Math question- Need help answering this

[Iceman]

Trying to figure this out. Wondering if someone can help me out here.

Say someone has a 2% edge. Basically winning at a 53.5% rate versus -110 lines. What is the pct chance he would be ahead/behind one unit after 2,000 bets?

Not sure I worded this correctly but me and another guy were talking about this earlier and trying to figure this out. Can anyone help me with this?

Basically is there a way to put a number on ones success rate (being ahead) after 2,000 plays if he is playing with a percieved 2% edge?

Thanks

[mathdotcom]

http://www.stat.tamu.edu/~west/applets/binomialdemo.html

Roughly speaking
n = 2000
p = 0.52
Prob X is at least 1000

96.5%

[Iceman]

http://www.stat.tamu.edu/~west/apple...omialdemo.html

Roughly speaking
n = 2000
p = 0.52
Prob X is at least 1000

96.5%

Wow really that high? I knew it was high but not that high. If you are correct that could be very reassuring. Thank you very much.

Anyone else? Does this sound/look about right?

[donjuan]

Roughly speaking
n = 2000
p = 0.52
Prob X is at least 1000

96.5%

LOL?

[melcon]

Hi,

obviously 1000 wins are not enough at odds of -110 (which is 1.90909 in decimal odds).

So you ask to have a positive result after these 2000 games.

As a Win pays you 0.90909 net return and a loss -1 you ask how you have to choose number of wins at least so that.


W * 0,90909 - L > 0

as you are expected to win OR lose you can replace L by N - W with N being the number of games you are betting on.

W * 0,90909 - (N-W) > 0

<=> W*(1,90909) > N
<=> W > N / 1,90909


with N = 2000 W has to be 1048 to show a small profit.


Now you can use the calculator above to calculate the result with

N = 2000
p = 0.535

at least 1048...

This turns out to be 84,34%.


Hope that helps