Please check my BlackJack Math Logic

Can you help me with a BlackJack math question?

I am sure it is an elementary percentage conversion question but I want to make sure my logic and equations are correct.

I believe that I may not be using the statistics in Appendix 4 "Standard Deviation in BlackJack" in wizardofodds.com correctly.

I want to express the number of BlackJack Losses in a row in an easy to understand sentence (stated below).

"You will experience X losses in a row approximately 1 in every Z hands"
X = any number of losses in a row to be computed
Z = the approximate number of hands it takes to incur that loss in a row

Appendix 4 "Standard Deviation in BlackJack" data that I am using:
Probability of a Win 42.42%
Probability of a Loss 49.09%
Probability of a Push 8.48%

I don't want to count pushes so to determine the actual percentage without pushes, I used the below equation:
49.09/(42.42 + 49.09) = .5364
Answer: 53.64%

Now to determine the possibility of x number of losses in a row.
For the example of 10 losses in a row, I used the equation:
.5364^10 = .000250
Answer: .025%

To convert the percentage to odds I used the equation:
.00025/100 = 400,000

Does this mean I can express the result as:
"You will experience 10 losses in a row approximately 1 in every 400,000 hands"?

Is this correct?
Please correct any mistakes I made in my logic and/or formulas.
Thank you,

It's 1/0.00025, which should just give you 4000 hands.

And then your final statement is a common but important mistake. Search for 'streak' in this forum, and you will find dozens of people who have asked similar questions, and received detailed responses.

Quote: Byron

I don't want to count pushes so to determine the actual percentage without pushes, I used the below equation:
49.09/(42.42 + 49.09) = .5364
Answer: 53.64%

agree. close enough

Quote: Byron

Now to determine the possibility of x number of losses in a row.
For the example of 10 losses in a row, I used the equation:
.5364^10 = .000250
Answer: .025%

wrong
start over
what calculator did you use?

I get about 0.0019719127 in win7 calc
and 1 in 507.12 attempts

Quote: Byron

Does this mean I can express the result as:
"You will experience 10 losses in a row approximately 1 in every 400,000 hands"?

streaks are in attempts
the average length (number of hands or trials) of each attempt for 10 in a row is
2.152778445
(1+p+p^2+p^3+...+p^9)
507.12*2.15 gets the average

1,091.7210
average number of hands (not rounds) to lose 10 in a row
so does (1-(p^run))/(q*(p^run)) where q=1-p

average does not mean 100%, or even 50% - the median is 759

use the 2nd calculator here to see the distribution
Streak Calculators
also looks to be very accurate (one can change the value of p in the Excel calculator)

Sally has some data in tables in this thread
on-average-how-many-trials-will-it-take

why do you want this information?
Good Luck

Thanks for that correction to my equation.

Your comments:
streaks are in attempts
the average length (number of hands or trials) of each attempt for 10 in a row is
2.152778445
(1+p+p^2+p^3+...+p^9)
507.12*2.15 gets the average

1,091.7210
average number of hands (not rounds) to lose 10 in a row
so does (1-(p^run))/(q*(p^run)) where q=1-p

My question:
Thank you for the above formulas, final questions....
P=probability which is my case is 0.0019719127 , correct?
What would the value of "run" =? Is that my example of 10?

Just to quote me, have the closing tag [ / q ] before your reply

and everything to me is in blue

Quote: Byron

My question:
Thank you for the above formulas, final questions....
P=probability which is my case is 0.0019719127 , correct?
What would the value of "run" =? Is that my example of 10?

No
P<> 0.0019719127
p= 0.5364 (probability of a loss)

yes, run = 10 = the length of the streak. your example of losing 10 in a row

p^10 = 0.0019719127
or
1/ p^10 = 507.12 = the average number of attempts (not hands)

hope this all helps