So a street con sets up a small stand and challenges people to play a game.

He offers 500 dollars if you win but if you lose you have to pay him 250.

He has the 4 aces on a table and asks you to shuffle them and put them on the table side by side.

then you have to pick 2 cards and both of them have to be the same color, if you can do this 2 out of 4 times then you win. He claims that it's a fair game since there is a "50%" chance that you will pick up both same colored cards and since you only need to get 2 out of 4 times correct then it's a fair game.

so the question is-

What is the probability that you will pick both colored cards on your first try?

What is the probability that you will pick up both colored cards in your first and second try? Winning the game by not making a single mistake.

what is the probability of you winning the game even if you make 2 mistakes, thus getting 2 out 4 tries correct?

Is this a fair game?

The way I figured is that since the first card you will pick will be either red or black then you have a 100% chance of picking either but your second card there will be 3 cards left. So there is only a 1 in 3 chances of picking the same colored card so I guess 33.33%?

So the chance of winning the game is only 33.33%

If you try to get the same 2 colored cards on your next try would the percentage of that be half of 33.33%?

Winning without making a mistake seems like a slim chance so it's not a fair game.

Overall winning without making a mistake would be half of 33.33%?

That's the part where I'm stuck.any input would be much appreciated thank you.