e-mail to webmaster.
Let's suppose I play the 3-digit lottery game (pick 3). The game has a total of 1,000 combinations. Thus, any particular pick-3 combination has a probability of 1 in 1,000 (we write it 1/1,000). I also mention that all combinations have an equal probability of appearance. Also important - and contrary to common belief: the past draws do count in any game of chance and Pascal demonstrated that hundreds of years ago. Evidently, the combinations have an equal probability, but they appear with different frequencies. Soon as I choose a combination to play (for example 2-1-4) I can't avoid asking myself: "Self, how many drawings do I have to play so that there is 99.9% degree of certainty my combination of 1/1,000 probability will come out?"
My question dealt with three elements:
The Fundamental Formula of Gambling (FFG) is an historic discovery in theory of probability, theory of games, and gambling mathematics. The formula offers an incredibly real and practical correlation with gambling phenomena. As a matter of fact, FFG is applicable to any sort of highly randomized events: lottery, roulette, blackjack, horse racing, sports betting, even stock trading. By contrast, what they call theory of games is a form of vague mathematics: The formulae are barely vaguely correlated with real life.
Substituting DC and p with various values, the formula leads to the following, very meaningful and useful table. You may want to keep it handy and consult it especially when you want to bet big (as in a casino).
Number of Trials N Necessary For An Event of Probability p to Appear With The Degree of Certainty DC:
DC Ї |
p= .90 |
p= .80 |
p= .75 |
p= .66 |
p= 1/2 |
p= 1/3 |
p= 1/4 |
p= 1/6 |
p= 1/8 |
p= 1/10 |
p= 1/16 |
p= 1/32 |
p= 1/64 |
p= 1/100 |
p= 1/1,000 |
10% | - | - | - | - | - | - | - | - | - | 1 | 1 | 3 | 6 | 10 | 105 |
25% | - | - | - | - | - | - | 1 | 1 | 2 | 3 | 4 | 9 | 18 | 28 | 287 |
50% | 1 | 1 | 1 | 1 | 1 | 1 | 2 | 3 | 6 | 7 | 10 | 21 | 44 | 68 | 692 |
75% | 1 | 1 | 1 | 2 | 2 | 3 | 4 | 7 | 11 | 13 | 21 | 43 | 88 | 137 | 1,385 |
90% | 1 | 2 | 2 | 2 | 3 | 5 | 8 | 12 | 17 | 22 | 35 | 72 | 146 | 229 | 2,301 |
95% | 1 | 2 | 2 | 3 | 4 | 7 | 10 | 16 | 22 | 29 | 46 | 94 | 190 | 298 | 2,994 |
99% | 2 | 3 | 3 | 4 | 7 | 11 | 16 | 25 | 34 | 44 | 71 | 145 | 292 | 458 | 4,602 |
99.9% | 3 | 4 | 5 | 6 | 10 | 17 | 24 | 37 | 52 | 66 | 107 | 217 | 438 | 687 | 6,904 |
Let's try to make sense of those numbers. The easiest to understand are the numbers in the column under the heading p=1/2. It analyzes the coin tossing game of chance. There are 2 events in the game: heads and tails. Thus, the individual probability for either event is p = 1/2. Look at the row 50%: it has the number 1 in it. It means that it takes 1 event (coin toss, that is) in order to have a 50-50 chance (or degree of certainty of 50%) that either heads or tails will come out. More explicitly, suppose I bet on heads. My chance is 50% that heads will appear in the 1 st coin toss. The chance or degree of certainty increases to 99.9% that heads will come out within 10 tosses!
Even this easiest of the games of chance can lead to sizable losses. Suppose I bet $2 before the first toss. There is a 50% chance that I will lose. Next, I bet $4 in order to recuperate my previous loss and gain $2. Next, I bet $8 to recuperate my previous loss and gain $2. I might have to go all the way to the 9th toss to have a 99.9% chance that, finally, heads came out! Since I bet $2 and doubling up to the 9th toss, two to the power of 9 is 512. Therefore, I needed $512 to make sure that I am very, very close to certainty (99.9%) that heads will show up and I win . . . $2! Very encouraging, isn't it? Actually, it could be even worse: it might take even 10 or 11 tosses until heads appear! This dangerous form of betting is called a Martingale system. Beware of it! Normally, though, you will see that heads (or tails) will appear at least once every 3 or 4 tosses (the DC is 90% to 95%). Nevertheless, this game is too easy for any player with a few thousand dollars to spare. Accordingly, no casino in the world would implement such a game. Any casino would be a guaranteed loser in a matter of months! They need what is known as "house edge" or "percentage advantage". This factor translates to longer losing streaks for the player, in addition to more wins for the house! Also, the casinos set limits on maximum bets: the players are not allowed to double up indefinitely.
Dice rolling is a more difficult game and it is illustrated in the column p=1/6. I bet, for example, on the 3-point face. There is a 50% chance (DC) that the 3-point face will show up within the first 3 rolls. It will take, however, 37 rolls to have a 99.9% certainty that the 3-point face will show up at least once. If I bet the same way as in the previous case, my betting capital should be equal to 2 to the power of 37! It's already astronomical and we are still in easy-gambling territory!
Let's go all the way to the last column: p=1/1,000. The column illustrates the well-known 3-digit lottery game. It is extremely popular and supposedly easy to win. Unfortunately, most players know little, if anything, about its mathematics. Let's say I pick the number 2-1-4 and play it every drawing. I only have a 10% chance (DC) that my pick will come out winner within the next 105 drawings! The degree of certainty DC is 50% that my number will hit within 692 drawings!
We don't need to analyze the lotto games. The results are, indeed, catastrophic. If you are curious, simply multiply the numbers in the last column by 10,000 to get a general idea. To have a 99.9% degree of certainty that your lotto (pick-6) ticket (with 6 numbers) will come out a winner, you would have to play it for over 69 million consecutive drawings! At a pace of 100 drawings a year, it would take over 690,000 years!..
Taken from SALIU.COM without permission! - IS
Comments: