suppose there two identical lottery tickets named A and B . A man thinking to get two tickets, either one from each or both same type
what way he will have best wining chances, is them have same chance?
from my knowledge it is same $P(A)=\frac 1x+\frac 1x$ and $P(B)=\frac 2x$ where $x$ is the all number of permutations of a ticket
advance statistical answer expected. thanks
This is my interpretation of the question (please correct me if I'm wrong).
There are two different lotteries, lottery A and lottery B. Any given ticket from each lottery has probability $\frac1x$ of winning. A man has a choice to buy one ticket for each lottery, or two tickets for lottery B. Which gives him a greater chance of winning?
If we are talking about chance of winning, the answer is to buy two from the same lottery. That has probability $\frac2x$ of winning, as you identified. But if he has one ticket for each lottery, the two events aren't mutually disjoint -- it's possible to win both tickets -- so you can't just add the probabilities as $\frac1x+\frac1x$. Instead you have to add the probabilities and then subtract the probability that both events occur, which, since they are independent, is $\frac1x\cdot\frac1x=\frac1{x^2}$. Thus the probability of winning in this case is $\frac1x +\frac1x -\frac1{x^2} = \frac2x-\frac1{x^2} < \frac2x$.
However, note that (assuming each lottery has the same jackpot) while the one-from-each strategy gives him a lower probability of winning, it also gives him a small ($\frac1{x^2}$) probability of winning twice as much, which is impossible under the two-tickets strategy. If we're computing his expected winnings, these two effects will cancel each other out, and the expected winnings will be the same.