Calculating the probability of guessing 3 card draws out of 6 My friend guessed the correct card drawn from a full deck of 52, 3 times out of 6 guesses (card not replaced each time).
I'm trying to work out the probability. My thinking is that if you guessed right the first 3 out of 6 times it would be:
1/52 (correct) * 1/51 (correct) * 1/50 (correct) * 48/49 (wrong) * 47/48 (wrong) * 46/47 (wrong)
But then there are 20 different ways to choose 3 from 6 so I multiply the outcome by 20. Does that give me the approximate answer?

In case I'm not being clear:
There are 52 cards and a guess is made and a card drawn.
Now there are 51 cards and a different guess is made and a card is drawn.
This happens 6 times and out of those 6 times, the correct guess is made 3 times. What are the chances of that?
 A: In your example the first three guesses are right and the last three are wrong. But if the first three guesses were wrong and the last three were right, then you'd get a (slightly) different answer because the probabilities are all a bit different from each other. So you can't just multiply by 20 and get the correct answer (although it won't be horribly wrong for your particular question).
Instead you need to compute the probability of each of the 20 possibilities and add them up. The answer I get is
\begin{equation}
\frac{151999}{977208960} \approx 1.55544 \times 10^{-4} . 
\end{equation}
This is about as likely as guessing 12 or 13 coin flips in a row.
Additional detail
Here is some additional detail on how I thought about the problem. Let
\begin{equation}
y = (y_1,\ldots,y_6) \in \{0,1\}^6
\end{equation}
be a collection of independent Bernoulli variables:
\begin{equation}
p(y) = \prod_{i=1}^6 \textsf{Bernoulli}(y_i|\theta_i) ,
\end{equation}
where $p(y_i=1|\theta_i) = \theta_i$ and
\begin{equation}
\theta_i = \frac{1}{53-i} . 
\end{equation}
There are $2^6 = 64$ points in the probability space. Evaluate $p(y)$ at each of the points for which $\sum_{i=1}^6 y_i = 3$ and add the results up.
A: Hypergeometric distribution. The probability of drawing all three of three specified cards in six draws without replacement is a hypergeometric probability:
$$\frac{{3\choose 3}{49\choose 3}}{{52\choose 6}}.$$
This can be evaluated in R using the hypergeometric PDF
function dhyper as follows:
dhyper(3, 3,49, 6)
[1] 0.0009049774

choose(3,3)*choose(49,3)/choose(52,6)
[1] 0.0009049774

Simulation using sampling without replacement.
This probability is a little too small to simulate gracefully,
but a million iterations give a reasonable approximation by simulation.
set.seed(2022)
deck = 1:52
Three.of.6 = replicate(10^6, sum( sort(sample(deck, 6))[1:3]==1:3 )==3)
mean(Three.of.6)
[1] 0.000891       # aprx 0.0009
2*sd(Three.of.6)/1000
[1] 5.967267e-05   # aprx 95% margin of simulation error.

The logical vector Three.of.6 has a million TRUEs and FALSEso, its mean is the proportion of its TRUEs.
Thus the simulation gives $0.00089 \pm 0.00006,$ which
includes $0.0009.$
Note: I do not completely understand your attempted solution,
but there is more than one way in which to choose the three
required cards. Also, I think you may be mixing ordered
and unordered outcomes.
In the hypergeometric formula,
numerator and denominator both count unordered outcomes. By contrast, my simulation keeps track
of order.
Among the pages linked as 'Related' in
the margin of this question, you may find the two
on "poker" to be useful.
