Odds Of Losing 7 Blackjack Hands In A Row

Out of those 100 times, the dealer had a blackjack 8 times. I did have a blackjack twice. The two times that I got a blackjack, the dealer also got a blackjack. Out of the 100 plays, the dealer got a 10 in his visible card 53 times. Out of those 53 times, 35 the dealer had another 10 in his hand. In 11 situations the dealer had a 19 on his hand.

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• Odds Of Losing 7 Blackjack Hands In A Row
• Odds Of Losing 7 Blackjack Hands In A Row One

• Out of those 100 times, the dealer had a blackjack 8 times. I did have a blackjack twice. The two times that I got a blackjack, the dealer also got a blackjack. Out of the 100 plays, the dealer got a 10 in his visible card 53 times. Out of those 53 times, 35 the dealer had another 10 in his hand. In 11 situations the dealer had a 19 on his hand.
• As I always say all betting systems are equally worthless so flying by the seat of your pants is just as good as flat betting over the long term. When I said the probability of losing 8 hands in a row is 1 in 173 I meant that starting with the next hand the probability of losing 8 in a row is 1 in 173.
• So the odds of losing 7 hands in a row is 1.09% not great odds but certainly not impossible Though the odds of winning a hand of blackjack is independent so If you've lost 6 in a row the chance you lose the next hand is still 52.5%.

This may seem very elementry, but it has been a lifetime since Prob/Stat class.
1. What are the odds of losing 4,5, and 6 hands of BJ in a row?
2. What are the odds of losing 4,5 and 6 pass/don't pass bets in craps in a row?
3. If I know the probability of a win (say 48%) how do I figure this for 4,5,6 or for that matter any number of trials in a row?Simple: It's just the odds of one occurrence to the power of the number of occurrences you're concerned with.
Of course, with BJ, it's complex because of pushes, splits and doubles.
But if the assumption is you'll lose 52% of the time, then 52& ^ 4 = 7.3%, 52% ^ 6 = 1.9%, etc.

Simple: It's just the odds of one occurrence to the power of the number of occurrences you're concerned with.
Of course, with BJ, it's complex because of pushes, splits and doubles.
But if the assumption is you'll lose 52% of the time, then 52& ^ 4 = 7.3%, 52% ^ 6 = 1.9%, etc.

I thought that was it but it seemed too simple.

I thought that was it but it seemed too simple.

And the answer is for only when n=4 or for 4 trials.
The next question to follow should be (using the blackjack lose rate): Probability of losing 4 hands in a row in 5 trials? How about 10 trials?
The math of streaks is a very challenging quest when p is NOT equal to .5, like in a fair coin toss, and all the recursive, matrices, Fibonacci number formulas that all deal with a fair coin toss probability will not help us in finding an exact answer for a biased coin toss.
BruceZ at http://forumserver.twoplustwo.com/25/probability/successes-row-904091/ has a perl script to calculate the chance of at least a run in n trials.
Better yet,
He repaired an online calculator using javascript that can be found HERE that a friend of mine placed in a html page, and it is 100% accurate up to n=1,000,000 (that is as high as I have tested). BruceZ (IMHO) is a rare math genius good guy that does a great job on explaining how to do something.
Probability of losing lose 4 BJ hands in a row in 5 trials? 10.82119168%
How about 10 trials? 27.7326854715615%
How about 100 trials? 98.4214302644146%
How about a ML Baseball team that has a 50% win rate, in 162 games how about a win streak of 5 or more? 93.4519474125220%
6 or more? 72.7712414312032%
How about a ML Baseball team that has a 60% win rate,(the best teams) in 162 games how about a win streak of 5 or more? 99.7768554358947%
6 or more? 96.6318323647770%
A calculator or script is a very useful tool. It saves time so computer simulations are not needed.
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Odds Of Losing 7 Blackjack Hands In A Row 4

Soon to be in a new thread
But the streaks that are calculated are an 'at least 1' calculation.
At least 1 streak of 5 or more in 162 games.
I am working on the math to solve the below question.
What about 2 streaks or more?
3 streaks or more? n streaks or more.
Example. a fair coin toss. n=10. Probability of at least 1 run of 2 or more heads is exactly 880/1024 or 85.9375%
Formula is 1-F2[12]/2^10 where F2[12] is the 12th Fibonacci 2-step number. (144)
Probability of at least 2 runs of heads 2 or more in 10 flips.
Answer 368/1024 or 0.359375
I have no formula for the answer, yet. Just crunched the numbers in Excel after placing all possible sequences (2^10) in lexicographical order.
A photo of my Excel sheet is HERE so one can see what i am trying to explain.
It is a great exercise in math and when I get enough information on it I will post a new thread about it.

