Has anyone considered martingaling the make them all bet?
April 26th, 2018 at 7:30:12 PM
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I saw actual odds of making a make them all was roughly 190-1 wit payoffs of 175-1. With the low $1 bet with say a $700 bankroll you could cover a lot of rolls. Almost enough to say you could hit it twice if probabilities play out. Anyone ran a simulation on this with say a 2.5k bankroll?
$170 ($1 for 170 rolls)
$110 ($2 for 55 rolls) (225 rolls total)
$400 ($4 for 100 rolls) (325 rolls total)
$170 ($1 for 170 rolls)
$110 ($2 for 55 rolls) (225 rolls total)
$400 ($4 for 100 rolls) (325 rolls total)
April 26th, 2018 at 8:44:56 PM
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How much time do you want to spend on this? According to WoO, average rolls until a 7-out is 8.525510 (though I think A/T/S loses on a 7-comeout?)... at a brisk 15sec per roll, you're spending 6 hours on your first 170. Chop off a sixth for the 7-comeouts, and you're still looking at five hours.
On top of that, you're still dealing with massive variance. And, if you're you're also playing the main game, and not just a 1-4 bet every come out, dealing with the requirement of an even larger bankroll.
Seems to me you'd be better off doing the same kind of thing on roulette, where the HE is smaller and resolutions faster.
On top of that, you're still dealing with massive variance. And, if you're you're also playing the main game, and not just a 1-4 bet every come out, dealing with the requirement of an even larger bankroll.
Seems to me you'd be better off doing the same kind of thing on roulette, where the HE is smaller and resolutions faster.
April 27th, 2018 at 3:33:15 AM
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Your examples don't suggest to me Martingaling. Do you mean that if you would lose a dollar, next time two dollars is bet, then four dollars, etc? If so you don't get 170 rolls for $170 etc.
You could 'hit it twice' if you get lucky, you wouldn't want 'the probabilities to play out'
You could 'hit it twice' if you get lucky, you wouldn't want 'the probabilities to play out'
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
April 27th, 2018 at 6:01:38 AM
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Quote: odiousgambitYour examples don't suggest to me Martingaling. Do you mean that if you would lose a dollar, next time two dollars is bet, then four dollars, etc? If so you don't get 170 rolls for $170 etc.
You could 'hit it twice' if you get lucky, you wouldn't want 'the probabilities to play out'
That kind of reminds me of a Fresh Prince Of Bel-Air episode where Hillary says,"I bet on all of the horses running in the race." Will says"Hillary, one of the horses has to win!" Hillary says"That's the point!" I wondered if Hillary had a good point by playing all of the horses. I mean I myself play all 17 numbers in the Lucky Money game knowing I'll win something. Hillary would win sonething by playing all the horses." On a website someone thought along the lines of what I was thinking and said"Hillary is being smart by playing all of the horses." The reply was,"Not really. Will is saying only one horse will win meaning playing all the horses is a stupid idea. Let's say Hillary puts $50 on all 10 horses running. Lucky Runner wins the race, but since she bet on each horse she still loses more than she won. She would have been better off just picking only one horse and hope Lucky Runner was the one she picked."
In both The Hunger Games and in gambling, may the odds be ever in your favor. :D
"Man Babes" #AxelFabulous
"Olive oil is processed but it only has one
ingredient, olive oil."-Even Bob, March 27/28th. :D The 2 year war is over! Woo-hoo! :D
I sometimes speak in metaphors. ;) Remember this. ;)
Crack the code. :D 8.9.13.25.14.1.13.5.9.19.14.1.20.8.1.14! :D
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April 27th, 2018 at 6:07:41 AM
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Quote: NathanThat kind of reminds me of a Fresh Prince Of Bel-Air episode where Hillary says,"I bet on all of the horses running in the race." Will says"Hillary, one of the horses has to win!" Hillary says"That's the point!" I wondered if Hillary had a good point by playing all of the horses. I mean I myself play all 17 numbers in the Lucky Money game knowing I'll win something. Hillary would win sonething by playing all the horses.
It would be easier to just skip betting on the race, and just pay the book their vig for nothing.
Hedging bets is never a good idea, unless there is another component to your play that makes it +EV.
April 27th, 2018 at 6:10:16 AM
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Quote: gamerfreakIt would be easier to just skip betting on the race, and just pay the book their vig for nothing.
