will it take?

tuttigym

link to original post

I get:

$10 Martingale: 81.33% (simulated)

$200 Martingale: 81.36% (simulated)

$200 flat (no odds): 82.123% (calculated)

link to original post

Thanks, I knew i could count on you to provide an answer.

Wait a minute...there's something wrong with that...

If your initial bet is your target, a Martingale should be better than flat betting, since the expected number of bets needed, and, as a result, the amount exposed to the per-bet HE, is smaller.

I have corrected my original post.

The base bet for a Martingale that wins as much as $200 flat betting is $100.

Quote:tuttigym

I am assuming these bets are all PL only?

Yes.

Quote:tuttigym

On the $10 Marty sim, how many PL wagers, on average, would it take to reach that $200 win?

Same question for the $200 Marty?

The $200 flat would achieve the win on one roll, correct?

So, the realty questions are: Would you play like that? and Do you believe that any "educated" player would play like that?

I also would like to know how your answers above relate to the wild algebraic posts above.

p.s. I personally do not regard computer sims to be reflective of actual play. (Kinda just my thing.)

tuttigym

link to original post

Expected number of bets needed for:

$5 Martingale: 85.25

$100 Martingale: 4.609

$200 Martingale: 2.283

Flat Bets of $200: 5.094

Quote:ThatDonGuyQuote:tuttigym

A reality check if you will:

A player walks up to the $10 minimum craps table with a bank roll buy-in of $1000. His goal is to win $200, i.e., leave the table with $1200. Using all of the above "math," what should be his wagering approach? How long will it take?

tuttigym

link to original post

I get: (corrected}

$10 Martingale: 81.33% (simulated)

$200 Martingale: 82.574% (simulated)

$200 flat (no odds): 82.123% (calculated)

link to original post

I thought about this answer. Does the 80+% figures you posted mean that one would win $200 80+% of the time or lose $1000 80+% of the time?

And are you saying (posting) that the lengthy algebraic formula only relates to a Martingale method of gambling at craps? The equation has so many components, and the Martingale method has basically only two components or options, so how can that complicated equation possibly reflect the simplicity of a Martingale method of gambling?

tuttigym

$1000 bankroll

$10 table minimum

$200 goal

assumed

$1000 table maximum (or higher)

passline without odds, to be able to achieve the $200 goal with a 7 or 11, as well

probabilities of passline bet

50.707070% for losing

49.292929% for winning

also assumed

The goal has to be achieved on consecutive tries of winning Martingales, because otherwise not enough bankroll will be left, neither to recoup the loss nor to achieve the $200 goal on top of that.

Martingale of length 1 = a single passline, 1 try

$200 bet on 1 passline

49.29% probability of winning the passline = winning $200

Martingale of length 2, 1 try

$200 bet on 1st passline

$400 bet on 2nd passline, if 1st passline has been lost

74.29% probability of winning any of the two passlines = winning $200

Martingale of length 3, 2 tries

$100 bet on 1st passline

$200 bet on 2nd passline, if 1st passline has been lost

$400 bet on 3rd passline, if 2nd passline has been lost

86.96% probability of winning 1 try = winning $100

Martingale of length 4, 4 tries

$60 bet on 1st passline

$120 bet on 2nd passline, if 1st has been lost

$240 bet on 3rd passline, if 2nd has been lost

$480 bet on 4th passline, if 3rd has been lost

93.39% probability of winning 1 try = winning $60

Martingale of length 5, 7 tries

$30 bet on 1st passline

$60 bet on 2nd passline, if 1st has been lost

$120 bet on 3rd passline, if 2nd has been lost

$240 bet on 4th passline, if 3rd has been lost

$480 bet on 5th passline, if 4th has been lost

96.65% probability of winning 1 try = winning $30

78.77% probability of winning 7 consecutive tries = winning $210

Martingale of length 6, 14 tries

$15 bet on 1st passline

$30 bet on 2nd passline, if 1st has been lost

$60 bet on 3rd passline, if 2nd has been lost

$120 bet on 4th passline, if 3rd has been lost

$240 bet on 5th passline, if 4th has been lost

$480 bet on 6th passline, if 5th has been lost

98.30% probability of winning 1 try = winning $15

From the comparison it is clear that, while for longer Martingales a higher amount is exposed to a potential loss, the probability of losing a single Martingale becomes smaller, the longer the Maringale. Or, put the other way and as shown above, the probability of winning a single Martingale increases with each step added.

