Probability Question on picking up tiles

Question

A game involves picking up to $4$ tiles out of a bag that contains $20$ tiles. The bag contains $4$ tiles with the number $1$ printed on them, $6$ with $2$ on them, $3$ with $3$ on them and $7$ with the number $4$ printed on them.

You pick the tiles one at at time. If the $1^{\text{st}}$ tile you choose has a $1$ on it then you win. If the $2^{\text{nd}}$ tile has a $2$ then you win and similarly for the remaining choices. After each choice, the tile you picked is returned to the bag.

What is the probability of winning this game?

So far I have: $\frac{4}{20} + \frac{6}{20} + \frac{3}{20} + \frac{7}{20}$ as the tiles are returned to the bag. Not sure if I'm on the correct track though

Answer 1

There are four ways to win:

1. You draw a 1 on the first draw, with probability $4/20$.
2. You fail to draw a 1 on the first draw, with probability $16/20$, and then draw a $2$ on the second draw with probability $6/20$. Since these choices are independent, the probability that this occurs is $\frac{16}{20} \cdot \frac{6}{20}$.
3. You fail to draw a 1 on the first draw, fail to draw a 2 on the second draw, and then draw a 3 on the third draw. What is the probability this occurs?
4. You fail to draw a 1 on the first draw, fail to draw a 2 on the second draw, fail to draw a 3 on the third draw, and then draw a 4 on the fourth draw. What is the probability this occurs?

These four events are mutually exclusive and exhaustive. Add their probabilities to determine the probability that you win.

Answer 2

The problem with the way you're doing it is that, for example, the probability of winning on the second draw is not $\frac{6}{20}$ but $\frac{16}{20}\cdot\frac{6}{20}$ because in order to win on the second draw, you must lose on the first draw.

An easy way to do this problem is to notice that the probability of winning is $1$ minus the probability that you lose on all four draws. That is $$1-\frac{16\cdot14\cdot17\cdot13}{20^4}$$

Answer 3

Would you be able to find the probability of winning if the chosen tiles were not placed back into the bag again?