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I haven't seen quite this scenario on a card drawing problem on here. I'm trying to figure out the probabilities for a card game I'm developing. There are 3 separate decks with 15 cards each. In each deck there are 2 'white cards' let's say and we're interested in drawing those. So if I draw 3 cards from each of the decks, then what is the probability that I draw at least 1 'white card' and not one of the other 13?

I think I calculated the probability for one of the decks so I'll put my work here for someone to check it.

Probability of drawing at least 1 white card when drawing 3 cards from one deck:

First, I computed the probability of drawing exactly 1 white card when drawing 3 cards from one deck, which is as follows.

$P(W_1) = (_3C_1) \left(\frac{2}{15}\right)^1 \left(\frac{13}{15}\right)^2$ $P(W_1) = 0.3004$

Then, I computed the probability of drawing 2 white cards when drawing 3 cards from the deck.

$P(W_2) = (_3C_2)\left(\frac{2}{15}\right)^2 \left(\frac{13}{15}\right)^1$ $P(W_2) = 0.0462 $

So then the probability of drawing at least 1 white card is,

$P(W)= P(W_1) + P(W_2) = 0.3466$

So how do I go about incorporating the other 2 decks into my equation? What's the probability of drawing a white card when drawing 3 cards from each deck? Thanks.

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    $\begingroup$ Do you want at least one white card total? Or do you want at least one white card for every deck? $\endgroup$
    – Zubin Mukerjee
    Commented Nov 12, 2019 at 3:46
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    $\begingroup$ At least one white card total. $\endgroup$
    – Braden
    Commented Nov 12, 2019 at 3:49
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    $\begingroup$ OK. In that case, there is a very quick trick using what some people call complementary counting. I've shown how to use this for your problem, in my answer :) $\endgroup$
    – Zubin Mukerjee
    Commented Nov 12, 2019 at 3:56

2 Answers 2

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Notice that the probability of drawing at least one white card is the same as $1$ minus the probability of drawing no white cards:

$$ P\left(\text{at least $1$ white card}\right) = 1 - P\left(\text{no white cards}\right)$$

The probability that no white cards are drawn when taking three cards from a single deck is:

$$P\left(\text{no white cards from a single deck over three draws}\right) = \left(\frac{13}{15}\right)\left(\frac{12}{14}\right)\left(\frac{11}{13}\right)$$

In order to get no white cards over all nine draws, it's necessary to get no white cards on the three draws from each of the three decks:

$$P\left(\text{no white cards}\right) = \left(\left(\frac{13}{15}\right)\left(\frac{12}{14}\right)\left(\frac{11}{13}\right)\right)^3$$

Therefore,

\begin{align*} P\left(\text{at least $1$ white card}\right) &= 1 - P\left(\text{no white cards}\right)\\\\ &= 1 - \left(\left(\frac{13}{15}\right)\left(\frac{12}{14}\right)\left(\frac{11}{13}\right)\right)^3 = \boxed{\frac{32227}{42875}} \approx 0.75165 \end{align*}

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    $\begingroup$ Note: This is for drawing one card from each deck, not for three cards from each deck". $\endgroup$
    – Graham Kemp
    Commented Nov 12, 2019 at 4:14
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    $\begingroup$ @GrahamKemp Thanks for the correction - I had misunderstood the question. I've fixed the answer :) $\endgroup$
    – Zubin Mukerjee
    Commented Nov 12, 2019 at 4:33
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    $\begingroup$ Nice this is the method my friend used as well! Thank you for taking the time to post. $\endgroup$
    – Braden
    Commented Nov 12, 2019 at 4:39
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So I was messing around with it more and talking to a friend about it. Can anyone tell me if this is correct...

$P(W) = 1 - P(W')$

$ = 1 - \left(\frac{^{13}C_3}{^{15}C_3}\right)^3 $

Which comes out to $P(W) = 0.752$

It just seems quite high to me.

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  • $\begingroup$ That's the same answer I got for 9 cards being drawn, total $\endgroup$
    – Zubin Mukerjee
    Commented Nov 12, 2019 at 4:34

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