Lottery chance (odds) calculation

Question

We have two lottery game with 1:1000 odds to win. I have money to buy 20 tickets. Q1. If I spend my money on one of these games only then I have 1:50 chance to win. Am I right? Q2. If I spend my money to buy 10-10 tickets and play both of games then I have 1:100 chance to win on the first and the same chance to win on the second game. But what is my chance to win at least one game this time? I have only 1:100 or 1:50 like in question Q1? If I split my money I split (reduce) my chance too?

Answer 1

Q2
$$1 - \left[0.99\right]^2 = 1 - 0.9801 = 0.0199 < 0.02 = \frac{1}{50}.$$

$\underline{\text{Explanation of discrepancy between Q1 and Q2}}$

Re the strategy in Q2, your overall probability of winning at least one of the two contests has been reduced from $(0.02)$ to $(0.199)$. This reduction is offset by the (very remote) possibility of your winning both contests.

Answer 2

Since @user2661923 has given an answer, another way to think this problem is:

At least win one game, means: "A=win 1st lost 2nd", "B=lost 1st win 2nd", "C=win both"

$$P(A)=0.01\cdot 0.99,~~~P(B)=0.99\cdot0.01,~~~P(C)=0.01\cdot0.01$$

$$P(A)+P(B)+P(C)=0.0199$$

Another interesting fact is if you compare the expectation for these two strategies. Suppose you win $X$ amount money if you win any one game

Plan (1): $E_1=0.02X$

Plan (2): $E_2=0.01\cdot0.09X+0.09\cdot0.01X+0.01\cdot0.01(X+X)=0.02X$

So the expectations are the same for the two plans.