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Mateusz

All-In against Martingale

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Ok guy's, I'll finally explain to you somehow why all in is always better than playing martingale.

I'll come with an example, we have a 100$ and go to the casino. Our plan is to make 200$, and we choose roulette with 18 reds/blacks and 1 green, and we want to use 1 of those 2 ways:

1. All in red/black and hope for the best (0.4864864864... - those are the chances)

2. We choose 0.25$ and we double after loss, reset to base after win (but what are the chances of that?)

So I asked my friend if he could make me a simulation, and he did, a simulation where 10000 people comes to play that strategy and we have the results.

here's the site: http://www.bogus.tk/stawianie/ (it's in poland but you should see how it work anyway)

The first column shows how the game of the first person looked.

The second column gives us the results of the 10000 games, and you can refresh many times but it will always be around 40% winnings 60% loses

That means the strategy gives u only 40% chance to get another 100$, which is REALLY small compared to the 48.6%

I still don't know how that works out but the amount of the winning bets needed + house edge causes the chances, cuz when I changed started pot to 500$, and the goal 1000$, chances went to 37%, still I'd love to know the exact mathematical formula.

It will probably look similar for other payouts than 2x, it's all about house edge so next time you use this strategy, think of it twice.

 

 

Small 10k giveaway (on primedice) = theoretically if u follow this martingale strategy and bet 0.25$, increase by 2 after every loss and return to base after win, you can NEVER get 200$ and NEVER lose all money, how is that possible? If u can explain me that I'll give you 10k on PD

 

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if you gonna stop after 8 loss in a row you will have some money to leave because you cant handle 9th roll with your left bankroll so you wont loose all your money

i assumed you didnt win any till that

you have 100 which will equal 400 chips as 0.25  

1+2+ 4+ 8 +16 +32+ 64 +128=255 your loss 145 your bankroll at the end and you going home with 38.25 

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1 minute ago, Mateusz said:

You don't stop, you increase per lose and back to base by win, and you never win, but you are close to that.

You don't go home untill you loss all/win 200$

problem is you cant increase at some point so what you gonna do returning base bet at this point?

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3 minutes ago, Mateusz said:

You just go all in what's left

if we going all in if we cant increase how come we cant loose all our money ? what will happen that all in showed as loss also?

i think i dont get what you asking

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10 minutes ago, Mateusz said:

Our plan is to make 200$, and we choose roulette with 18 reds/blacks and 1 green, and we want to use 1 of those 2 ways:

1. All in red/black and hope for the best (0.4864864864... - those are the chances)

2. We choose 0.25$ and we double after loss, reset to base after win (but what are the chances of that?)

So I asked my friend if he could make me a simulation, and he did, a simulation where 10000 people comes to play that strategy and we have the results.

Excellent experiment. To notice:

- Starting with 100 dollars and 0.25 bets, you can fail 8 times in a row [(1-0.4864)^8], the chance of it happening from the beggining is 0.0048%, but it increase the more you play. Easy mode: Take a coin and try to take "Tails" 8 times in a row: chances are 0.0039%. But a streak of 8 Tails in 1000 bets chance is the stunning 86%

- House edge is not about value, but times played. 1% house edge means you are losing 1% per bet. The more TIMES you play, the more you lose.

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22 hours ago, Xylber said:

 House edge is not about value, but times played. 1% house edge means you are losing 1% per bet. The more TIMES you play, the more you lose.

 

Wow, didn't know this fact. Thank you for the explanation. By the way, as per your theory if we play 100 bets then the chances of losing 100% right..??

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20 minutes ago, akki785 said:

Wow, didn't know this fact. Thank you for the explanation. By the way, as per your theory if we play 100 bets then the chances of losing 100% right..??

dont know what you mean, but by 2 experiments I think that the chances in martingale goes down and down by the amount of wins u need to double starting capital

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22 minutes ago, akki785 said:

Wow, didn't know this fact. Thank you for the explanation. By the way, as per your theory if we play 100 bets then the chances of losing 100% right..??

Depends on what exactly you are asking. You do not lose 100% of the value you have when you start, but surely 100% of the value you should be winning without house edge:

A) Percentages do not sum up, but multiply. Heads/Tails has a 50% chance, 2 Heads in a row is not 100%, but 25% (50%*50%). So no, your bank account is not "0" after 100 bets.

BUT

B) Imagine a game like roulette with 100 numbers, 1% house edge. If you bet 1 coin at 1 number, if you win, it should pay "100 coins" (fair), but it pays "99 coins" (house edge). Then you bet 100 times, and you win those 100 times, it means you end up having 100bets*99payment= 9900 Coins (house edge 1%), instead of 10000 (fair), so, you have 100 coins LESS. Did you see you lose "100%" of your initial value, even if you won them all?

