Blackjack strategy: Flawed, or flawless?

Okay, I know that there is supposedly no strategy to beat casino blackjack short of card counting.

So I am certain this strategy has a flaw, but I can't think of what that flaw may be, so tell me the error in this.

Say that I want to win $100,000 at the casino blackjack table, after only starting with $100.

So bet it all each time. If you win 10 in a row, you can walk away with $102,400.

Here's my reasoning, You have a 49% chance of winning a hand, where the house edge is 51%

Ergo, you have a mere 4.9% chance of winning ten hands in a row.

Rounded up. you should expect to win 10 hands in a row approximately once every 20 tries.

So I go into a casino tonight with $100, and bet it all each win. 95% chance I leave the building down $100 bucks.

But then I return tomarrow. And the next day, and the next day, until I win 10 in a row, then I cash out.

I'll then have, on average, spent about $2000 to get at that $102.400 win.

Sure, I might have "lost" a lot more theoretically when I bet it all after a 2-3 hand win streak and THEN lose, but at the end of the day, my bank statement will only show that I'm down a $100 dollar ATM withdrawal for the day.

Is there an error in my math?

Of course this requires a casino with a high roller room, where table maximums go beyond $50k, but these do exist.

I don`t know the exact number,but the odds of winning ten hands in a row are much higher than once every 20 times.

Quote: Clayfighter

You have a 49% chance of winning a hand, where the house edge is 51%

Is there an error in my math?

Yes. Your odds of winning a hand is just over 42%, not 49%

You're playing a game with a negative expectation. Nothing you do with a betting strategy, short of counting and knowing when YOU have the advantage, will work in the long run.

Quote: 21forme

Yes. Your odds of winning a hand is just over 42%, not 49%

You're playing a game with a negative expectation. Nothing you do with a betting strategy, short of counting and knowing when YOU have the advantage, will work in the long run.

Yeah he has a far more severe problem then that. If you use his math with a 42% chance that would give him 4.2% chance so say 1/25 so losing 2400 to win 100k.

The problem is you don't just divide by 10 you have to take the number to the 10 power. 49% is a fine number to use since it is close to the number won if you exclude ties. This gives us .49^10 so .079% chance or roughly 1/1253. So 1253 attempts to win 1024X suddenly seems like a far worse offer.

Yeah, I was going to say, re: the math.

winning 1 in a row coin flip: 50% chance.
winning 2 : 25% chance
winning 3: 12.5% chance
etc. to winning 10 in a row.

There is a huge error in this logic. You are not going to turn a negative EV game into a positive one by progressive betting. Please remember there is a table maximum and you only need to lose a certain amount of times before you will need to bet above the table limit to have a chance of getting your money back. You may win for a while but eventually you will hit a losing streak that bankrupts you.

Quote: davethebuilder

You are not going to turn a negative EV game into a positive one by progressive betting. . .. You may win for a while but eventually you will hit a losing streak that bankrupts you.

I totally agree. And the OP has his maths completely wrong in any case.

Proof that Progressive betting doesn't work:-
https://wizardofvegas.com/member/oncedear/blog/#post1330

$:o)

Ah!

https://wizardofvegas.com/forum/gambling/betting-systems/21359-debunking-roulette-marty-with-pictures/

Quote: Clayfighter

Rounded up. you should expect to win 10 hands in a row approximately once every 20 tries.

Is there an error in my math?

That's about as wrong as it gets.

Quote: Clayfighter

Here's my reasoning, You have a 49% chance of winning a hand, where the house edge is 51%

Ergo, you have a mere 4.9% chance of winning ten hands in a row.

Is there an error in my math?.

Yes. A few errors. The less important error is that you win well less than 49% of your blackjack hands.
The more important error is that since each hand is an independent event, you need to multiply each likelihood to figure out the odds of a streak of any given length. So to win 10 events in a row, even if you did have a 49% of winning each event, is .49 to the 10th power. I think that is around 1 in 5000

You may win $100,000 if you have the luck,the time, the patience,the money and find a table game that has $100 minimum to $51,200 maximum. It may take you few days to get the 10 in a row streak of wining the $102400, or at 102,000 th day and make $ 0.
Am I right?

I would recommend switching tables after each win. Why stop at 10? I would win 1 hand at 30 different tables.

Also why come back the next day. If you lose your 100, then pull out another 100 and try again. Do thst until you get a streak of 10.

