Lucky Red Casino

Probability of losing x # of hands in a row

Discussion in 'General' started by SBT, Nov 21, 2011.

  1. SBT

    SBT Member

    I am new to this endeavor. I am wondering, has any blackjack player or writer calculated the probability of losing a certain number of hands in a row (given proper execution of Basic Strategy)? I am aware that one cannot simply raise .5 to the exponent of the number of hands, because the dealer has the edge and this is not like flipping a coin. I have played bj in a casino about 60-70 hours so far. I would estimate that losing five or six hands in a row would be rare, but it happens to other players and it has happened to me a couple of times.
     
  2. Southpaw

    Southpaw Well-Known Member

    Flash cites the figure often on here, and it would be really easy to look up probably on WoO or using Casino Verite products.

    I believe that the player will win something like 43-43% of hands.

    Spaw
     
  3. tthree

    tthree Banned

    If you bar pushes I think it is about 47% wins/decision on average.
     
  4. sagefr0g

    sagefr0g Well-Known Member

    Attached Files:

  5. Jacob

    Jacob Well-Known Member

    Thanks be to Froggie for the link.

    :1st:
     
  6. Sucker

    Sucker Well-Known Member

    Perhaps I'M missing something - but there appears to be something wrong with that chart. :confused:

    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%)?
     
  7. 21gunsalute

    21gunsalute Well-Known Member

    Sorry, but it's not that rare at all to lose 5 or 6 hands in a row. I'd dare say it happens to me at least once an hour, and sometimes much, much more than that. I've lost 15 or more hands in a row 3 times in the same week.
     
  8. sagefr0g

    sagefr0g Well-Known Member

    i sure don't know, can't even remember where i got the image, it was a long time ago, lol.
    one thing maybe could make a difference, the zero count is not depicted far as the win/loss/tie/bj percentages. zero tc carries a house edge, could that make up the difference? perhaps some of those bj's are ties?
    what else is strange, i thought in the long run positive counts attributed a small but measurable higher number of wins than negative counts, that also is not the case in the image.
    edit: ahhh, ok i got the image from this site:
    (Dead link: http://www.bjstats.com/bjch.asp)
    (Dead link: http://www.bjstats.com/8100961108305XXX.jpg)
    ........
     
  9. sagefr0g

    sagefr0g Well-Known Member

    for the OP

    errhh here's another link i think pertains to your post:
    (Dead link: http://www.bjstats.com/8102861108305XXX.jpg)
    created from this link:
    (Dead link: http://www.bjstats.com/bjch.asp)
    .........
     
  10. iCountNTrack

    iCountNTrack Well-Known Member

    Winning/pushing/losing frequencies are not too useful figures in a game like BJ because they are different payouts because of doubles, splits, double after splits, blackjack. So it is more useful if you get the probabilities of each possible round outcome. So for example for a game with only one split allowed and DAS, the possible outcomes are -4, -3, -2, -1, 0, +1, +1.5, +2, +3, +4 . the following data is for 6D, S17, 1 split, DAS; Player hand 8,8; dealer upCard 6

    Code:
    player's Hand  8,8
    dealer's upCard  6
     
    player's probabilities for standing
    p_-1 = 0.578747564318
    p_0 = 0
    p_+1= 0.421252435682
    p_+1.5 = 0
    EV for standing= -0.157495128637 ± 0.987519764084
     
    player's probabilities for doubling
    p_-2 = 0.690684064282
    p_0 = 0.0441173812318
    p_+2= 0.265198554486
    EV for doubling= -0.850971019592 ± 1.76050526807
     
    player's probabilities for hitting
    p_-1 = 0.690684064282
    p_0 = 0.0441173812318
    p_+1= 0.265198554486
    EV for hitting= -0.425485509796 ± 0.880252634033
     
    player's probabilities for splitting
    p_-4 = 0.00419012222521
    p_-3 = 0.0650328867493
    p_-2 = 0.260197245456
    p_-1 = 0.0767357312541
    p_0 = 0.0743673854602
    p_+1= 0.0497634029576
    p_+2 = 0.334266924866
    p_+3 = 0.123229425568
    p_+4 = 0.0122168754629
    EV for splitting= 0.32786365993 ± 2.08656032979
    
     
  11. kewljason

    kewljason Well-Known Member

    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. :confused:
     
  12. SBT

    SBT Member

    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.
     
  13. kewljason

    kewljason Well-Known Member

    I couldn't even tell the most hands I have lost in a row. :confused: 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! :eek::mad::mad::laugh: 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. :laugh:
     
  14. SBT

    SBT Member

    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.
     
  15. caramel6

    caramel6 Well-Known Member

    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?
     
  16. Sonny

    Sonny Well-Known Member

    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-
     
  17. tthree

    tthree Banned

    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.
     
  18. Sonny

    Sonny Well-Known Member

    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-
     
  19. Cardcounter

    Cardcounter Well-Known Member

    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.
     
  20. iCountNTrack

    iCountNTrack Well-Known Member

    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
     

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