Under the given conditions of a $1000 bankroll and the goal of winning $200, 7 consecutive winning Martingales of max. 5 steps with a unit of $30 has the highest probability (78.77%) of succeessfully winning (at least) $200 to the player's original $1000.

Quote:tuttigymQuote:ThatDonGuyQuote:tuttigym

A reality check if you will:

A player walks up to the $10 minimum craps table with a bank roll buy-in of $1000. His goal is to win $200, i.e., leave the table with $1200. Using all of the above "math," what should be his wagering approach? How long will it take?

tuttigym

link to original post

I get: (corrected}

$10 Martingale: 81.33% (simulated)

$200 Martingale: 82.574% (simulated)

$200 flat (no odds): 82.123% (calculated)

link to original post

I thought about this answer. Does the 80+% figures you posted mean that one would win $200 80+% of the time or lose $1000 80+% of the time?

And are you saying (posting) that the lengthy algebraic formula only relates to a Martingale method of gambling at craps? The equation has so many components, and the Martingale method has basically only two components or options, so how can that complicated equation possibly reflect the simplicity of a Martingale method of gambling?

tuttigym

link to original post

The 80+% is the probability of winning $200.

As for my "lengthy algebraic formula," the lengthiness of it is the proof. Any formulas in there are quite simple. Also, they apply to any game with even-money bets.

Quote:ThomasKMy idea for analyzing the given setup is comparing Martingales of different lengths, searching for the optimal approach.

$1000 bankroll

$10 table minimum

$200 goal

assumed

$1000 table maximum (or higher)

passline without odds, to be able to achieve the $200 goal with a 7 or 11, as well

probabilities of passline bet

50.707070% for losing

49.292929% for winning

also assumed

The goal has to be achieved on consecutive tries of winning Martingales, because otherwise not enough bankroll will be left, neither to recoup the loss nor to achieve the $200 goal on top of that.

Martingale of length 1 = a single passline, 1 try

$200 bet on 1 passline

49.29% probability of winning the passline = winning $200

Martingale of length 2, 1 try

$200 bet on 1st passline

$400 bet on 2nd passline, if 1st passline has been lost

74.29% probability of winning any of the two passlines = winning $200

Martingale of length 3, 2 tries75.62% probability of winning 2 consecutive tries = winning $200

$100 bet on 1st passline

$200 bet on 2nd passline, if 1st passline has been lost

$400 bet on 3rd passline, if 2nd passline has been lost

86.96% probability of winning 1 try = winning $100

Martingale of length 4, 4 tries76.06% probability of winning 4 consecutive tries = winning $240

$60 bet on 1st passline

$120 bet on 2nd passline, if 1st has been lost

$240 bet on 3rd passline, if 2nd has been lost

$480 bet on 4th passline, if 3rd has been lost

93.39% probability of winning 1 try = winning $60

Martingale of length 5, 7 tries

$30 bet on 1st passline

$60 bet on 2nd passline, if 1st has been lost

$120 bet on 3rd passline, if 2nd has been lost

$240 bet on 4th passline, if 3rd has been lost

$480 bet on 5th passline, if 4th has been lost

96.65% probability of winning 1 try = winning $30

78.77% probability of winning 7 consecutive tries = winning $210

Martingale of length 6, 14 tries78.66% probability of winning 14 consecutive tries = winning $210

$15 bet on 1st passline

$30 bet on 2nd passline, if 1st has been lost

$60 bet on 3rd passline, if 2nd has been lost

$120 bet on 4th passline, if 3rd has been lost

$240 bet on 5th passline, if 4th has been lost

$480 bet on 6th passline, if 5th has been lost

98.30% probability of winning 1 try = winning $15

From the comparison it is clear that, while for longer Martingales a higher amount is exposed to a potential loss, the probability of losing a single Martingale becomes smaller, the longer the Maringale. Or, put the other way and as shown above, the probability of winning a single Martingale increases with each step added.