Hedging bets is never a good idea, unless there is another component to your play that makes it +EV.
I have edited my comment to add the other people's comments. ;)
In both The Hunger Games and in gambling, may the odds be ever in your favor. :D
"Man Babes" #AxelFabulous
"Olive oil is processed but it only has one
ingredient, olive oil."-Even Bob, March 27/28th. :D The 2 year war is over! Woo-hoo! :D
I sometimes speak in metaphors. ;) Remember this. ;)
Crack the code. :D 8.9.13.25.14.1.13.5.9.19.14.1.20.8.1.14! :D
"For about the 4096th time, let me offer a radical idea to those of you who don't like Nathan -- block her and don't visit Nathan's Corner. What is so complicated about it?" Wizard, August 21st. :D
April 28th, 2018 at 4:28:20 PM
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Although I've seen this bet won twice (never by me-I'd never touch it) I've always been curious about how they calculate the odds. A simple factorial doesn't account for the same number being rolled more than once, which will almost always happen.
The best way to achieve victory on that bet is not to make it and don't think about using a double up system (or any system) with it.
The best way to achieve victory on that bet is not to make it and don't think about using a double up system (or any system) with it.
"I should have bet black." - Winston Churchill .
April 28th, 2018 at 10:05:09 PM
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I have seen it twice too in Las Vegas. Shooter had $10 on it at Binions. I was too late to the party!Quote: Lucca3927Although I've seen this bet won twice (never by me-I'd never touch it)
many ways actually. there are a few threads at WoV showing how.Quote: Lucca3927I've always been curious about how they calculate the odds.
some use a Markov chain, some list all the permutations and calculate from there, others use inclusion-exclusion
for both ways to calculate this (long way is all #s in a subset B4 the 7 or the short way a 7 B4 the subset of #s)
that using a computer is fast and simple. here from BruceZ
> ##################################################################
> # Probability of rolling a subset of numbers before a single number
> ##################################################################
> start_time <- Sys.time()
> options(scipen=999)
>
> numbers = c(2,3,4,5,6,8,9,10,11,12,7) # Last must occur only after all others in any order
>
> in_36 = ifelse(numbers <= 7, numbers-1, 13-numbers) # Ways to make each number
> i = length(in_36)
> p = 0
> for (j in 1:(i-1)) { # Last number before j numbers
+ terms = combn(in_36[1:(i-1)],j) # Matrix w/combos of j numbers in C(i-1,j) columns
+ for (k in 1:ncol(terms)) { # Sum each column, compute and add probabilities
+ p = p + (-1)^(j+1) * in_36/(in_36 + sum(terms[1:j,k]))
+ }
+ }
> end_time <- Sys.time()
> time <- end_time - start_time
> time
Time difference of 0.1562321 secs
> p=1-p
> p
[1] 0.005257704
> a=1/p
> a # 1 in
[1] 190.1971
> a-1 # odds against
[1] 189.1971
> ##################################################################
> # Probability of rolling a subset of numbers before a single number
> ##################################################################
> start_time <- Sys.time()
> options(scipen=999)
>
> numbers = c(2,3,4,5,6,7) # Last must occur only after all others in any order
>
> in_36 = ifelse(numbers <= 7, numbers-1, 13-numbers) # Ways to make each number
> i = length(in_36)
> p = 0
> for (j in 1:(i-1)) { # Last number before j numbers
+ terms = combn(in_36[1:(i-1)],j) # Matrix w/combos of j numbers in C(i-1,j) columns
+ for (k in 1:ncol(terms)) { # Sum each column, compute and add probabilities
+ p = p + (-1)^(j+1) * in_36/(in_36 + sum(terms[1:j,k]))
+ }
+ }
> end_time <- Sys.time()
> time <- end_time - start_time
> time
Time difference of 0.1718872 secs
> p=1-p
> p
[1] 0.02635391
> a=1/p
> a # 1 in
[1] 37.94503
> a-1 # odds against
[1] 36.94503
The question also goes back to the 1960s and it was done then too.
i agreeQuote: Lucca3927A simple factorial doesn't account for the same number being rolled more than once, which will almost always happen.
