Probability of losing x # of hands in a row

[kewljason]

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.

[SBT]

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 five-hand 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.

[kewljason]

I couldn't even tell the most hands I have lost in a row. I don't even begin to think about something like that until I have lost probably 10ish and then it might hit me, "hey I've lost a bunch in a row"! But was it 10? or 8? or 12? or maybe only 6 but seems like 10 because of splits, DD. Or maybe it's only 5 or 6 but seems like more because I am max betting and the dealer is magically pulling 6 card 20's with a big plus count! In reality, as long as the count is still good, how many I have lost in a row is totally irrelevant to me. Although I might begin to consider that I am being cheated which is extremely unlikely, but it will cause me to take a closer look at everything, which isn't bad.

I am wondering where you are going with this SBT. My first thought when I saw this topic is that you were heading towards a topic better fitted for the voodoo forum.

[SBT]

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.

[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?

[Sonny]

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-

[tthree]

Originally Posted by Sonny

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-

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.

[Sonny]

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-

[Cardcounter]

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.

[iCountNTrack]

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 x-1 hands and lose them, the probability of losing each of the previous x-1 is 1 because we now have the information, so the probability on the xth hand is 1^1(x-1)*0.49 = 0.49