If I bet 1$ 1000 times, what are the chances......

E

Eli M

Member
#1
#1
I know everyone says the martingale strategy won't work but can you please explain it to me if I do it this way?

I have a casino where the table minimum is 1$ and the max is 5000$.

I bet 1$ and double every loss. Now what are the chances that I lose 10 hands in a row before winning 1023 times?

If the number is below 50% then I'll have 1023$ bankroll sessions using 1$ bets and doubling losses until I have 2046$ in my bankroll. Once I get to 2046$ I cashout and restart at 1023$. It seems to me the chances of getting 10 loses in a row are lower than winning 1023 times.
 
M

Meistro

Well-Known Member
#2
#2
"Now what are the chances that I lose 10 hands in a row"

An approximate answer would be roughly .5 to the 10th power or .1%. Actually that's not quite accurate because your chance of winning any hand in particular is actually less than 50%. But it doesn't matter, the martingale system is not a long term winning strategy. So long as you are making negative expectation bets the casino is going to make money off you in the long run. What I recommend is your learn how to count cards, it's not really that difficult.

The martingale strategy will certainly create the illusion of being a winning player, because it is very easy to generate many small winning sessions using this betting system, but eventually you are going to experience catastrophic losses that outstrip your wins. Good luck.
 
Z

ZeeBabar

Banned
#3
#3
What happens when you bet $1 on your first hand, you get a pair of Aces, you split and lose both, You have lost $2 so next hand, you bet $4 and you get a 9/2 versus a 6, you double and lose the $8. Third hand, you have $16, you get a pair of 7's against a 5, you split, get another 7, split and get another 7 (four 7's), you get a 4 on one and you double, you get a 3 on another and you double again, you get an Ace on the third 7 and you double again and you get a 2 on the 4th 7 and you double. Now you have 4 hands, all doubled and the dealer has a 6 under, hits it for a 21 and you lose all hands and doubles. You have just lost $64 after 3 rounds.

Now, how much will you bet on the 4th round? $128? Can you see the problem?
 
E

Eli M

Member
#4
#4
Meistro said:
"Now what are the chances that I lose 10 hands in a row"

An approximate answer would be roughly .5 to the 10th power or .1%. Actually that's not quite accurate because your chance of winning any hand in particular is actually less than 50%. But it doesn't matter, the martingale system is not a long term winning strategy. So long as you are making negative expectation bets the casino is going to make money off you in the long run. What I recommend is your learn how to count cards, it's not really that difficult.

The martingale strategy will certainly create the illusion of being a winning player, because it is very easy to generate many small winning sessions using this betting system, but eventually you are going to experience catastrophic losses that outstrip your wins. Good luck.
Click to expand...
I don't think you're understanding me. The reason why I do 1023$ sessions and cashout at 2046$ is because at 1023$ I need to lose 10 hands to lose all my bankroll. There will be more sessions that I will reach 2046$ and cashout than losing 10 times in a row. So you're wrong in the long run I will be beating the casino.
 
M

Meistro

Well-Known Member
#5
#5
Eli M said:
I don't think you're understanding me.
Click to expand...
Look, I hate to break it to you, but your system isn't novel nor is it effective. Throw your money away if you want, but the only way to make money gambling is to make bets with a positive expectation. If the odds are not in your favour, then the more you play the more you expect to lose. And by making mostly tiny wagers you all but guarantee you will end up a loser, because of the law of large numbers.
 
E

Eli M

Member
#7
#7
Meistro said:
Look, I hate to break it to you, but your system isn't novel nor is it effective. Throw your money away if you want, but the only way to make money gambling is to make bets with a positive expectation. If the odds are not in your favour, then the more you play the more you expect to lose. And by making mostly tiny wagers you all but guarantee you will end up a loser, because of the law of large numbers.
Click to expand...
It's easy to say it's not going to work. I'm looking for mathematical proof but I'm not hearing any logic from you.
 
M

Meistro

Well-Known Member
#8
#8
Here is the mathematical proof.

Expectation = Action * Edge.

Assuming you play like a typical blackjack player you will have a disadvantage of 1.5%. So in the long run, you will expect to lose 1.5% of everything you bet. Varying your bet size won't change this except that the more you bet the more you stand to lose.

So if you bet $5000 you will have an expected loss of $5000 * 1.5% = $75.

There you have it, absolute mathematical proof that your system does not work. Now I believe we can all rest easy, knowing that no one will ever advance this theory again.
 