here sample 100 attempts with the rolls
> bonusCraps
2 3 4 5 6 7 8 9 10 11 12
[1,] 0 0 0 0 1 1 0 0 1 0 0
[2,] 0 0 0 0 0 1 0 0 0 0 0
[3,] 1 0 0 2 0 1 1 4 1 2 0
[4,] 0 1 0 1 1 1 1 1 0 0 0
[5,] 0 0 0 0 2 1 2 0 1 0 0
[6,] 0 0 0 0 0 1 0 0 0 0 0
[7,] 0 1 1 1 1 1 1 2 0 0 1
[8,] 0 0 0 0 0 1 0 0 0 0 0
[9,] 0 0 1 2 0 1 0 0 2 0 0
[10,] 0 0 0 0 1 1 2 0 0 0 0
[11,] 0 0 0 0 0 1 0 0 0 0 0
[12,] 0 0 0 4 0 1 1 1 0 0 0
[13,] 2 0 0 0 3 1 1 1 1 0 0
[14,] 0 0 2 2 4 1 2 1 4 0 0
[15,] 0 0 0 0 0 1 0 1 0 0 0
[16,] 0 0 0 1 1 1 0 0 0 0 0
[17,] 0 0 1 1 0 1 0 0 0 1 0
[18,] 0 2 2 4 6 1 6 2 1 1 1
[19,] 1 0 2 1 0 1 2 0 0 1 0
[20,] 0 0 0 0 1 1 0 0 1 0 0
[21,] 0 0 0 0 0 1 1 0 0 0 0
[22,] 0 1 3 1 3 1 2 2 1 3 0
[23,] 0 1 1 1 5 1 3 0 0 1 1
[24,] 1 1 0 0 0 1 1 0 0 1 0
[25,] 1 0 0 0 0 1 1 0 0 0 0
[26,] 0 1 2 5 3 1 4 2 2 0 0
[27,] 2 2 1 1 4 1 3 1 0 0 0
[28,] 0 0 1 0 3 1 0 0 1 0 0
[29,] 0 1 1 0 1 1 2 1 0 1 1
[30,] 0 0 0 0 0 1 0 0 0 0 0
[31,] 0 0 0 1 0 1 0 0 0 0 0
[32,] 1 0 1 1 4 1 1 3 1 0 0
[33,] 0 1 1 0 1 1 0 1 0 0 0
[34,] 0 0 0 0 1 1 0 0 1 0 0
[35,] 0 0 0 0 0 1 1 0 0 0 0
[36,] 0 1 0 0 4 1 1 2 0 0 1
[37,] 1 0 0 1 0 1 0 0 0 0 0
[38,] 0 0 0 0 0 1 0 0 0 0 0
[39,] 0 0 1 3 2 1 2 1 0 1 0
[40,] 0 0 1 0 1 1 1 0 0 1 0
[41,] 0 2 0 0 0 1 0 0 0 0 0
[42,] 0 0 0 0 0 1 0 0 0 0 0
[43,] 1 2 0 0 2 1 1 2 0 1 0
[44,] 0 0 0 1 0 1 0 0 0 0 0
[45,] 0 0 0 1 0 1 1 0 0 0 0
[46,] 0 1 0 0 0 1 2 0 0 0 0
[47,] 1 0 1 0 0 1 0 1 0 0 1
[48,] 0 1 2 2 2 1 6 1 1 1 1
[49,] 0 0 0 0 0 1 0 0 0 1 1
[50,] 1 0 1 1 0 1 0 0 0 0 0
[51,] 0 0 1 1 2 1 3 0 0 1 1
[52,] 1 1 1 0 4 1 2 0 0 0 0
[53,] 0 0 0 1 0 1 0 0 0 0 1
[54,] 0 0 2 0 0 1 1 0 1 1 0
[55,] 0 0 0 1 1 1 2 1 1 1 0
[56,] 0 0 0 1 0 1 1 1 1 0 0
[57,] 0 0 0 0 0 1 0 1 0 0 0
[58,] 0 0 1 0 0 1 0 0 0 0 0
[59,] 0 0 0 1 0 1 0 0 0 0 0
[60,] 0 0 0 1 2 1 0 0 0 1 0
[61,] 1 0 0 0 1 1 1 0 0 1 0
[62,] 0 1 1 0 1 1 1 1 0 0 0
[63,] 1 0 2 2 1 1 1 5 2 0 1
[64,] 0 0 0 0 0 1 0 1 2 0 0
[65,] 0 0 0 0 0 1 0 1 0 0 0
[66,] 0 0 0 0 0 1 0 0 1 0 0
[67,] 0 1 1 1 0 1 2 1 0 1 1
[68,] 0 0 0 0 0 1 0 1 0 0 0
[69,] 0 0 0 0 1 1 0 0 0 0 0
[70,] 0 0 0 0 0 1 0 0 1 0 0
[71,] 0 2 0 1 1 1 3 2 1 0 0
[72,] 0 0 0 0 0 1 0 1 0 1 0
[73,] 0 1 2 1 7 1 0 2 1 2 1
[74,] 1 0 3 2 3 1 3 3 1 1 1
[75,] 0 0 0 0 0 1 0 0 0 0 0
[76,] 0 0 2 0 1 1 1 0 0 0 0
[77,] 0 0 0 0 2 1 0 0 1 0 0
[78,] 0 0 0 0 0 1 0 0 0 0 0
[79,] 0 0 0 0 1 1 0 0 0 0 1
[80,] 0 0 1 0 2 1 3 3 0 1 0
[81,] 0 2 1 2 0 1 1 0 0 0 0
[82,] 0 0 0 0 0 1 0 0 0 0 0
[83,] 0 0 0 0 0 1 0 0 0 0 0
[84,] 0 0 1 1 0 1 0 0 0 0 0
[85,] 2 1 0 1 1 1 2 1 1 0 0
[86,] 0 1 0 0 0 1 1 0 0 0 0
[87,] 0 0 0 0 0 1 0 0 0 0 0
[88,] 0 0 0 0 1 1 0 0 0 0 0
[89,] 0 0 2 0 0 1 0 1 0 1 0
[90,] 0 0 0 1 1 1 0 0 0 0 1
[91,] 0 0 0 0 1 1 0 0 0 0 1
