Blackjack - Double up or go broke probability

[beachbum]

I've been having a heated discussion with some friends about this. Say you have $200 and want to either double your money or go broke. What is the probability you'll double up? Assume a standard 0.5% house edge for blackjack.

What is the optimal bet size to do so? $50 per hand, $25, the full $200? If you bet less than $100 to start, should you decrease your bet size at all when your bankroll drops below x number of bets? Any help or information source would be greatly appreciated.

[dtbog]

[ QUOTE ]
either double your money or go broke.

[/ QUOTE ]

This is just intuition, so don't take it as correct.. it's just my musing.

You're assuming an 0.5% house edge, so I take that to mean that you're ignoring the technique of counting cards. If so, then I argue that your chance of doubling up is 49.5%, and your chance of going broke is 50.5%. This you can achieve by betting $200 at once, and hoping to win.

(note that you have to subtract a miniscule percent for the fact that you can't split with this method, so you might lose hands that you would otherwise win.)

Here's my argument: placing only one bet, you're getting a single 0.5% house advantage, and your experiment is complete... you've either won or lost. You've wagered exactly $200, on a -0.5% proposition.

If you use ANY OTHER betting method -- betting $100 at a time, $5 at a time, whatever... unless you win or lose every bet you make, you'll be forced to make more than $200 in wagers, so you'll be conceding part of your profit back to the house, thereby coughing up part of your profit and making it more difficult to achieve the desired result -- namely, doubling your money.

Since variance is your friend here (you're trying to beat the odds), take the maximum amount of variance. The more money you wager, the more likely it becomes that statistics will predict the result of your experiment--- in this case, that you go broke.

Anyone agree?
-DB

EDIT: This would be the *maximum* probability of doubling your money, IMO. Different betting strategies would yield different results.

[Zoltri]

Great question! I have often wondered about this as well.
I dont have the answer but im sure the sharps in here do.

[beachbum]

Correct, ignore counting cards.

The "one and done" argument is great, but I don't believe the house edge is only 0.5% for a single hand where you can't split or double down. It might be more like 5% or more.

Also, if you have $200 and bet $100, now you can double or split. However, you can't split then double either card, or resplit since you only have 2 bets in your bankroll. So then what is the house edge for betting $100 with a bankroll of $200? 1%, 2% ?? This is where it gets tricky to optimize your bet size and figure out what the greatest % chance to double up is.

[pzhon]

If you can simplify the game to be a slightly biased coin-toss, you should bet the whole 200. You want to decrease the expected amount of house advantage you pay, and since the total variance is roughly constant, you want to maximize the variance per dollar bet, which is done by maximizing the size of your bets.

One problem with using blackjack is that the ability to double down or split (or less importantly, both) is factored into the house advantage. If you can't double down or split, you pay roughly 2% more, and you won't want to double down or split if this means you overshoot your target. Another problem with blackjack is that if you get blackjack, you might overshoot, and if this adds no value the house advantage is greater.

Given these complications, for actual blackjack you should bet at most 100 since betting 200 more than doubles the house advantage. I'm not sure what the optimal strategy is, but this type of problem comes up with sticky bonuses and online blackjack tournaments.

[charlie_t_jr]

[ QUOTE ]
The "one and done" argument is great, but I don't believe the house edge is only 0.5% for a single hand where you can't split or double down. It might be more like 5% or more.


[/ QUOTE ]

I'm not exactly sure what you mean here...but the house edge for a particular game is X%, that includes the rules of the game(what splitting and doubling is allowed).

That X% is off the top 1st hand dealt. The house edge changes as more cards are dealt, but you don't know that because you're not counting. So every hand as far as the non-counter is concerned has a X% disadvantage.

EDIT: After rereading your post, we may not be talking about the same thing...sorry.

[MicroBob]

[ QUOTE ]
One problem with using blackjack is that the ability to double down or split (or less importantly, both) is factored into the house advantage.

[/ QUOTE ]


Correct.

