tito295
Posted by tito295
Apr 22, 2011

modified stragedy with martindale betting

I would like someone to run this through a computer suppose you never split cards nor double down in blackjack, but play according to the rest of the basic stragedy what would be the outcome using the martindale betting system

Comments

teddys
teddys Apr 22, 2011

bankroll reduces to zero.

AZDuffman
AZDuffman Apr 23, 2011

once is a typo, twice it is an error. not "martindale" but "martingale." The only way martingale works is infinate bankroll, infinate time, and no betting limits by the other side.

teddys
teddys Apr 23, 2011

The "Martindale" system is a where you hire a highly-rated lawyer who will charge $400/hr. to tell you your system doesn't work.

guido111
guido111 Apr 23, 2011

http://wizardofodds.com/askthewizard/blackjack-houseedge.html (about half-way down)



What is the added house edge in blackjack if players are never allowed to double nor split?

Answer:

No splitting costs the player 0.58%.

No doubling costs the player 1.47%.

No splitting nor doubling costs the player 1.91%. April 3, 2005



Ouch. Hard to throw away almost 2% edge.



No need to run your idea thru a computer, has already been done by many others.

At another site where the Wizard has used their simulation software, see:



What does a Martingale session look like?

http://www.blackjackincolor.com/useless4.htm



Sure, the martingale will have many, many small wins and a few very large losses that can and will wipe out all the gains from the winning sessions.

Good Luck

rdw4potus
rdw4potus Apr 25, 2011

No one else has jumped on this, so I'll go there. This idea really is a stragedy...

tito295
Posted by tito295
Apr 18, 2011

probabilaty

Simple question, I would like to know, if I never split in blackjack but follow the rest of the basic chart, how many times will the house win ten games in a row, out of 1 billion hands I would like to know number of times not percentage

Comments

rdw4potus
rdw4potus Apr 18, 2011

750,000 times.

FleaStiff
FleaStiff Apr 18, 2011

Plan on playing a billion hands anytime soon?

buzzpaff
buzzpaff Apr 20, 2011

1,217.45362098761

tito295
tito295 Apr 21, 2011

How did you come up with your number?

guido111
guido111 Apr 23, 2011

To expand on 'rdw4potus' answer of 750,000 times.



1 billion hands:

the average number of a dealer win streak of 10 or more: 755,590.77

the average number of "exactly" a dealer win streak of 10: 358,905.6

the average number of a dealer win streak of 11 or more: 396,685.15



The expected number of total dealer runs is 249,375,000.276 [=p^2+(p*q*n)] (using p=.525 for a dealer win, q=.475 for a player win, n=# of hands)



I can quickly calculate, in a spreadsheet, the # of runs for exactly 1 thru 9 and subtract that total from the total # of dealer runs to arrive at the 755,590.77 average.



Example: # of runs of length 1 = p*q^2*n (qpq)

# of runs of length 2 = p^2*q^2*n (qppq)

# of runs of length 3 = p^3*q^2*n (qpppq) and so on.

Formula: r= run length [= p^r*q^2*n]

This arrives at very accurate results for high values of "n".

There is an exact formula from the WoO site that I can post if you would like to see.

1	118453125.3

2 62187890.7

3 32648642.58

4 17140537.34

5 8998782.094

6 4724360.595

7 2480289.31

8 1302151.886

9 683629.7397





Another method would be to multiply the # of total dealer runs (249,375,000.276) by:

0.00302994 for 1 billion hands

0.00302991 for 1 million hands

0.00302967 for 100,000 hands

0.00302721 for 10,000 hands

0.00300270 for 1,000 hands

This is a combination of 2 formulas used in Excel if one wants to use percentages.