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#1
October 18th, 2011, 09:04 PM
 RollingStoned Member Join Date: May 2011 Posts: 15
Doubling bet per lost hand for an eventual win? (5/10/20/40/80\$)

Without counting cards....would the following be a (somewhat?) profitable theory to winning money at blackjack? --

1) Bet 5.
a) If you win, you just made +5 bucks, repeat step 1.
b) If you lose, you just lost 5 bucks, go to step 2.

2) Bet 10
a) if you win, you just lost 5 on first hand and won 10 on second hand (net +5). Go to step 1.
b) if you lose, go to step 3.

3) Bet 20
a) if you win, go to step 1. You just lost 5 on first hand, 10 on second hand, and won 20 on third hand (-5 - 10 = -15, + 20 = 5, for a net income of +5).
b) if you lose, go to step 4.

4) bet 40
a) etc. go to step 1
b) etc. go to step 5

etc. etc.

Obviously it gets a little sketchy down the road where you're betting 80 or 160 bucks on a hand....and if you go to the casino with 155 or 315 (which is probably the amount I'd be going with).....if you get unlucky in 5 or 6 hands in a row, you're ****ed. But then again, what are the chances you'll lose 5 or 6 hands in a row?

So yeah, my question is basically what are the chances you'll lose 5 or 6 in a row? And is the theory proposed above worth it? I'd be using basic blackjack rules (hit against 5,6 if im < 12. hit against 2,3,4 if im < 13. hit against 10,A if im < 16. hit against 7,8,9 if im < 17).

Not sure what to do about doubling/splitting, either (if I'm using the above proposed strategy). Thoughts, ideas, tweaks, or anything I should know before doing this, or if this is even a good/legit strategy? I figured I'd come here n ask first before I do it.
#2
October 18th, 2011, 09:08 PM
 jaygruden Senior Member Join Date: Aug 2011 Location: Anytown, USA on the East Coast Posts: 279

No....This is Martingale. It's a fallacy system. You will lose.
#3
October 18th, 2011, 09:10 PM
 jaygruden Senior Member Join Date: Aug 2011 Location: Anytown, USA on the East Coast Posts: 279

#4
October 18th, 2011, 09:40 PM
 RollingStoned Member Join Date: May 2011 Posts: 15

I know it's a fallacy. But I'm not banking on 'the next hand' will be a winner. I'm banking on 'one of the next 5 hands' will be a winner.

I've never been super-great at statistics, buuuuuut:
The chance of a tails (loss) is 0.5*0.5*0.5*0.5*0.5 which equals 0.03125 (or about 3%). So there's a 97% chance you'll throw a heads in at least 1 of the 5?

I can see how it can be a fallacy (to double your bet every hand until you win), but I can't figure it out in my head to how it will make you lose. Can someone do the math (or tell me what the math is that needs to be done) ?
#5
October 18th, 2011, 10:35 PM
 shadroch Executive Member Join Date: May 2006 Location: NYC Posts: 6,696

There are literally a million articles on this. Every gambler seems to stumble upon this at some point, and the sad thing it usually works just well enough to get you in serious trouble.
Lets use your numbers- 5,10,20,40,80,160,320. If you lose five hands in a row, you will be wagering 320 just to break even. How hard do you think it is to lose 5 hands in a row?
Lets put that aside for a minute and concentrate on the hands you win. Lets suppose you do this for a few hours and find you've won 20 sequences and are up \$100. Everything is going great. you can't lose., but you do. Then you lose a second hand, then a third, and a fourth. You've suddenly lost 75% of your winnings, but you are not worried because you know the chances of losing five hands in a row are slim. The problem is the cards don't know you have lost four hands in a row and are going to play out just like every other hand- with you as a slight underdog.
You are placing a big bet at the wrong time for all the wrong reasons.
#6
October 18th, 2011, 11:34 PM
 AC232323 Member Join Date: May 2010 Posts: 67

Quote:
 Originally Posted by RollingStoned I know it's a fallacy. But I'm not banking on 'the next hand' will be a winner. I'm banking on 'one of the next 5 hands' will be a winner. I've never been super-great at statistics, buuuuuut: The chance of a tails (loss) is 0.5*0.5*0.5*0.5*0.5 which equals 0.03125 (or about 3%). So there's a 97% chance you'll throw a heads in at least 1 of the 5? I can see how it can be a fallacy (to double your bet every hand until you win), but I can't figure it out in my head to how it will make you lose. Can someone do the math (or tell me what the math is that needs to be done) ?
For simplification purposes let's assume that you have a 50/50 chance of winning (yes i know this isn't a right but the math with splits/doubles/etc is hard to explain so work with me). On average, how many hands do you have to play to expect to lose n in a row?
According to http://people.ccmr.cornell.edu/~ginsparg/INFO295/mh.pdf, the odds are (.5^(-n)-1)/.5. Plugging 5 in for n tells us that if we played 62 hands, we could expect to have a streak of losing 5 in a row.
Let's look at a typical table game setting. For my local casino, I have a 10 min bet and a 2000 max bet. On the xth hand you will be betting 10*2^(x-1). So on the 5th hand you would be betting 160. So, what happens if you get unlucky and lose a few more? Well, on you're 8th hand you will be betting 1,280. How many hands do you have to play to lose 8 in a row? A mere 510. And at that point you can't double your bet anymore. How much have you lost? \$2,550. And this was assuming a 0% house edge!
#7
October 19th, 2011, 03:56 PM
 21gunsalute Executive Member Join Date: Aug 2009 Location: Area 51 Posts: 1,144