Along with the whole BJ 3:2 pay-off thing.

just because the house is at a 0.5% advantage does NOT mean that you have a 50.25%-49.75% chance of winning an individual hand.

I believe the chances on an individual hand in BJ is something like:
42% - win
9% - push
49% - lose

some of your wins will be double-ups, and some will be 3:2 pay-off for BJ. This makes up much of the difference and gets you to a 0.5-1% disadvantage (depending on the individual game).


If you are just playing a slightly-biased coin-toss game where you win exactly 1-unit 49.5% of the time and lose exactly one unit 50.5% of the time then your best chance for doubling-up would be to bet it all on a single-bet.

The chances of doubling your income should be proportional to your bet-size (I think). As the bet-size goes smaller your chances of doubling your whole $200 roll goes smaller also.
Again...this is how I THINK it works....and I'm not ENTIRELY positive.

Breaking it down to smaller bets essentially lets the odds catch up to you.

Anyway in the slightly-biased coin-toss game (which is much simpler to work with than BJ), That 49.5% on EACH hand will be more likely to grind you down and hurt your chances of winning the smaller your bet-size gets.

If you bet in $100 units then your chances will be just slightly lower for a double-up....because now you need to win 2 hands at 49.5% instead of just 1.
If you bet $10 units then your chances of doubling-up get even smaller.

If you bet $1 units then doubling-up for a 200 unit increase when you're playing at a disadvantage on EVERY hand will be very difficult. Probably around 98% chance against you I would guess.
If you bet in units of just $0.01 then you would have virtually zero chance of doubling that $200 as a 20k unit profit while playing at a 1% disadvantage is highly unlikely. Probably somewhere around the 0.0001% range of coming through for you if you bet that small.

[beachbum]

[ QUOTE ]
just because the house is at a 0.5% advantage does NOT mean that you have a 50.25%-49.75% chance of winning an individual hand.

I believe the chances on an individual hand in BJ is something like:
42% - win
9% - push
49% - lose

some of your wins will be double-ups, and some will be 3:2 pay-off for BJ. This makes up much of the difference and gets you to a 0.5-1% disadvantage (depending on the individual game).


[/ QUOTE ]

Right, and that's what makes the question in the subject line so intriquing. Also, the question of what size your bet should be to optimize your doubling up percentage, which obviously will be under 50%.

Picture an x-y graph where the vertical y-axis is double up percentage, and the horizontal x-axis is bet size. This graph is just a representation of our problem. I imagine this graph would be non-linear and would be some type of curve that peaks out at a certain x,y point. In calculus, if you took the derivative of the equation of the graph it would be the point where it equals 0. (just trying to illustrate as best I can)

Now at this x,y point, the y-value gives us the highest possible double up percentage and the x-value gives us the optimal bet size to accomplish this. Just trying to find this point. Any other thoughts, or know any people/resources that have delved into this?

[Izverg04]

As you reduce the size of the bets, you encounter two competing factors. House advantage HA is reduced. Wagered amount WA is increased as you zig-zag more. We want to minimize WA*HA.

I believe betting 100% on 1 BJ hand has ~4% HA. Betting 50% on 1 BJ hand allows you to double down, and this removes most of the disadvantage. This bet has HA of ~0.75%. I believe both of the numbers are in the right ballpark, starting with HA of 0.5% when you don't have any of these limitations.

Clearly then, you should never bet more than 1/2 of your balance.

Also, it is clear that you should not bet less than 1/4 of your balance, because the gain in HA is just miniscule and can't counteract any increase in WA.

Whether you bet 1/2, 1/3 or 1/4 is just academic -- the difference will be too small to be interesting.

So, you start with $100 and always bet 1/2 of your balance (or 1/2 the distance to the target, whatever is smaller). You'd need a simulation to figure out how much you wager on average. $200 seems to be in the right ballpark. So you lose approximately $200*0.75%=$1.5 to house advantage, and you double up to $200 49.25% of the time. Should be about right.

