I just wondered how I would work out the chances of losing 9-10 consecutive hands at Blackjack, assuming a 6-deck shoe, surrender allowed and dealer stands on all 17's

Is there an easy way to work this out?

50/50

You'll have to look up the exact house edge with these rules, but let's say you lose 51%. The answer is then 0.51 ^ 10 = 0.11%

well this is a sort of a tricky question. you dont figure it out by knowing the house edge. im going to guess you lose 47% of the hands you play since there are times where you will bust first and the dealer will bust but youre the one who ends up losing. to make up for this, the casino gives you the option of splitting or doubling and paying 3:2 on blackjack. so im gonna guess its (.53)^9 for 9hands or (.53)^10 for 10hands.

Since you're likely thinking of betting $5, then bet $10 if you lose, then $20 if you lose that, etc thinking it's easy money do it. The odds of losing 10 straight hands is ridiculously low. Go for it buddy

In my experience, every time you sit down to play blackjack.

Good point. Plus the times you make up money by splitting and doubling down.

It's a lot easier to lose 10 consecutive bets than to lose 10 consecutive hands -- do you consider a split two hands?

Previous posts have pretty much nailed the math on the head, but in my experience I can tell you the odds are actually a little different:

Probability of Losing 10 in a Row = (BAC*(Bankroll/BankAccount)) * (NOTISIWNPBJA)

BAC = Blood Alcohol Concentration
Bankroll = Amount Willing to Gamble
Bank Account = Total Amount in your checking + Savings account
NOTISIWNPBJA = Number of Times I Said I Would Never Play BlackJack Again

So for a typical 22 year old, with a BAC of 0.160 (twice the legal limit), a bankroll of $1000 with only $500 in his bankaccount (so 1500 total), who has said "Never again!" 3 times, the Probability of losing 10 hands in a row is:

= 0.160 * (1000/1500) * 3
= 0.32, or 32% chance he loses 10 in a row.

One of my former card-counting buds lost 16 hands in a row [in AC ftl] and quit forever. It was pretty lol in a variety of ways.

It's really hard to lose 16 in a row.

No. The house edge does not equal the probability of losing a hand.

On average you lose about 48% of hands. Losing 10 in a row is therefore approximately .00065 = .065%

Should happen once or twice a year to a serious card counter.

what is the chance of losing 21 bj hands in a row? that happened to me before thus making me quit playing blackjack. i swear i thought it was rigged

sometimes I feel like playing by the book really ****s me up more than it has helped me..

Just a question, if playing BlackJack, you only lose 48% of the hands you play and sometimes they pay you 3 to 2 on your wins, but your losses are flat, how is there a house edge? Am I missing something here? Seems like losing 51 or 52% of your hands played would make more sense.....someone please splain to me.

You get paid 3-2 on your blackjacks, not on every win

You will never win over 50% of the hands -- on average, or on specific counts.

Where did you get that number from willyc?

I am no expert, but if you lose 48%, then you have 52% to be split up between wins and pushes. So, it would make the most sense that you win a fair bit less than you lose and push the rest. This would explain why the house would still have an edge despite paying you 3:2 on some of your wins, etc.

-RMJ

1 in 9 or 10

Player win streaks are short and infrequent, while dealer winning streaks are long and frequent.

obviously, which is why I said "sometimes"

the guy who mentioned the pushes, I forgot about that, hard to believe it is as high as 4% of the time or so, but I don't know the maths so I won't argue.

i guess a way to figure this out would be to run a simulation seeing how many individual hands you win both counting splits as additional hands and not counting splits as additional hands and the number is somewhere in between.

I didn't really know either, I just inferred that pushes had to play a large enough role. Anyway, I was curious enough to google for an answer. No idea how legit, but the 48% figure lines up with what the other guy said...

"The casino usually wins 48% of blackjack hands on average, players win 43%, and 9% of hands are a push. A player’s 5% disadvantage in win frequency is offset by payouts for naturals, and extra profit on splits and doubles."

-RMJ

100% if you tried to lose every hand

I ran a couple of large samples using the percentages above (48%L, 43%W, 9%T - I have no idea if these numbers are accurate).

Anyway, I ran 11 samples of 25K hands, and the number of series of 10 consecutive losses within that sample ranged from a minimum of 6 to a max of 27. Not a lot of testing, but seems pretty likely over that size sample the odds of losing 10 in a row at one point are 100%.

I cut a second sample down to only 2500 hands, which I think is about what someone who plays only a few times a year would actually play. I ran 100 samples at this size:

n=100 (samples)
mean= 2.1 (series)
stdev=2.63
min=0, max=12

I should run more, but the likelihood is that even if you play recreationally only a few times a year you will still in all probability have this happen to you at least once.