I just lost 13 hands of blackjack in a row and wondered what the chance for that is. I followed basic strat all the hands.
What is the probability that you will lose the next 13 hands? Roughly 1/(2^13), as Optimus Prime indicates. But I'm guessing you didn't sit down to play, immediately lose 13 hands, and quit. Instead, suppose that you plan to play for, say, the next 8 hours (100 hands per hour). What is the probability that you will see a losing streak of at least 13 consecutive hands at some point during those 8 hours? This probability is much higher, roughly once every 20-ish such 8-hour sessions. The following figure shows this probability as a function of the number of hours of (future) play (where I'm making a cocktail-napkin simplification by treating a loss as a coin flip, which is conservative, i.e., losing streaks are slightly more common than this):
streak.jpg
A handy rule of thumb: for a sequence of n hands, the expected length of the longest consecutive losing streak is roughly lg n (where the logarithm is base 2).
i sat down. lost 13 hnds and left. didnt win a single hand.
Now, when you lose 13 hands in a row and all are at positive True Count, that would be tough on a person. I dont count how long my consecutive hand losses are but once, the dealer remarked that I had not on a single hand in a DD game until the very last one and then, at a minimum bet, I won the last hand before the shuffle. The penn was 0.05 so, a lot of hands lost
Go here: http://www.beatingbonuses.com/calc_streak.htmOriginally Posted by kingzzz21
And here: https://sites.google.com/view/krapstuff/home
Note that, for a RESOLVED hand of blackjack, where ties are excluded from the outcome, the probability of a loss is more like 0.52.
Don
Wow! I recently, playing with another person, and all of us, dealer and the two of us got BJ. The first guy had opted for even money, I did not insure figuring the odds of all 3 getting a BJ was remote. I pushed, and everyone thought I was stupid for not taking up "sure" money.
Your decision should be about the count not the cards on the table. Just so you know the odds of predicting all 3 players getting a BJ is a tiny fraction of the odds that the third player has a T under an ace given that the other two have a BJ:
Odds of all 2 players and the dealer with an ace up getting a BJ off the top of a 2 deck game:
(2*(8/104)*(32/103)) * ((2*(7/102)*(31/101)) * ((6/100)*30/99) = (0.0477968)*(0.0421277)*(.0181818) = 0.0000366 or 0.00366%
The odds the dealer has BJ with an ace up given the other 2 hands played got a BJ off the top of a 2 deck game:
30/99 = .303030303 or 30.03%
Your poor statistical logic was only off by a factor of 10,000 or 4 significant digits.
Last edited by Three; 09-11-2018 at 06:16 AM.
Explain the difference between your two highlighted statements. I am a simple HiLo player, need a bit more explanation. In both cases, all 3 people (dealer and players) getting cards have a BJ. Why are the percentages different?
On the first hand of a 2 deck game, the RC at the point where the dealer asks for insurance is a Minus 5. Are you saying I should have insured my BJ? Here is an opportunity to display your greatness and explain to all of us how you would have played the hand. Just keep it to a single paragraph.
The low odds are for predicting all three getting a BJ before the hand is dealt. The quite probable number is the odds of the dealer having a BJ with an ace up given that that the two players already have a BJ and the dealer already has an ace up. Since you know that is the case when you make the decision, the only thing that needs to be calculated is the odds the dealer has a T in the hole. It doesn't matter that you will rarely be making that decision. Once the cards are dealt you are making that decision. Predicting this event before any cards are dealt brings into play the rarity of thew two players having a BJ and the dealer having an ace up.
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