[Blackjack]

It's 48% You, 52% house when you're playing all in one hand due to the edge you lose from not being able to split, dd, or surrender.

Blackjack

[cardcounter0]

Not being able to double down is not a disadvantage at all.
When you double down you actually DECREASE the chances of winning your hand --- you only get one card on a double down, remember?

You have 11, the dealer shows a 9. Basic Strategy is to double down, because in the long run you will WIN MORE MONEY, but you actually reduce the chances of winning your hand. Case in point -- you double down and get a 3 for 14. What are you chances of winning now? In the long run, you win more money by doubling -- but we are interested in winning the hand and doubling our money in the short run. Wouldn't it be nice in the above situation if we could hit our 14 again and try to better our chances at winning this particular hand?

The effects of splitting is neglible. First place, you would have to get a pair to be able to split. So only a very small percentage of hands is going to be splittable. Secondly, many hands have a very very small difference. If you get a 2,2 vs a dealer's 6. What is the difference between playing one hand of 4 vs 6 vs playing two hands of 2 vs 6? Not much.

Some splits are defensive in nature. 8,8 vs 10? Splitting only helps you lose less money in the long run. In the context of winning the most hands (not the most money) in the short run, it would be better to just hit than split.

And double down after splitting? See above.

So just bet the entire $200 on one hand, and play to win that hand.

[beerbandit]

what about playing two hands for 100 each?

not back-to-back hands but on one round


cheers

[cardcounter0]

No, playing multiple hands reduces variance. When trying to double your money in the short run, variance is your friend.

Extend the multiple hand concept out. Hypothetically Play 200 hands at $1 each. Now the only way you double your money is the percentage of the time that the dealer busts (hopefully you haven't busted any of your 100 hands first, or you still don't double your money).

Now compare the dealer bust percentage that doubles your money to the percentage of time you win a single $200 hand.
By playing two hands of $100 each you will run into lose one, win one situations, forcing you to give the house edge additional shots at your money.

[pzhon]

[ QUOTE ]
Not being able to double down is not a disadvantage at all.

[/ QUOTE ]
It means the house advantage is much larger than normal, more like 2.5% than 0.5%. Then there is the problem of blackjacks only paying 1:1 because of the overshoot, which adds another couple percent to the house advantage.

[ QUOTE ]
When you double down you actually DECREASE the chances of winning your hand --- you only get one card on a double down, remember?


[/ QUOTE ]
This is already factored into Izverg's and my analyses.

[ QUOTE ]

So just bet the entire $200 on one hand, and play to win that hand.

[/ QUOTE ]
Wrong. Try computing some actual probabilities that these strategies reach $400 or more, starting with $200. It is much better to wager $100/hand than $200, and wagering $100/hand isn't optimal.

[cardcounter0]

"Then there is the problem of blackjacks only paying 1:1 because of the overshoot, which adds another couple percent to the house advantage."

I have no idea what this means. Are you saying that if you get a blackjack with a $200 bet you suffer because you have more than doubled your money? Never heard the house say, sorry sir, you were trying to double your money, we are only going to pay you 1:1 on this blackjack.

"Try computing some actual probabilities that these strategies reach $400 or more, starting with $200."

The single $200 bet is easy to calculate. You double your money approximately 48% of the time. No way can making a series of -EV bets exceed that. The slight increase in house edge due to not being able to double down or split, is offset by the fact you aren't giving the house multiple shots at your money.

"It is much better to wager $100/hand than $200, and wagering $100/hand isn't optimal."

Optimal? What are you talking about? The optimal bet is $0. Every wager you place is -EV, you can make 400 $1 wagers, or 200 $2 wagers, or any combination you want -- the house edge takes a chunk out of each bet.

To get the most gain out of variance (and giving the house edge the smallest shot at your money) 1 bet of $200 is optimal to double your money.