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1 minute ago, Xylber said:

Depends on what exactly you are asking. You do not lose 100% of the value you have when you start, but surely 100% of the value you should be winning without house edge:

A) Percentages do not sum up, but multiply. Heads/Tails has a 50% chance, 2 Heads in a row is not 100%, but 25% (50%*50%). So no, your bank account is not "0" after 100 bets.

BUT

B) Imagine a game like roulette with 100 numbers, 1% house edge. If you bet 1 coin at 1 number, if you win, it should pay "100 coins" (fair), but it pays "99 coins" (house edge). Then you bet 100 times, and you win those 100 times, it means you end up having 100bets*99payment= 9900 Coins (house edge 1%), instead of 10000 (fair), so, you have 100 coins LESS. Did you see you lose "100%" of your initial value, even if you won them all?

Exactly, let's say you win 100 games in this way so you won 9900coins, because of house edge, and then you lose 100 games, and you are on -100 comparing to starting capital.

That works the same way in martingale, that's why chances to get your capital doubled goes down by the amount of wins you need to double it, cuz you do more and more bets with house edge

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9 minutes ago, Xylber said:

Depends on what exactly you are asking. You do not lose 100% of the value you have when you start, but surely 100% of the value you should be winning without house edge:

A) Percentages do not sum up, but multiply. Heads/Tails has a 50% chance, 2 Heads in a row is not 100%, but 25% (50%*50%). So no, your bank account is not "0" after 100 bets.

BUT

B) Imagine a game like roulette with 100 numbers, 1% house edge. If you bet 1 coin at 1 number, if you win, it should pay "100 coins" (fair), but it pays "99 coins" (house edge). Then you bet 100 times, and you win those 100 times, it means you end up having 100bets*99payment= 9900 Coins (house edge 1%), instead of 10000 (fair), so, you have 100 coins LESS. Did you see you lose "100%" of your initial value, even if you won them all?

Good explanation..

Clearly understood..

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On 1/19/2018 at 11:17 AM, Xylber said:

House edge is not about value, but times played. 1% house edge means you are losing 1% per bet. The more TIMES you play, the more you lose.

 

Well, it IS about value.

More specifically, it's about the wagered amount. You can expect a loss of (house edge)% * wagered amount when you are betting.

A house edge of 1% will mean that you statistically lose 1% of your total wagered amount. Regardless of whether you are betting it all at once or whether you are betting it in smaller increments.

The only issue with martingale is that you will still approach 0 eventually and because some people find sanctity in the strategy, they may wager large amounts.

Additionally, there is huge variance in the martingale strategy. Wagering a fixed amount means that you usually approach the .99 * wagered amount value and are centered around that with little deviations. However, when you are playing the martingale strategy there are wild fluctuations and the swings can potentially end up with a player losing their entire bankroll.

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3 hours ago, actmyname said:

Well, it IS about value.

More specifically, it's about the wagered amount. You can expect a loss of (house edge)% * wagered amount when you are betting.

A house edge of 1% will mean that you statistically lose 1% of your total wagered amount. Regardless of whether you are betting it all at once or whether you are betting it in smaller increments.

The only issue with martingale is that you will still approach 0 eventually and because some people find sanctity in the strategy, they may wager large amounts.

Additionally, there is huge variance in the martingale strategy. Wagering a fixed amount means that you usually approach the .99 * wagered amount value and are centered around that with little deviations. However, when you are playing the martingale strategy there are wild fluctuations and the swings can potentially end up with a player losing their entire bankroll.

Well, I was expecting a counterargument like that. And of course, it is the way you say it. The reason I stated "The more TIMES you play, the more you lose" is because this thread is called "Martingale": the more times you bet on a Martingale, there is more chances of getting a losing streak. If your bankroll is enough, and the times you play is low, you have an extremely high chance of winning.

I will take the chance to add more info to the thread:

The chances of losing a coin toss (50% win/50% lose) 2 times in a row in 2 rounds is 25%:
OX
XO
XX
OO

The chances of losing a coin toss (50% win/50% lose) 2 times in a row in 3 rounds is 37.5%:
OOO
OOX
OXO
XOO
OXX
XOX
XXO
XXX

The chances of losing a coin toss (50% win/50% lose) 2 times in a row in 4 rounds is 50%:
OOOX OOXX XOXO XXOX
OOXO OXOX XXOO XXXO
OXOO XOOX OXXX XXXX
XOOO OXXO XOXX OOOO

And when you bet with Martingale, if you want to get that .99 variation calculated, you must consider each "double bet" as a separated universe: You will end up having all "1" bet in one table, each "2" in other, "4", "8", etc.

Edited by Xylber

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Does the site have a chart posted anywhere for the games probabilities? Or is this all data from players personal testing? I would liketo get more educated on how the house edge work, and also having an idea on the % of greens per bet and whatnot.

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