# 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?

Lottery chance (odds) calculation
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?
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pawia6g
Step 1
Q2
$1-{\left[0.99\right]}^{2}=1-0.9801=0.0199<0.02=\frac{1}{50}.$
Step 2
$\underset{_}{\text{Explanation of discrepancy between Q1 and Q2}}$
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.
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robbbiehu
Step 1
At least win one game, means: "$A=\text{win 1st lost 2nd}","B=\text{lost 1st win 2nd"},"C=\text{win both}$"

$P\left(A\right)+P\left(B\right)+P\left(C\right)=0.0199$
Step 2
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): ${E}_{2}=0.01\cdot 0.09X+0.09\cdot 0.01X+0.01\cdot 0.01\left(X+X\right)=0.02X$
So the expectations are the same for the two plans.