You are trying to come up with a EV maximizing solution, but the maximum EV solution is the not best way to double your money. Besides, two $100 hands is wrong also. Since you could split 3 times to 4 hands, and the maximum EV play could result in double downs on all 4 hands, you should never bet more than 1/8 your entire bankroll to maximize EV -- or in this case $25.

[pzhon]

[ QUOTE ]
"Then there is the problem of blackjacks only paying 1:1 because of the overshoot, which adds another couple percent to the house advantage."

I have no idea what this means. Are you saying that if you get a blackjack with a $200 bet you suffer because you have more than doubled your money?

[/ QUOTE ]
Relative to a bet with the same house advantage and no overshoot, yes. Because of the overshoot, you don't double up as much as you might expect from a game with a house advantage that is supposed to be so low.

It's like trying to double up with video poker. Even though the house advantage may be low, that depends on hitting the jackpot a small fraction of the time, which would overshoot. If you have to try to double up using casino games, avoid ones that would force you to overshoot.

[ QUOTE ]
"Try computing some actual probabilities that these strategies reach $400 or more, starting with $200."

The single $200 bet is easy to calculate. You double your money approximately 48% of the time. No way can making a series of -EV bets exceed that. The slight increase in house edge due to not being able to double down or split, is offset by the fact you aren't giving the house multiple shots at your money.

[/ QUOTE ]
Cutting the size of your bets by a factor of 2 is easily worthwhile when it decreases the house advantage by a factor of 10. Do the math.

[ QUOTE ]
"It is much better to wager $100/hand than $200, and wagering $100/hand isn't optimal."

Optimal? What are you talking about?

[/ QUOTE ]
The original poster asked about maximizing the probability of reaching at least $400. That probability is what I am talking about optimizing. You suggested the simple strategy of betting everything on one hand. I claim that betting $100/hand is better than your strategy. I do not claim it is optimal: I do not claim that there is no other strategy better than betting $100/hand.

[ QUOTE ]
Besides, two $100 hands is wrong also.

[/ QUOTE ]
If it is better than your strategy, it shows your strategy is not optimal. That is all it was intended to do.

[ QUOTE ]
Since you could split 3 times to 4 hands, and the maximum EV play could result in double downs on all 4 hands, you should never bet more than 1/8 your entire bankroll to maximize EV -- or in this case $25.

[/ QUOTE ]
Once again, it is worth betting only half of what you have to decrease the house advantage. Once again, it is not worth betting only 1/4 of what you have to enable you to double down after splitting.

If you are still convinced that betting $200 on one hand is optimal, would you care to make a wager on whether I can demonstrate a strategy which reaches $400 more frequently to the satisfaction of, say, BruceZ, or BillC, or the Wizard Of Odds?

[cardcounter0]

What ever expert you choose make sure they are aware of the CA risk-adverse work done by Steve Jacobs over at bjmath.com

The following plays should not use basic strategy, do not double down or split on these hands when your bet amount (f) is bigger than the following fractions of your total bankroll.

4,4 v. 5 Hit if f > 0.591%
5,3 v. 5 Hit if f > 0.981%
4,4 v. 6 Hit if f > 1.27%
A,3 v. 4 Hit if f > 1.27%
A,7 v. 3 Hit if f > 1.57%
A,4 v. 4 Hit if f > 1.63%
A,5 v. 4 Hit if f > 1.71%
5,3 v. 6 Hit if f > 1.86%

...

One of the last double-downs to disappear would be 11 vs. 6.

...

9,2 v. 6 hit if f > 24.41%
8,3 v. 6 hit if f > 25.11%
7,4 v. 6 hit if f > 25.75%
6,5 v. 6 hit if f > 26.65%

See? With your suggested $100 (f = 50.00%) bet amount, doubling down is not the optimal play -- so it really doesn't matter that you are able to double down and you think this gives you an advantage -- it makes it worse!

Flay away! The number your strategy needs to beat is .478 probability of doubling bankroll. Of course you are going to be splitting and doubling on the above hands, since this is the whole reason you are making the smaller bets, right?

