Fundamental theorem of gambling?

Question

i'm very confused on the fundamental theorem of gambling. It is

N= log(1-DC)/log(1-P)

DC= degree of certainty, N=number of trials, and p= probability

could someone show me how this works? thanks!

Answer 1

I don't know about any theorems but here is the way gambling goes!  Play games with the best odds, Craps and some blackjack games are usually the best!  Avoid the slots like the plague.  Do not take more money than you want to lose!  Leave checkbooks, credit cards, etc at home!  Why playing craps or blackjack, you need your wits!  Avoid the free flowing booze.  Tip your dealer every now and then...they are trying to make a living.

Answer 2

Here is a link to Saliu's page. He is the one who claims to have conceived this formula and that it applies to any gambling event. He has some examples here:

http://www.saliu.com/formula.htm

Personally, I do not see the application as practical.

Gambling is not that easily reduced to a formula, maybe for the lottery of some other event which is pure chance, but I do not think you can apply it to poker for example, or to sports betting such as football games.

For example here is a site for Football picks that uses a different approach.

http://www.football-free-picks.com

Answer 3

You bet, you win or lose, you go home, with or without money.  If you bet the house and lose, you have nowhere to go.

Answer 4

It's simple, but not that practical.

A little algebra will help the math take on a meaning:

N = log(1-DC) / log (1-P)

N log (1-P) = log (1-DC)

(1-P) ^ N = (1 - DC)

Now, if P is the probability of a hit, then 1-P is the probability of a miss.

And, for any probability X, X^N is the probability of X happening N times in a row.  (if the probability of rolling a 1 is 1/6, then the probability of rolling two 1s (er, snake eyes) is (1/6)^2, or 1/36)

Back to the equation:

(1-P) ^ N = (1 - DC)

If P is 1/6, and n is 2, then:

5/6 ^ 2 = 1 - DC

DC = 11/36.  That's the odds that you won't roll a 1 in two tries.  So what?