BlackJack odds of hitting goals

"Thanks Wizard for your ongoing guidance in the gambling world."

Can you please help me prove a friend wrong.

His plan,

$1500 bank roll
Flat bet starting with $100 pot at $10 per hand playing basic strategy.
If win $100 then walk away as a complete session.

If lose $100 increase bank to $200 and flat bet $20 until up or down $200
Same system as above applies. If win then take +$100 profit and walk or increase to $400 and $40
Then the end result is $800 and $80 flat bet until again up $100 or bust out at $1500.

He would like to believe that he can make $100 more times then he would bust for $1500.

I told him he would win more rounds then he would lose but over the long haul he would fall into the same house odds as always.

Can you please help me show the math to back this up.

Adam

You're both right.

He will absolutely win $100 more often than he will lose $1500. But, on average, he will not win $100 more than 15x as often as he loses $1500. So, on average, he will lose money in the long run.

I think most of the regulars on the WoV site would tell you that his system is foolish, and it doesn't work. I disagree with this. I make no bones about the fact that this system does not change the house advantage. But this doesn't tell the whole story. Variance is important too. By increasing his bet, he is increasing his variance relative to the amount already lost, which significantly affects his ability to reach a breakeven or $100 win point. This changes the shape of the distribution, and can provide a greater likelihood of reaching his goal.

If his goal is really to maximize his chances of winning $100 while risking $1500, his system is reasonable. He is much more likely to reach his goal using this system than if he just flat bets $10 per hand until he either wins $100 or loses $1500.

But again, just to be clear, none of this changes the house edge.

As for the "math", I can try to ballpark it. I think we can all agree that, due to the house edge, each of his little mini-session is more likely to lose the stake than double it. For instance, when he starts with a bankroll of $100 and $10 bets, he is more likely to lose the $100 than double it. I will estimate his chance of doubling at 45%, and losing at 55%, but that's pretty much just rectal extraction.

The only way he will lose his entire bankroll is if he loses at $100, then at $200, then at $400, then at $800. That's 4 losses in a row, which has a probability of 0.55*0.55*0.55*0.55 = 9.15%. Any other outcome (90.85%) results in a $100 win. So, on average, his system will return (-1500*0.0915 + 100*0.9085) = -$46.41 per session.

It sounds like a variation of a Martingale system.

In the Martingale system, you double you bet on each loss, until you finally win one unit. The risk is running thru the bankroll or hitting the table limit.


It sounds like you and your friend are ready to risk $1500 trying to win $100.

Oh, sure, you'll succeed, and succeed often. But you will, on average, succeed less than 15 times for each failure.


I.E. It will be a net loss.

Quote: vegasrvp

I told him he would win more rounds then he would lose but over the long haul he would fall into the same house odds as always.

All betting systems have the same fundamental flaw. Cards and dice and steel balls have no memory of what happened before, only people do. Each event is relatively independent (not completely independent for blackjack, which is why card counting is a reasonable thing to do). What determines the house advantage is the extra 0 in roulette, or the rules of the particular blackjack game. The results of previous bets do not matter.

More to the point, the reason that people believe in these systems is that they are unable to calculate the odds of a losing streak that will wipe them out. Thousands of psychological experiments have shown that almost everyone in the world underestimates these odds. The casino plays on this mistake in the human psyche, by catering to this belief. The devestating wipe out comes far more often then people think. The mathematics of this calculation is quite sophisticated (above most undergraduate college math classes) which partly explains the difficulty in accepting this result.

The assumption that previous results will affect future results in gambling is often called the "gamblers fallacy" since it happens so often in the mind of a gambler.

Basically: Yes, you are both right. You will win $100 quite often with that strategy, with an occasional loss of $1500.

My opinion: $100 is a trivial amount for someone who can support a $1500 bankroll. Personally, I have a completely opposite strategy. I overplay my bankroll, busting most of the time, but every once in awhile I'll triple-quadruple it. I would recommend maybe starting at $25/hand and going for $200-250. Also, if you're willing to do this, and you're in Vegas, you could put $1k down on a -500 favorite to win $200... and the experience will probably last longer.

I'm starting to wonder if my friend might be onto something.

I have decided to play this a little on the wizard of odds site using his recommendations for strategy. I have played 23 times with the goal of going up ten units.

Funny results: I am 23 for 23.

There is no matter as to how many units or the value you use. I have just tried to disprove his theory. I can't.

In the 23 times I have played I have only progressed past the $10 flat bet section a few times and got to the $80 only once.

Does anyone have any math to prove this is either good or bad?

I have to say I am now interested in looking at this further.

Based on the little math I know he has to win 15 times for every bust. With 23 straight wins not only am I profitable but I would have to lose twice in the next 7 rounds to break even over 30 sessions.