[BlueBear]

If you want to double it up with the higher probability, bet it all at one go and play that hard optimally (without splitting or doubling).

Same for any other game, if your sole intention is to double up with the smallest risk, bet your session bankroll all at one go. Betting a smaller percentage of your session bankroll multiple times, would only increase your risk of going bust.

[Izverg04]

pzhon, I admire your patience. Well, admire is not the right word, it's just you do what I can never do -- keep calmly arguing with a brick wall. I suspect you'll keep posting in this thread, while I'd be only capable of berating people at this point /images/graemlins/smile.gif I'll never become moderator on any forum.

[pzhon]

I rather hoped you would state an amount to be wagered. Oh well.

[ QUOTE ]
What ever expert you choose make sure they are aware of the CA risk-adverse work done by Steve Jacobs over at bjmath.com

[/ QUOTE ]
That is completely irrelevant. What is being optimized on this thread is not the logarithm of the bankroll, the Kelly Criterion. The problem is to try to maximize the probability of reaching at least $400, starting with $200.

[ QUOTE ]
6,5 v. 6 hit if f > 26.65%

See? With your suggested $100 (f = 50.00%) bet amount, doubling down is not the optimal play -- so it really doesn't matter that you are able to double down and you think this gives you an advantage -- it makes it worse!

[/ QUOTE ]
Do you believe that applies here? It's obviously right to double down in this situation, when you are trying to reach $400. Your quote is out of context. Want to bet $1k against $10k on it?

[ QUOTE ]

Flay away! The number your strategy needs to beat is .478 probability of doubling bankroll.

[/ QUOTE ]
First, I don't accept that figure, since according to the assumptions I made, you only win 47.5% of the non-ties. If you tie, you will just bet it all again, right? If you tell me which rule set gives you 47.8% wins among the non-ties, then I may adjust the rest of my figures, but the fair comparison is with 47.5%.

Second, I beat 47.8% anyway. Here is the strategy I assumed:

<ul type="square">
At $50, bet it all, and burn $25 if you hit blackjack.
With probability 52.5%, bust; 47.5%, go to $100.

At $100, bet it all.
With probability 52.5%, bust; 42.5%, go to $200; 5%, go to $250.

At $150, bet $100.
With probability 52.5%, go to $50; 42.5%, go to $250; 5%, go to $300.

At $200, bet $100.
With probability 4.5%, bust; 43.5%, go to $100; 9%, stay at $200; 32.5%, go to $300; 4.5%, go to $350; 6%, win.

The above is the only questionable step. These numbers come from rounding data from a simulation conservatively. It gives a house advantage of 1.25%, which is too high for blackjack with no doubles after splits, and no resplits.

At $250, bet $100.
With probability 52.5%, go to $150; 42.5%, go to $350; 5%, win.

At $300, bet $100.
With probability 52.5%, go to $200; 47.5%, win.

At $350, bet $50.
With probability 52.5%, go to $300; 47.5%, win.
[/list]

This system succeeds 47.95% of the time, which is better than if you bet everything until you don't push. Once again, this system is not optimal. It involves ignoring the blackjacks at $50, and it still involves overshooting too much. Minor tweaks improve it to over 48.1%, more than a half percent better than betting $200/hand.

A reason this is better is that when overshoots don't count, the house advantage/variance ratio is too large by your method. By wagering $200/hand, your variance is almost 4 times as large as my variance from wagering $100/hand. However, your house advantage paid is 5% of $200, 8 times as large as my 1.25% of $100 on the first hand. Your ratio of house advantage/variance is twice as large as mine.

[tylerdurden]

[ QUOTE ]
Not being able to double down is not a disadvantage at all.
When you double down you actually DECREASE the chances of winning your hand --- you only get one card on a double down, remember?

[/ QUOTE ]

If the goal is to win ONE single hand, then doubling down is bad. However, that's not the goal. The goal is to double the bankroll.