If 6 cards are drawn from a deck of cards with replacement. What is the probability that the outcome is a club at least once. I thought it would be 13/52. But I was wrong.
[Math] card drawing with replacement
probability
Related Solutions
The term without replacement means that the deck is different for each draw so the probability of getting a heart on the first draw is $\frac{13}{52}$ as there are 13 hearts in a full deck of 52 cards.
Now given you have already drawn a heart on your first draw the probability of getting a heart on your second draw is $\frac{12}{51}$ as there are now only 12 hearts left in the pack and the pack has 51 cards remaining.
So the probability of hearts on both your first and second draw is $\frac{13}{52} \cdot \frac{12}{51} = \frac{156}{2652} = \frac{1}{17}$
I'm sure you can see from here how you would calculate three hearts.
The term with replacement means that the card is put back and mixed up again after each draw so the probability of drawing a heart on the second draw is $\frac{13}{52}$ because you are still drawing from a full pack of cards.
For two hearts the probability is $\frac{13}{52} \cdot \frac{13}{52} = \frac{169}{2704} = \frac{1}{16}$ and for three hearts ...
Someone else asked a similar question a little while after this one. I'm modifying and reposting my answer from there to here. I've also added an example since it's early in the morning and the baby is still sleeping.
The short answer is $$\frac{{80\brace52}52!}{52^{80}}.$$
Explanation: Suppose we have a deck of $k$ cards, say $C=\{c_1,c_2,\ldots, c_k\}$, and we draw, with replacement, $n\geq k$ of them. Let $(d_1,\ldots, d_n)$ be the sequence of draws (so each $d_j\in C$). I'll call this a drawing.
Obviously there are $k^n$ possible drawings.
In how many drawings is each card chosen at least once? I'll call these good drawings. Well, let $(d_1,\ldots, d_n)$ be a drawing and suppose we have $k$ buckets labelled $\{c_1,c_2,\ldots, c_k\}$. Place $d_j$ into bucket $c_i$ when $d_j=c_i$. Then we know that the drawing is good if and only if each bucket is non-empty.
Now, suppose we have $k$ identical buckets and place symbols $\{D_1,\ldots, D_n\}$ into the buckets, making sure that each bucket gets at least one symbol. We make the convention that two such placements are the same if the buckets can be rearranged to get one from the other. By the definition of the Stirling number of the second kind, there are ${n\brace k}$ ways of doing this. Next, label the buckets with labels $\{c_1,\ldots,c_k\}$. There are $k!$ ways of doing this.
For each of the above placements and labellings, we can then define a good drawing $(d_1,\ldots, d_n)$ by setting $d_j=c_i$ if and only if $D_j$ was placed in bucket labelled $c_i$.
Thus there are exactly ${n\brace k}k!$ good drawings and so the probability that a drawing is good is $$\frac{{n\brace k} k!}{k^{n}}.$$
Example with a deck of $k=2$ cards, say $\{0,1\}$, and $n=3$ draws. In this case a drawing is simply a binary sequence of length 3, and clearly only $(1,1,1)$ and $(0,0,0)$ are not good drawings, so the probability should be $3/4$.
Now let $B_1$, and $B_2$ be the buckets. We want to put symbols $D_1,D_2,D_3$ into these and so there are ${3\brace 2}=3$ possibilities:
$D_1\in B_1, \{D_2,D_3\}\in B_2$ (which we consider the same as placing $D_1\in B_2$ and $\{D_2,D_3\}\in B_1$).
$D_2\in B_1, \{D_1,D_3\}\in B_2$,
$D_3\in B_1, \{D_1,D_2\}\in B_2$.
If $B_1$ is labelled $1$ and $B_2$ is labelled $0$ we get the drawings $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$. The other labeling gives the drawings $(0,1,1)$, $(1,0,1)$ and $(1,1,0)$. As noted above, this is indeed all good drawings in this case and the probability of getting a good drawing is $$\frac{{3\brace 2}2!}{2^3} = \frac{3}{4}.$$
Best Answer
If you don't draw a club at least once, you draw zero.
The odds of not drawing a club on a particular draw is $\frac{3}{4}$. Since you replace the card each time, each of the draws is from a full deck, and so the events are independent. Thus, the chance of drawing a nonclub 6 times is
$$P(\text{no clubs}) = \underbrace{\left(\frac{3}{4}\right) \cdots \left(\frac{3}{4}\right)}_n = \left(\frac{3}{4}\right)^6 = \frac{729}{4096},$$
and so the chance of drawing a least one club in $6$ draws is
$$P(\text{at least $1$ club}) = 1 - P(\text{no clubs}) = 1 - \frac{729}{4096} = \frac{3367}{4096} = 0.82202\ldots .$$