The peril to the gambler, is not that he has the odds against him, but that his pocket is not infinitely deep. Let's take the example of a simple gamble process. Each time the bid is 1. You win, your pocket gains 1, you lose, your pocket must give up 1. The gambling mechanism has an advantge of alpha to the gambler versus the house.
Now if the game is fair, no matter how many times the game is played (N), the expectation of the total gain through N plays is always 0. However, the expectation value of the absolute value of the gain is determined by the Central Limit Theorem (CLT), that is, sqrt(N). Since this can be either positive or negative, we have about 1/2 the chance of getting something close to sqrt(N), or -sqrt(N). So in a fair game, if you play N times, N being a sufficiently large number (say at least 100), you have about 1/2 chance of winning sqrt(N). However, before that, your gain might have gone to the otherside first, and the likelyhood that your maximum loss (before your fortune turns) is also sqrt(N). In other words, in order to run long enough to win sqrt(N), your pocket needs to be at least as deep as sqrt(N) to avoid being wiped out.
In a favorable game, alpha>0, E(N)=alpha N. Your likely best and worst case scenarios are alpha N +/- sqrt(N). So if you play long enough you accumulate your gains and eventually it is virtually impossible for you to lose. This threshold can be find by alpha N=sqrt(N), or N=1/alpha^2. For example, if your favor is 2%, you need to play 2500 times to be on safe ground. But before you reach this safe ground, you still need to have reserve to handle possible worst situations. The worst situation is solved by taking d(alpha N-sqrt(N))/d N = 0, or N=1/(2 alpha)^2. You would need to have 1/(4 alpha) to pass this vulnerable phase of play.
In an unfavorable game, the longer you play, the more likely you are going to lose. However, just as the reverse of the previous paragraph, there is a sweet spot N=1/(2 |alpha|)^2, where you have a reasonable chance of winning as much as 1/(4|alpha|). To play more than this number of games is likely to reduce your possible winning.
Now if the game is fair, no matter how many times the game is played (N), the expectation of the total gain through N plays is always 0. However, the expectation value of the absolute value of the gain is determined by the Central Limit Theorem (CLT), that is, sqrt(N). Since this can be either positive or negative, we have about 1/2 the chance of getting something close to sqrt(N), or -sqrt(N). So in a fair game, if you play N times, N being a sufficiently large number (say at least 100), you have about 1/2 chance of winning sqrt(N). However, before that, your gain might have gone to the otherside first, and the likelyhood that your maximum loss (before your fortune turns) is also sqrt(N). In other words, in order to run long enough to win sqrt(N), your pocket needs to be at least as deep as sqrt(N) to avoid being wiped out.
In a favorable game, alpha>0, E(N)=alpha N. Your likely best and worst case scenarios are alpha N +/- sqrt(N). So if you play long enough you accumulate your gains and eventually it is virtually impossible for you to lose. This threshold can be find by alpha N=sqrt(N), or N=1/alpha^2. For example, if your favor is 2%, you need to play 2500 times to be on safe ground. But before you reach this safe ground, you still need to have reserve to handle possible worst situations. The worst situation is solved by taking d(alpha N-sqrt(N))/d N = 0, or N=1/(2 alpha)^2. You would need to have 1/(4 alpha) to pass this vulnerable phase of play.
In an unfavorable game, the longer you play, the more likely you are going to lose. However, just as the reverse of the previous paragraph, there is a sweet spot N=1/(2 |alpha|)^2, where you have a reasonable chance of winning as much as 1/(4|alpha|). To play more than this number of games is likely to reduce your possible winning.
选择“Disable on www.wenxuecity.com”
选择“don't run on pages on this domain”
