How to calculate session odds with standard blackjack

Hello all,

I am new to the forum, so forgive me if this question has been asked before.

I was wondering if someone could explain how to figure out my odds of winning a certain number of bets before going bust if I am betting flat at $10 with say $200 at the table. So to be more specific, playing by the book at standard 6 deck blackjack table. Whatever rules are most common in Vegas (I don't know them off hand). And also more specifically, being up from my starting balance by 3 bets, 4 bets, etc up to 10 bets. Is that enough information to figure out a calculation?

Thanks in advance!

Hi Bigbrian, and welcome to the forums!

Eh, you've given enough information for something =p. Given what you've stated, let's take a "generic" .5% HE blackjack game. You're going to flat bet $10 per hand and you want to know how many hands you have before you go bust, starting with $200. Well, that's simple. You just need to calculate your Expected Value (EV) after X number of hands and when your EV is -200 you'll be bust...

Normally counters do this the other way around to find their "N0" which is their mathematical break even point... Where even with variance/luck they WILL NOT be down money. So let's get a crackin...

Avg Advantage = -.5%
Avg Bet = $10
OriginalSD = 1.1 * AvgBet = 1.1 * 10 = 11
SD(x hands) = Sqrt(x) * OriginalSD

So what's your EV after, let's say 1,000 hands?

EV(1,000 hands) = TotalWagered * HouseEdge = (NumHands*AvgBet) * HouseEdge = (1000 * 10) * (-.005) = -50

So we could mathematically figure that if you play 4x that number of hands you'll hit your -200 point of being bust:

EV(4,000 hands) = (4000 * 10) * (-.005) = -200... This is the point where if the game worked out mathematically perfect, and there was no luck, that you would bust.... 4,000 hands. I'm sure you realize though that QUITE OFTEN when someone buys in for $200 they go bust the same night (within a couple hours). However, I've seen it both ways where someone buys in for $200, never bets more than $10, and wins $500.

So it's pretty tough to say "how long will I last with $200" due to the fact the game carries so much variance/luck. Anything can, and will, happen in the short run. It's only when you get to the long run (N0, typically 1,000 hours of play) that you can start to phase out this variance and you're left with your EV.

You can also check out this table: https://wizardofodds.com/games/blackjack/appendix/12/ It gives your chance of losing your bankroll completely after a certain number of hands based on your betting units.

https://wizardofvegas.com/member/oncedear/blog/#post1370

Ignores house edge, but mind crushingly simple estimate under most real world conditions.

Quote: Bigbrian000

Is that enough information to figure out a calculation?

Thanks in advance!

yes, it will be very close
here is Alan Krigman's Excel in Google

https://goo.gl/D0pJvD

one has to know about expected value and variance to use it
<<< >>>
here is what I got using it

up $X B4 ruin
# of rounds played not considered.
that requires a different method of calculation

up $30
85.826444%

up $40
81.891506%

up $50
78.272400%

up $60
74.932682%

up $70
71.841311%

up $80
68.971675%

up $90
66.300839%

up $100
63.808920%

up $200 (doubled!)
45.778006%


hope this helps out sum

why the interest in this??

Sally