# Computing probability with a biased deck

I have been struggling with this question.

Let's say we have $m$ decks. One deck contains all (52) red cards. The rest of the decks contain 26 red and 26 black cards.

Given that I draw a card and it's red, what is the probability it was drawn from the biased deck (containing all red cards)?

My approach: apply Bayes' rule.

$P(d=b| c=r) = \frac{P(c=r|d=b) P(d=b)}{\sum_i(P(c=r|d_i) P(d_i)}$

But, my answer is in terms of $m$. Is that correct?

• Think for a moment about the qualitative difference between the situation where $m=1$ and $m=1\,000\,000$. Doesn't that convince you the answer must depend on $m$? – whuber Oct 21 '16 at 19:45
• It should depend on $m$. Clearly as $m$ gets larger the probability should tend to zero, for instance. – dsaxton Oct 21 '16 at 19:48
• Are we assuming that the deck is selected randomly (uniformly) from the set of all decks (this seems to be assumed, but should be mentioned). – Juho Kokkala Oct 22 '16 at 6:38

You are exactly correct! Imagine if you had a million decks, the chance that it came from the biased one (1/million) is tiny compared to the 1/2*(1-1/million) $\approx$ 1/2. The total red cards in the others dominate.