Let's say the odds of winning in a single game is P, where P is between 0 and 1 (eg. 0.42 for Blackjack, which has roughly 42% chance of winning a hand).

How do you calculate the probability you will be ahead after N games? Assuming the same bet each game and an endless bankroll?

Sum of probabilities for successful trials above n/2 for n trials with a binomial distribution. You can implement this in a spreadsheet or other programme of your choosing easily.


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Thanks for the reply but I still don't know how to calculate that. Is there a formula or algorithm?

The formula for the binomial distribution is well-known and is available in virtually all spreadsheets and computer programming languages. This gives the prob of observing exactly X successes out of N trials (where X can take on values 0,1,2,...,N) when the prob of success of each trial is given by P. It is typically called something like BINOM.

As the above poster said, what you seek is the sum of the binomial distribution values for all X greater than N/2. Unfortunately, there is no "closed form" formula for this. However, many spreadsheets and computer programming languages have a special expression for the sum of binomial probs from 0 to Y. It is typically called something like CUMBINOM.

Since you want the probs for values greater than N/2, you seek the complement sum probs ranging from N down to N/2. This is simply one minus the sum of the probs ranging from 0 to [N/2]-1. So you'd want something like 1-CUMBINOM.

To summarize, what you seek is a simple sum of binomial probs. This is typically available as the complement of an expression given in many spreadsheets and programming languages.

Hope that makes sense.

There are many ways to find the probability of a net win after N trials and not just 1.

IF you are looking to use Blackjack, one can to use a formula by D. Schlesinger from Blackjack Attack (I use it in Excel) that uses variance and normal distribution or it can be found in a calculator over at qfit.com.

What game exactly are you looking for or does it have to be a game?

another said this
"To summarize, what you seek is a simple sum of binomial probs.
This is typically available as the complement of an expression given in many spreadsheets and programming languages."

Just 1 method that works for some games and a very incomplete answer, imo.

One can use the normal distribution, simulations and transition matrix type solutions just to name a few more methods, not to mention simulations.

You really need to be more specific as to the type of win and game and IF exact probabilities are required or not.

The below answers using a binomial probability solution assume
equal $$$ bets every time, same probability of winning each time
and only a win or lose result
and enough $$$ to make each and every wager.

Here, a few examples I did real fast using a program
as doing this by long hand, for me, is a waste of time and very boring.
closer look here:
even money bets payoff
00 Roulette
0.2650235 (51+ wins)
51 wins is N/2 +1 because 50 wins and 50 losses results on a 0 net gain.
you asked for "the probability you will be ahead after N games?"
1k trials: 0.0447959 (501+ wins)

using a calculator here:
binomial probability calculatorilities

enter:
n=100
k=51
p=18/38

result:
P: 51 or more out of 100
0.265023450545
****
2 to 1 bets payoff
00 Roulette
0.3357928 (34+ wins)
100/3 = 33 wins. that still results in a net loss.(66 + -67 = -1)
N/3 +1 = 34 wins. (68 + -66 = 2)
back to the calculator
n=100
k=34
p=12/38

result:
P: 34 or more out of 100
0.335792839527


Both results in agreement with my results from my program(s) I use.
<<<<< >>>>>
I can go on without possibly answering your question, so I ask you to be way more specific.
As one can see the pass line bet at Craps with odds is even more fun to calculate.

Oh yes, I forgot that with blackjack, one can also take the result distribution (like in video poker) and use a transition matrix to calculate the result, as one can win 1 , 1.5 or 2 or more units each round as well as pushing and losing 1 or 2 units each round. way more challenging to calculate. many just simulate many sessions of play to get a good approximation.

Computers do well computing if the program is written properly, imo.

I also like to know the probability of being 'down' after N trials
because
ending up 'even'
after N trials
sometimes 'feels' like a 'win'

hope this helps more than it hurts

That's exactly what I'm looking for, thanks! Trying to show someone the futility of betting in -ev games in the long term, or why the house always wins.

well, have fun with it! The basic idea is the longer one plays against a house edge the less chance they have of coming out even or ahead after N trials.

I think single 0 Roulette has this to offer.
max 35 lifetime spins
n=35
k=1
p=1/37
P: 1 or more out of 35
0.616710268465 (not bad!!)
(flat betting each spin)

should be easy to see that 1 win makes for a 1 unit net lifetime win
35 + -34 = 1
exactly 2 wins results in a 37 unit win
probability about 0.2440675
or close to 1 in 4

hey, i say you got the 1 win, go for it!
have fun!