In Texas Hold’em does flipping the river before the turn change the game? [duplicate]

Solution 1:

There are two cases here:

Case 1: First card chosen is a queen

$$\frac{4}{52}*\frac{3}{51}=\frac{1}{221}$$

Case 2: First card chosen is not a queen.

$$\frac{48}{52}*\frac{4}{51}=\frac{16}{221}$$

Adding both the cases, we get $\frac{17}{221}$ = $\frac{4}{52}$ = $\frac{1}{13}$

Solution 2:

Think about it this way: Shuffle a deck of cards randomly. The probability of drawing a queen as your second card is the same as the probability that the second card in the deck is a queen, which is clearly 4/52.

Solution 3:

A slightly more intuitive way of looking at this:

The probability that the second card is a queen should be the same as the probability that the second card is an ace, and the same as the probability that the second card is a 2 etc. There are $ 13 $ possibilities for the card number/letter, so the answer is $ \frac{1}{13} $

Solution 4:

You can draw a pair of cards by drawing the first card, then drawing the second card. Let's call these cards A and B. You're interested in the probability that card B is a queen.

Now consider a different experiment: draw a pair of cards as before, but this time call the first one card B, and the second one card A. I claim that these two experiments are identical. The reason is that for any two cards X,Y, the probability to draw X then Y is the same as the probability to draw Y then X.

The second experiment makes it clear that the probability that card B is a queen is 4/52, since there are 4 queens out of 52 cards.