E

Eli M

Member
#9
#9
ZeeBabar said:
What happens when you bet $1 on your first hand, you get a pair of Aces, you split and lose both, You have lost $2 so next hand, you bet $4 and you get a 9/2 versus a 6, you double and lose the $8. Third hand, you have $16, you get a pair of 7's against a 5, you split, get another 7, split and get another 7 (four 7's), you get a 4 on one and you double, you get a 3 on another and you double again, you get an Ace on the third 7 and you double again and you get a 2 on the 4th 7 and you double. Now you have 4 hands, all doubled and the dealer has a 6 under, hits it for a 21 and you lose all hands and doubles. You have just lost $64 after 3 rounds.

Now, how much will you bet on the 4th round? $128? Can you see the problem?
Click to expand...
I didn't mention that you should forget about doubling and splitting. Only hitting and standing. The casino will have a bigger edge on me but not enough to cause 10 hands in a row losses very often.
 
E

Eli M

Member
#11
#11
Meistro said:
Here is the mathematical proof.

Expectation = Action * Edge.

Assuming you play like a typical blackjack player you will have a disadvantage of 1.5%. So in the long run, you will expect to lose 1.5% of everything you bet. Varying your bet size won't change this except that the more you bet the more you stand to lose.

So if you bet $5000 you will have an expected loss of $5000 * 1.5% = $75.

There you have it, absolute mathematical proof that your system does not work. Now I believe we can all rest easy, knowing that no one will ever advance this theory again.
Click to expand...
Do you read what I write or you just repeat the same thing over and over again. I'm not talking only about the martingale strategy, there's a reason why I do the 1023$ sessions. I know obviously that if i use only the martingale strategy I will end up by losing.
 
E

Eli M

Member
#13
#13
Meistro said:
I will admit I am a bit unclear about what you mean by the $1023 session thing. Could you please elaborate?
Click to expand...
Let's say you have 102,300$ and split it into 100 different bankrolls/sessions of 1023$ each. Each sessions goal is to reach 2046$ and cashout by betting 1$ and doubling you losses which means you would need a 10 in a row losing streak to lose a session. In some sessions you will have a losing streak of 10 in a row but in most sessions you won't. I don't have math to prove that in most sessions I won't hit a 10 losing streak but thinking logically it seems the probability of reaching 2046$ is higher than getting a 10 losing streak which means I would win most my sessions so guaranteed profit. If you have an argument say it but provide some math or logic, not it just won't work.
 
M

Meistro

Well-Known Member
#16
#16
The fibonacci gives you more opportunities to get your money back before busting. Instead of 1, 2, 4, 8 you go 1, 1, 2, 3, 5, 8, 13 so you take 6 jumps to get to 8 units instead of 4.
 
E

Eli M

Member
#17
#17
Meistro said:
The fibonacci gives you more opportunities to get your money back before busting. Instead of 1, 2, 4, 8 you go 1, 1, 2, 3, 5, 8, 13 so you take 6 jumps to get to 8 units instead of 4.
Click to expand...
How is that a strategy? By betting 1 $ twice and 3$ after that you are not making back the money you have lost already. What you're basically doing is losing your money at a slower pace.
 
M

Meistro

Well-Known Member
#18
#18
No with the fobonacci you always break even so long as you eventually get a win, but you are not always going to start off with a loss. So what you do is you bet $1, and if you win, great, you keep betting $1. Then once you lose you go into the fibonacci. So you make your profits from the times that you win right away. And it gives you more chances to increase without going bust. Way better than the martingale.
 
M

Meistro

Well-Known Member
#19
#19
Martingale is for amateurs, Fibonacchi is the pro increase your bet when you lose progression. Would you rather have 15 shots to get your money back before going bust or 10?
 
gronbog

gronbog

Well-Known Member
#20
#20
Assuming that a push does not interrupt a streak or a sequence, the ratio of wins to losses when playing basic strategy is about 47/53 (wins/losses). This does not change much if you choose not to double or split, although the house edge against you does increase significantly.

So the probability of losing 10 rounds without winning is about 0.53^10 or 0.00174 or 0.174% or about one in about 572 sequences. The probability of NOT starting a 10 loss sequence is therefore 1 - 0.00174 or 0.9983 or 99.83%. The probability of NOT seeing a 10 loss sequence within 1025 rounds is therefore 0.9983^1025 or 0.1662 or 16.62%. The probability that you WILL see such a sequence is then 1 - 0.1662 or about 83.37%.

Your plan is not new and is doomed to failure.
 
You must log in or register to reply here.
Top