Your question is a simple random process problem, and the truth is that the reason gamblers lose everything is not necessarily because their winning rate is too low, but rather likely because they have too little capital and continue to bet. Therefore, as long as your money is insufficient, following the gambler to bet against them will also lead to losing everything.
For instance, a gambler has a 50% chance of winning or losing, but he starts with a relatively small capital, while the casino's funds are nearly infinite compared to him. For this gambler, as long as he keeps betting, the final outcome is almost certain to be losing all: if your money is insufficient, even if you bet against this gambler, your winning probability is only 50%. At this point, you are just another gambler; your final outcome will either be losing everything before this gambler does, or not losing everything before this gambler does, and then losing everything while continuing to bet against the next gambler.
In other words, as long as the gambling continues until one side loses everything, the determination of who ultimately loses or wins not only depends on winning rates but also largely on capital. Even if your winning rate is 50-50 against the opponent, if the opponent has infinite money, you will inevitably be the one who loses everything in the end.
How to understand this mathematical problem? We examine a simple mathematical model, suppose there is a gambler who starts with 100, bets 10 each time, and each bet has a 50% chance of winning. Winning 10 becomes 20, losing 10 results in nothing.
This process is actually a random walk, which can be simplified to starting at position 0 on a one-dimensional axis, where you can only move left or right. Each time you move, you randomly toss a coin: heads means move one step to the right, tails means move one step to the left, giving a 50% chance of moving left and a 50% chance of moving right, and reaching -10 means losing everything.
Next, I will explain with a relatively simple and straightforward method: if you keep walking, the probability of reaching -10 is 1; that is, if the money on the opposite side is infinite, and the gambler keeps betting, the probability of losing all is 1, i.e., almost surely losing everything.
First, we can easily prove: starting from any position x, there must exist a natural number k such that after walking more than k steps, the probability of having reached -10 at least once in the path is greater than 40%.
Why is this proposition correct? Because:
Assuming we start from 0, then
1: In this game, we can only take one step at a time, we cannot jump, thus if you reach any point to the left of -10, it indicates that you must have reached -10 at least once. This means that if the number of tails is more than the number of heads by more than 10 times, it necessarily implies you are to the left of -10. Since we cannot jump, this means you must have reached -10.
2: This is a game where the direction of movement is determined by tossing a coin, and the probability of the number of heads being greater than or equal to the number of tails is certainly 50%.
3: If we toss coins infinitely, the probability of the number of tails being exactly one more than the number of heads must tend to zero, because the growth of combinations is slower than the exponential function of 2.
4: Similarly, it can be easily concluded that as the number of coin tosses increases, the probabilities of the number of tails being exactly two, three,... up to nine more than the number of heads all tend to zero.
5: The probability of the number of tails exceeding the number of heads by more than 10 must equal 1 minus the probability of the number of heads being greater than or equal to the number of tails minus the probabilities of the number of tails being exactly one to nine more than the number of heads. Therefore, as the number of coin tosses increases, the probability of tails exceeding heads by more than 10 must approach 50%.
6: Since the probability of the number of tails exceeding the number of heads by more than 10 times tends to 50% as the number of coin tosses increases, then as the number of tosses increases sufficiently, the probability of tails exceeding heads by 10 must exceed 40%.
7: Once the number of tails exceeds the number of heads by more than 10 times, you must have reached -10. Therefore, starting from any position x, there must exist a natural number k such that after walking k steps, the probability of having reached -10 at least once is greater than 40%.
Likewise, assuming we start from +1, the probability of the number of tails exceeding the number of heads by more than 11 times will also tend to 50% as the number of coin tosses increases. Assuming we start from +2, the probability of exceeding by more than 12 times must also tend to 50%... and so on. Starting from any point, if we keep walking, the probability of reaching -10 must tend to 50%.
Therefore, starting from any position x, there must exist a natural number k such that after walking more than k steps, the probability of having reached -10 at least once in the path is greater than 40%. We can express k as a function of x, k(x).
Next, we can prove that starting from 0 (or any positive position), continuing infinitely will yield a probability of 1 for reaching -10.
Clearly, starting from 0, after walking k(0) steps, the probability of having reached -10 is greater than or equal to 40%. Assuming that after walking k(0) steps we have not reached -10, then this probability is less than 60%. Assuming after walking k(0) steps we stop at some point x1 to the right of -10, then starting from x1 and walking k(x1) steps, the probability of having reached -10 must again be over 40%. Assuming we have not reached -10, but instead stopped at point x2 to the right of -10, then if we continue walking k(x2) steps, there will inevitably be over 40% probability of having reached -10...
Similarly, the probability of not reaching -10 even if we continue infinitely must be less than or equal to (1-40%)(1-40%)(1-40%)..., thus the probability of not reaching -10 if we continue infinitely is zero.
Therefore, if we keep going infinitely, the probability of reaching -10 must be 1.
We know that unlike walking, a gambler who loses all their money cannot continue gambling, and we have proven that the probability of walking infinitely to reach -10 is 1. Therefore, if the casino has nearly infinite money relative to the gambler, the probability that the gambler will lose everything must be close to 1.
This proves that even if the gambler's wins and losses are 50-50, they will still end up losing everything. If you find a gambler who wins and loses at a 50-50 rate, then betting against them gives you no advantage in winning probabilities; your betting outcomes will also inevitably be 50-50. Your capital is not as much as the casino's, so if you keep playing like this, the outcome will likely result in losing everything.