Where I am getting stuck on disproving this theory is based on the odds of this game I am going to win between 45 and 49 of 100 hands. If I go on a bad streak the increase in betting to the next session will give me the ground to make it up when the cards turn back in my favor. I lose less when the dealer gets blackjack versus when I get it and also my ability to split and double when the cars are in my favor and betting higher make up the losses faster.

Here is the true question of this theory:

what are the odds of losing 10 units flat betting before winning 10 units 4 consecutive times? Where I cannot understand the math is when there is an increase in betting and the ability to make up ground the odds change in your favor very quickly.

I know this might be an extreme example but if you start out playing at $10 and lose 20 of 30 hands you are down 10 units. But if you win 20 of the next 30 you are back to a 50/50 win loss total but have won 10 units and are walking away.

PLEASE HELP ME UNDERSTAND???????

ADAM

Quote: vegasrvp

What are the odds of losing 10 units flat betting before winning 10 units 4 consecutive times? Where I cannot understand the math is when there is an increase in betting and the ability to make up ground the odds change in your favor very quickly.

As I said earlier...

It is very difficult to determine the odds of winning 10 before losing 10. I don't know if the odds are 40%, 45% or 49%. But since every individual bet carries a house advantage, I hope we can all readily agree that the odds are less than 50%.

So now your system boils down to flipping a biased coin 4 times in a row. If you get tails 4 times in a row, you lose 15 units. If you every get a head, you win one unit. For an unbiased 50/50 coin, this is an even bet. You get heads four times in a row 1/16th of the time, losing 15 units. You win 15/16th of the time, winning a total of 15 units.

For a biased experiment, you'll win (on average) less than 1/16th of the time and fail to break even in the long run.

The fact that you've won 23 times in a row is nice and lucky, but doesn't disprove the math. The fact that the odds are against you doesn't necessarily indicate that you'll lose during 23 trials, or 100 trials, or even 1000 trials. The variance is still fairly large relative to the negative expectation.

Your difficulty in understanding seems to come directly from the gambler's fallacy: "If I go on a bad streak the increase in betting to the next session will give me the ground to make it up when the cards turn back in my favor." You need to replace "when" with "if". The fact that you've lost one, two or three sessions does not increase the likelihood of winning during the next session. You are still playing the next session with a negative expectation. Every session has a negative expectation and a variance which are proportional to the size of the bet. Yes, if you get lucky in the fourth session you can quickly make up all your losses from the first three. But if you lose, you will more than double your losses. And the likelihood that you will lose is still (always) greater than the likelihood that you will win.

We can estimate the probability of making it through 23 sessions. Let's assume a generous game with a house edge of 0.28% (stands on 17, early surrenders). Let's simplify the game by saying you have a 49.86% chance of winning a given hand and earning 1:1 on your best, and a 50.14% chance of losing. This is equivalent to a 0.28% house edge.

Then the probability of losing the current bankroll before winning $100 is 0.514. We can get this result using Fermat/Huygen formula with p = .4986, q = .5014, and n = 10. At this point, it should be clear it's a Martingale strategy that survived 23 sessions.

To lose your entire bankroll (starting at $1500) before winning $100, you need four consecutive losing sets. The probability of a losing set is 51.4%. So the probability of four consecutive losses is (.514)^4 = 6.98%, and you have a 93.02% probability of winning $100 before losing your entire bankroll in a given session.

The probability of 23 consecutive winning sessions is (.9302)^23 = 18.9%.

So the winning streak is the result of some good, but not exceptional luck.

Nice first post, ckjy. Welcome to the board!

Quote: ckjy

Let's simplify the game by saying you have a 49.86% chance of winning a given hand and earning 1:1 on your best, and a 50.14% chance of losing. This is equivalent to a 0.28% house edge.

Great math formula. Thank you for sharing.

I beg to disagree about subtracting the house edge to arrive at a winning probability for the game of Blackjack.
Please correct me if I am wrong.

I know I saw these numbers on the Wizards of Odds site.

Net Win in Blackjack
Event/Probability
Win/.4243
Lose/.4909
Tie/.0848
sourse #1 https://wizardofodds.com/blackjack/appendix4.html
sourse #2 https://wizardofodds.com/ask-the-wizard/blackjack-probability/ (3rd question)

Net Win in Blackjack
Event /Probability
Win given no tie /0.4635
Loss given no tie /0.5365

I made a gross simplification which turned Blackjack into a coin flip. The net win in Blackjack can return a variable number of units depending on how you won, which prevents the use of the aforementioned formula. So I assumed the same house edge in a coin flip, for the rest of the calculations.

Quote: ckjy

I made a gross simplification which turned Blackjack into a coin flip. The net win in Blackjack can return a variable number of units depending on how you won, which prevents the use of the aforementioned formula. So I assumed the same house edge in a coin flip, for the rest of the calculations.

I kind of figured that. Hard to add in BJs, doubles and winning splits into it.
Nice to run computer simulations.

Are my win/lose with no tie figures, based off the Wizards win/lose/tie, correct?