Quote: GWAE

I would recommend switching tables after each win. Why stop at 10? I would win 1 hand at 30 different tables.

Also why come back the next day. If you lose your 100, then pull out another 100 and try again. Do thst until you get a streak of 10.

Don't be cruel to the newbie :o)
He just made a few mistakes. So long as he takes the sincere advice on board, we don't need to wish financial suicide on him, not even as an in joke.

So everyone has pretty much chimed in telling you that your math is wrong. Here you go:

One hand of blackjack... Win Percentage = 42%, Lose Percentage = 49%, Push Percentage = 9%.

Odds of winning 10 hands in a row = .42^10 = .000407, which is about 1 in 2500...

Thus, when you lose $100 2,499 times that'll cost you $249,900. Not quite worth making the ~$100k that 1 time in 2500.

Quote: Romes

So everyone has pretty much chimed in telling you that your math is wrong. Here you go:

One hand of blackjack... Win Percentage = 42%, Lose Percentage = 49%, Push Percentage = 9%.

Odds of winning 10 hands in a row = .42^10 = .000407, which is about 1 in 2500...

Thus, when you lose $100 2,499 times that'll cost you $249,900. Not quite worth making the ~$100k that 1 time in 2500.

Hi Romes,

Correct me if I'm wrong, but I think you short changed him with the pushes. Also you calculated to ^9 and not ^10. You probably short paid him on BlackJacks too where his win might make the 10 win objective moot.

Odds of winning 10 times in a row without encountering a push are as you said. But a push isn't a lose. If he want's to double on each win ( Let his bet ride) then he'll need 10 wins before encountering a lose which may, of course be more than 10 wagers in total after he has the occasional push.

Therefore for resolved bets excluding and ignoring pushes, Pwin= 42/(42+49) = 46.16% Plose=49/(42+49)=53.84%
Pwin10beforeLose=.4616^10=0.000473 or 1 in 2116 . . . . or very roughly about 1 in 20 (LoL)

Oh, and he will be well stuffed if he has 9 wins in a row and is dealt a pair of aces on the next hand. That would be just my luck.

Quote: OnceDear

Hi Romes,

Correct me if I'm wrong, but I think you short changed him with the pushes. Probably short changed him on BlackJacks too where his win might make the 10 win objective moot.

Odds of winning 10 times in a row without encountering a push are as you said. But a push isn't a lose. If he want's to double on each win ( Let his bet ride) then he'll need 10 wins before encountering a lose which may, of course be more than 10 wagers in total after he has the occasional push.

Therefore for resolved bets excluding pushes, Pwin= 42/(42+49) = 46.16% Plose=49/(42+49)=53.84%
Pwin10beforeLose=.4616^10=0.000473 or 1 in 2116 . . . . or very roughly about 1 in 20 (LoL)

Oh, and he will be well stuffed if he has 9 wins in a row and is dealt a pair of aces on the next hand. That would be just my luck.

Hey Once,
My assumption was 10 wins in a row, regardless of win amount. Thus, if he wins $100k on 9 wins in a row (due to blackjacks) it's erroneous and he'll play another hand. If you ignore ties then yes it would change to approximately 46.x^10.

I edited as you were replying.
How upset he would be if he had 9 winning blackjacks in a row and insisted on going all in on that tenth wager. . . and then being dealt two aces $:o)
I would bet money that if that occurred he could look around the table and shake hands with Alan.

Quote: Clayfighter

Okay,
Here's my reasoning, You have a 49% chance of winning a hand, where the house edge is 51%
Ergo, you have a mere 4.9% chance of winning ten hands in a row.
Rounded up. you should expect to win 10 hands in a row approximately once every 20 tries.
Is there an error in my math?

As others have pointed your math is completely wrong.
You confuse multiplication with exponenentiation.
10 consecutive time of an event with 49% prob is NOT 49% x 1/10 to make it 1 in 20 times
BUT 49%^10 (that's to the 10th power)which makes it around 1 to 1.250 times.

And an easy guide of how to quickly estimate a consecutive win prob for any win/Lose game.
For a fair game (ie 50:50 game) these are around
10 in a row is 1 in 1.000
20 in a row is 1 in 1.000.000
30 in a row is 1 in 1.000.000.0000
Basically these are binary calculations. 10 consecutive means 2^10 = 1024 (use 1.000 for aproximation)
So every 10 additional consecutives you add 3 zeros (in decimal)

And casino games are far from fair, so the numbers are higher than these.
So a 20 consecutive event in a casino is a rare even (1 a million event)
A 30 consecuitive event is a very rare event (in a billion event)
And a 40 consecutive event does not happen (in a trillion event)

Quote: AceTwo

As others have pointed your math is completely wrong.
You confuse multiplication with exponenentiation.
10 consecutive time of an event with 49% prob is NOT 49% x 1/10 to make it 1 in 20 times
BUT 49%^10 (that's to the 10th power)which makes it around 1 to 1.250 times.