Odds Of Losing 7 Blackjack Hands In A Row Chart

Probability of at least 2 runs of heads 3 or more in 10 flips.
Answer 56/1024 or 0.0546875
Probability of at least 2 runs of heads 4 or more in 10 flips.
Answer 5/1024 or 0.0048828125
Probability of at least 2 runs of heads 5 or more in 10 flips.
Answer 0/1024
Need at least 11 flips for it to happen once.

...He repaired an online calculator using javascript that can be found HERE that a friend of mine placed in a html page, and it is 100% accurate ...

I'm not sure what your point is, but using that calculator, I got the exact same results, albeit not rounded, as I reported in my post above.the basic question of chances of 'how many in a row all wins' or all losses is easy enough even for me to do, however, if you go to 'at least one win' or similar it is something different.
PS:
>3. If I know the probability of a win (say 48%) how do I figure this
1 minus .48 is .52, in case that went past anybody

I'm not sure what your point is, but using that calculator, I got the exact same results, albeit not rounded, as I reported in my post above.

I was not questioning your math. You are 100% correct. 4 BJ losses in a row in 4 hands. .52^4
But it is for only 4 hands.
How about at least 4 losses in a row in 10 hands?
20 hands?
100 hands?
It is just more complex math.
Streaks (or runs), either win or lose, exist.
The more trials, the longer the streaks, win and lose.
More trials, more streaks.

I was not questioning your math. You are 100% correct. 4 BJ losses in a row in 4 hands. .52^4
But it is for only 4 hands.
How about 4 losses in a row in 10 hands?
20 hands?
100 hands?
It is just more complex math.
Streaks (or runs), either win or lose, exist.
The more trials, the longer the streaks, win and lose.
More trials, more streaks.

This calculator will raise things to the power of x quite nicely for very large numbers of trials; I like it anyway
http://web2.0calc.com/

I was not questioning your math. You are 100% correct. 4 BJ losses in a row in 4 hands. .52^4
But it is for only 4 hands.
How about at least 4 losses in a row in 10 hands?
20 hands?
100 hands?
It is just more complex math.
Streaks (or runs), either win or lose, exist.
The more trials, the longer the streaks, win and lose.
More trials, more streaks.

What I see is the probability of getting exactly 4 losses in a row in 4 hands is .52^4 just like DJ posted or a binomial distribution formula will also work.
The probability of getting at least 1 run of 4 losses or more in a row in 10 trials is 27.7326854715615% using the streak calculator.
How about getting a run of exactly 4 losses in a row in 10 trials?
It is a different question and should have a different answer.
A quick 1 million, 10 flip sim shows
0 849208 84.92% for NO exact run of 4
1 147086 14.71%
2 3706 0.37%
What this does not show are the runs of 5,6,7,8,9 and 10.
So I guess if a gambler 'fears' a losing streak of exactly 4 in a row, he better also fear the streaks of 4 'or more' in a row.

This may seem very elementry, but it has been a lifetime since Prob/Stat class.
2. What are the odds of losing 4,5 and 6 pass/don't pass bets in craps in a row?

For the pass line:
losing 4 in a row in 4 trials 251/495^4 = 6.6111%
Taking it further...
losing 4 in a row or more, at least once, in 10 trials = 25.6268663234987%
losing 4 in a row or more, at least once, in 30 trials (hour at the craps table average)= 65.7730718927031%
losing 4 in a row or more, at least once, in 100 trials = 97.7367454127325%

Odds Of Losing 7 Blackjack Hands In A Rowing

So, Nope27 question, for another thread, is
losing 4 in a row or more, at least twice, in 100 trials
losing 4 in a row or more, at least 3 times, in 100 trials
I got it now. Yes, some interesting math.
Nope27, how did you arrange your 10 coin flips in lexicographical order? I had to look that one up.
Do you have VBA or a software for it?

Odds Of Losing 7 Blackjack Hands In A Row

You are then trying to calculate the probability of multiple k-streaks within n-trials.
Interesting concept.

Odds Of Losing 7 Blackjack Hands In A Row One

So playing Craps for 3 hours, how many runs of losses of 4 or more, at least twice, can one expect? Simulations would provide an accurate answer.

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