Quote:
 Originally Posted by RollingStoned Without counting cards....would the following be a (somewhat?) profitable theory to winning money at blackjack? -- 1) Bet 5. a) If you win, you just made +5 bucks, repeat step 1. b) If you lose, you just lost 5 bucks, go to step 2. 2) Bet 10 a) if you win, you just lost 5 on first hand and won 10 on second hand (net +5). Go to step 1. b) if you lose, go to step 3. 3) Bet 20 a) if you win, go to step 1. You just lost 5 on first hand, 10 on second hand, and won 20 on third hand (-5 - 10 = -15, + 20 = 5, for a net income of +5). b) if you lose, go to step 4. 4) bet 40 a) etc. go to step 1 b) etc. go to step 5 etc. etc. Obviously it gets a little sketchy down the road where you're betting 80 or 160 bucks on a hand....and if you go to the casino with 155 or 315 (which is probably the amount I'd be going with).....if you get unlucky in 5 or 6 hands in a row, you're ****ed. But then again, what are the chances you'll lose 5 or 6 hands in a row? So yeah, my question is basically what are the chances you'll lose 5 or 6 in a row? And is the theory proposed above worth it? I'd be using basic blackjack rules (hit against 5,6 if im < 12. hit against 2,3,4 if im < 13. hit against 10,A if im < 16. hit against 7,8,9 if im < 17). Not sure what to do about doubling/splitting, either (if I'm using the above proposed strategy). Thoughts, ideas, tweaks, or anything I should know before doing this, or if this is even a good/legit strategy? I figured I'd come here n ask first before I do it.
What are the odds you'll lose 5 or 6 hands in a row? I'd dare say it happens to me at least once every time I play. I've lost 15 or more hands in a row 3 times in the same week. Martingale is a terrible idea. It's bad enough that you can lose so much money playing, but let's look at the flip side: A "successful" Martingale progression will net you 1 unit or \$5 in your case. Why risk losing hundreds or thousands of dollars for a \$5 win?
#8
October 19th, 2011, 04:52 PM
 arrando Member Join Date: Feb 2011 Posts: 84

However, if you must play a progression, a losing progression would probably be better than a winning progression (but only slightly) based on the theory that, if you lost, say the last 3 hands, there is a good chance that those 3 hands involved a lot of small cards, thus giving you a more positive count that coincides with more of your money out. This is from wizardofodds.com
#9
October 20th, 2011, 12:35 AM
 RollingStoned Member Join Date: May 2011 Posts: 15

At this point I realize it's not a good theory...just wondering if my math is on or off.

Ah, thanks for the info guys. I did some math for it, and was wondering if this is correct:

I assumed you can have enough money for 8 hands, starting at \$5 at first hand and \$640 on the 8th hand.

So for this, I assumed that you have a 50% chance of winning any given hand, even though it's a bit lower than that...

So I did 0.5^8 which is ~~ 0.0039... or about 0.4%, meaning there's a 0.4% chance you'll get 8 losses in a row.

I then figured that the total amount of money you'd need for this would be \$1275. 5+10+20+40+80+160+320+640=1275.

100/0.4 = 250, so you will lose 8 hands in a row 1 out of 250 "rounds". (1 round = 8 hands, or 8 lost hands in a row.)

So from there I assumed that you'd win 249 rounds and lose 1 round. Each won round gives you +\$5, and a loss round gives you -\$1275. So I'm going to assume (at your best luck), you'll win your first 249 rounds and lose your 250'th round.

249 * 5 = 1245 winnings, plus what you started with (1275), so you'll be at 2520\$ after your 249th round. Your 250th round you'll lose \$1275. So you do 2520-1275 = 1245, the amount of money you'll have after your 250th round.

So basically after 250 rounds, you'll have a net loss of \$30.

Is my math right, or is it way off? (This is of course with the assumption that you have 50% chance to lose a hand, instead of the 53 or 54%..or whatever the house's edge is....and also assuming that at your best luck you'll win 249 "rounds" and THEN lose the 250th round.)

Yes? No? Close? Or way off?
#10
October 20th, 2011, 03:07 AM
 shadroch Executive Member Join Date: May 2006 Location: NYC Posts: 6,696

Way off.

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