[92,] 1 0 0 1 1 1 0 1 1 0 0
[93,] 0 0 0 0 0 1 0 0 0 0 0
[94,] 1 0 0 1 3 1 1 1 0 0 1
[95,] 0 1 0 0 0 1 0 0 0 0 0
[96,] 0 0 0 0 0 1 0 0 0 0 0
[97,] 0 2 0 0 0 1 1 1 1 1 0
[98,] 0 0 0 0 1 1 0 0 0 0 0
[99,] 0 0 0 0 1 1 1 0 0 0 0
[100,] 0 0 0 0 0 1 0 0 0 0 0
>
that takes the fun away from playing it!Quote: Lucca3927The best way to achieve victory on that bet is not to make it and don't think about using a double up system (or any system) with it.
There is a casino 3 miles from my house that has it.
time for some fun!
Sally
I Heart Vi Hart
April 29th, 2018 at 3:36:53 AM
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Quote: mustangsally
that takes the fun away from playing it!
There is a casino 3 miles from my house that has it.
time for some fun!
Sally
The fun is best when you WIN.
You are lucky with that three miles.
April 29th, 2018 at 4:11:31 AM
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I Agree. The FUN part is collecting the winnings.Quote: FleaStiffThe fun is best when you WIN.
You are lucky with that three miles.
April 29th, 2018 at 9:13:39 AM
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Quote: odiousgambitYour examples don't suggest to me Martingaling. Do you mean that if you would lose a dollar, next time two dollars is bet, then four dollars, etc? If so you don't get 170 rolls for $170 etc.
You could 'hit it twice' if you get lucky, you wouldn't want 'the probabilities to play out'
Sounds like a pseudo-Martingale - bet 1 until the total amount lost is less than what would be won on the next bet of 1, then bet 2 until the losses again exceed the value of a win, then (presumably, in this case) bet 3, and so on.
May 3rd, 2018 at 10:03:31 AM
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Quote: ThatDonGuySounds like a pseudo-Martingale - bet 1 until the total amount lost is less than what would be won on the next bet of 1, then bet 2 until the losses again exceed the value of a win, then (presumably, in this case) bet 3, and so on.
Yes. Not sure it would work or not. The amount of time at the table to cycle through is the issue i see but because of the numner of bets made before an increase i think it is the most likely to ne jit of a progressional/martingale type system to hit.
May 3rd, 2018 at 11:55:19 AM
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Last time I was next to an e-roulette game (brand escapes me), I noticed that it was reporting that one number hadn't come up in the last 250 spins. If I did my math right, this strategy, if you just happened to have the awful luck to catch that number, would've cost you a little over 8500.