And an easy guide of how to quickly estimate a consecutive win prob for any win/Lose game.
For a fair game (ie 50:50 game) these are around
10 in a row is 1 in 1.000
20 in a row is 1 in 1.000.000
30 in a row is 1 in 1.000.000.0000
Basically these are binary calculations. 10 consecutive means 2^10 = 1024 (use 1.000 for aproximation)
So every 10 additional consecutives you add 3 zeros (in decimal)

And casino games are far from fair, so the numbers are higher than these.
So a 20 consecutive event in a casino is a rare even (1 a million event)
A 30 consecuitive event is a very rare event (in a billion event)
And a 40 consecutive event does not happen (in a trillion event)

I'm in over my head here, pls forgive me.
Using the round numbers provided above, close snuff for discussion.
Looks like 20 in a row is one in a million, did I get that right?
(Pls God, let's not get into pushes at this point)
I think I can wrap my head around a number as large as a million, a big number.
So somewhere in a million whatevers a rare event should happen.
If we were rolling dice (no let's not roll dice).
If we were tossing coins ( that's better), this rare event would happen, on average, once every million toss's (sp).
I'm thinking that's not the best way to look at it.
The odds of this event happening in the first 20 toss, are about the same as the odds of this event happening in the last 20 toss out of a million, yes? (1 thru 20 vs 999,980 thru 1,000,00).
Maybe a better way to look at it is to up the numbers.
Let's say the odds of this event happening 10 times, in ten million toss, is the same. Is that OK?
Or 100 times in one hundred million toss, does that still work?
Now I can work with this data, every million toss or so this event will happen, cool. Rare event.
What I do not understand, and pls help me here;

How come I gotta wait so long for my rare event? I want it now ;-)

Quote: TwoFeathersATL

Looks like 20 in a row is one in a million, did I get that right?

Yes. 20 in a row on a 50:50 game (say coin toss) is one in million event. Exact number is 2^20=1.048.576

Quote: TwoFeathersATL

If we were tossing coins ( that's better), this rare event would happen, on average, once every million toss's (sp).
I'm thinking that's not the best way to look at it.
The odds of this event happening in the first 20 toss, are about the same as the odds of this event happening in the last 20 toss out of a million, yes? (1 thru 20 vs 999,980 thru 1,000,00).

No. If you toss the coin 20 times and then stop and consider this 1 event and if all are Heads is a Win. Then toss the coin 20 times and consider this the 2nd event. And do this event many times, on average you will Win 1 in every million events.
It is a different question what is the probability that if you observe 1 million trials, there will be at least a 20 consecutive streak.
The probability of that is around 91%.

Quote: TwoFeathersATL

What I do not understand, and pls help me here;
How come I gotta wait so long for my rare event? I want it now ;-)

You might get lucky and it happens immediately !!!

Quote: AceTwo

Yes. 20 in a row on a 50:50 game (say coin toss) is one in million event. Exact number is 2^20=1.048.576


No. If you toss the coin 20 times and then stop and consider this 1 event and if all are Heads is a Win. Then toss the coin 20 times and consider this the 2nd event. And do this event many times, on average you will Win 1 in every million events.

It is a different question what is the probability that if you observe 1 million trials, there will be at least a 20 consecutive streak.
The probability of that is around 91%.

I'll round numbers a bit.

Methinks you are a bit wrong Ace2

Roughly 1 million sets of 20 tosses (20,000,000 tosses in total) you'll expect to see all heads in roughly 1 set. I think we can agree on that.

If instead, you toss 1 million times and consider each toss ( except the last 19 ) the start of a set of 20, that is just 19 shy of 1,000,000 sets. Of those sets, you'd still expect one 'win' on average, Definitely not at least 20 such streaks in 1M tosses. Definitely.

Where you get 91% probability from escapes me. I assert you are mistaken sir.

I was trying to keep this simple, just wonder about the this and that of that and this...
No such luck.
I'll have to come back later, got to start the scallops for Sparkles now.
Later.