Under the given conditions of a $1000 bankroll and the goal of winning $200, 7 consecutive winning Martingales of max. 5 steps with a unit of $30 has the highest probability (78.77%) of succeessfully winning (at least) $200 to the player's original $1000.

link to original post

You forgot to take into account the fact that, in each one of those, a loss will lose a different amount. For example, in the "1 step" one, your potential loss is only $200, while in the 6-step one, a loss loses $950.

However, if you are trying to reach a particular target, a Martingale with as high of a starting bet as possible is probably the best approach, as you expose as little money as possible to the house edge per bet. The smaller the amount, the more steps it requires, which results in the total amount bet increasing, which results in the expected overall loss increasing as well. The best approach to win $200 with $1000 is probably 200-400-400 (then your next bet is 400 instead of 200, as you need 400 to reach your target of 1200).

The player's goal is to win 1 to 4 ($200 to $1000) which translates to probabilities of 4/5 for winning and 1/5 for losing in a fair game:Quote:ChumpChange

So you've got greater than a 1 in 5 chance of going bust. ...

E(fair game)=4/5*1 + 1/5*(-4) = 4/5 - 4/5 = 0

With a probability of 78.77%, slightly less than 80.00% (=4/5) this is a classical -EV game with a house advantage.

The 7 and 14 times of repeating the Martingales are included in the probability of 78.xx%. In some sense this "produces" that overall probability, although it "feels" as if the player were much more likely to lose than to win, playing that many passlines.Quote:... And you're trying to win 7 times or 14 times in a row. ...

The $1000 were given in the excercise. If I had to, I also would rather try it with the smallest table minimum I could find and risk only 31 minimum units to win 7.Quote:... I can see this turns out real bad unless $1,000 means nothing to you.

link to original post

It is worth noting that the 78.77% probability of achieving this goal will only work once in a lifetime. Any further attempt to repeat a first success will drasitcally decrease the chances of being successful, i.e. not losing the $1000.

1 succesful attempt: 78.77%

2 successful repeats: 62.04%

3 successful repeats: 48.87%

4 successful repeats: 38.49%

...

It is true that each of the approaches has a different amount to be lost in case of bad luck. The excercise asks how $1000 should be used in order to win an extra $200. My interpretation is that the full $1000, or any portion of it, may be used as a wager.Quote:ThatDonGuyYou forgot to take into account the fact that, in each one of those, a loss will lose a different amount. For example, in the "1 step" one, your potential loss is only $200, while in the 6-step one, a loss loses $950. ...

That is the reason why I also listed the 1 step Martingale, which effectively is a single passline bet. It only requires the $200 but it also only has a probability of 49.29% of being successful.Quote:... However, if you are trying to reach a particular target, a Martingale with as high of a starting bet as possible is probably the best approach, as you expose as little money as possible to the house edge per bet. ...

That is exactly the poperty of the listed approaches. But it is also true that the probability of being successful, i.e. not losing the high amount, increases together with the increased amounts and thus makes it more likely to achieve the goal.Quote:... The smaller the amount, the more steps it requires, which results in the total amount bet increasing, which results in the expected overall loss increasing as well. ...

This is an interesting approach. After the first $200 passline, the player alternates winning and losing $400 until either two consecutive passlines lose ($1000 lost) or two consecutive passlines win, achieving the goal of $1200. What technique do you use to calcualte the probabilities of winning and losing these sequences? Thanks in advance.Quote:... The best approach to win $200 with $1000 is probably 200-400-400 (then your next bet is 400 instead of 200, as you need 400 to reach your target of 1200).

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