
November 21st, 2011, 08:56 PM


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Quote:
Originally Posted by Sucker
Perhaps I'M missing something  but there appears to be something wrong with that chart.
According to this chart, in a negative deck you lose 46.7% of the time, and counting BJs you win 45.2% of the time. However; because of the fact that BJ pays 3:2, shouldn't the actual WIN figure for a BJ be counted as 6%, rather than 4%; therefore tipping the scale into the player's favor (47.2% to 46.7%)?

The numbers in this chart are different than I have seen from many other sources. Most notably that loss number, which usually is closer to the 49% range in other sources. Win number (win + BJ's) is also off (higher) than other sources I have seen.
Last edited by kewljason; November 22nd, 2011 at 07:57 AM.

November 22nd, 2011, 08:39 AM

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Join Date: Nov 2011
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Thanks for replies and information. I took .53 (dealer's win average) to the fifth power. Extrapolating, one has an approximately 4% chance of losing five hands in a row, in any fivehand sequence. That would mean that such a streak happens once per hundred hands, on average. Not sure if this is the proper way to approach the problem mathematically. But some of you have affirmed that you get such a streak once per hour, and longer streaks at times.

November 22nd, 2011, 08:55 AM


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November 22nd, 2011, 10:11 AM

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Join Date: Nov 2011
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Kewljason, yes, I was considering a voodoo strategy. I was considering raising my bet by one unit each time that I lost a hand. If one wins by the sixth bet progression, you are not behind in total losses. But one cannot protect oneself from randomness. Even though it is only a 2% likelihood of losing six hands in a row, that’s still risky. I would probably only try this system at $5 table, and only when the count was not low.

November 22nd, 2011, 11:19 AM

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Join Date: Aug 2009
Posts: 105


how many in a row
anyway, do someone knows how many hands in a row to lose when tc2, tc3, tc4?
I believe should be lower number of hands comparing with a negative counts?

November 22nd, 2011, 11:38 AM


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Location: Los Angeles, CA
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Quote:
Originally Posted by caramel6
anyway, do someone knows how many hands in a row to lose when tc2, tc3, tc4?
I believe should be lower number of hands comparing with a negative counts?

The percentages don't change that much in actual play. In fact, the probability of winning actually decreases as the count gets higher!
http://www.blackjackincolor.com/truecount5.htm
Sonny

November 22nd, 2011, 12:19 PM

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Join Date: Mar 2011
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Quote:
Originally Posted by Sonny

You get more blackjacks and doubles and splits and win a higher percantage of them at high counts. That is were your advantage is found.
Last edited by tthree; November 22nd, 2011 at 12:56 PM.

November 22nd, 2011, 12:31 PM


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Quote:
Originally Posted by tthree
You gat more blackjacks and doubles and splits and win a higher percantage of them at high counts. That is were your advantage is found.

Exactly right. Also, the probability of wins decreases because the probability of pushes increases. The more tens there are, the more likely everyone is to get a 20 and push.
Sonny

November 22nd, 2011, 01:50 PM

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Dealer wins 48 percent of the time player wins 44 percent of the time the other 8 percent it is a push. What is the number of losing hands that you are trying to figure out? Lets say that you throw out pushes you lose approximately 52.174% of your hands so take that number to the power of x and that is the probility of losing X hands in a row.

November 22nd, 2011, 06:58 PM


ChemMeister


Join Date: Oct 2008
Posts: 780


The odds of winning or losing x hands in a row is a totally useless figure. I , 0.49^(x) is the probability of losing x hands in a row BEFORE any hand is played, meaning before we have any information, however AFTER we play the first x1 hands and lose them, the probability of losing each of the previous x1 is 1 because we now have the information, so the probability on the xth hand is 1^1(x1)*0.49 = 0.49

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