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You have a deck with 10 red cards and 10 black cards. Then you add one more card -- unknown whether red or black.

The 21 cards are shuffled. You pick randomly and pull a red card. How does that change the probability that the unknown card is a red card?

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closed as off-topic by Michael Chernick, Sycorax, Peter Flom Aug 29 '18 at 11:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ +1. Welcome to our site. Some would deny that probability applies to the question, because not all unknown events have probabilities: sometimes its better just to admit they are unknown and see what that implies. (There nevertheless remain informative ways to analyze the situation.) Others will assert probability does apply, but in order to produce an answer most (if not all) of those will have to add an explicit assumption about what the probability was before a card was drawn from the new deck. $\endgroup$ – whuber Aug 28 '18 at 16:11
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Let us believe (for the duration of this answer) that probability does apply to this question. Bayes' Rule gives us the optimal mechanism for updating our prior (to picking a card) beliefs about the probability that the added card was red based upon our observed draw of a red card.

We'll denote our prior probability that the card is red by $\pi$ and our posterior (after seeing the results of the card we picked) probability by $\pi'$. The update from $\pi$ to $\pi'$ works as follows:

$$\pi' = {p(\text{draw red}|\text{red card})\cdot\pi \over p(\text{draw red}|\text{red card})\cdot\pi + p(\text{draw red}|\text{black card})\cdot(1-\pi)} $$

In this case, the probability that we draw a red card if the added card was red is $11/21$, and the probability that we draw a red card given that the added card was black is $10/21$. Substituting gives:

$$\pi' = {11\cdot\pi \over 11\cdot\pi + 10\cdot(1-\pi)} $$

where we have eliminated the division by $21$ from the top and the bottom through the usual process of cancellation. For concreteness, if we have a prior probability that the card is red of $\pi = 0.5$, the posterior probability that the card is red is:

$$\pi' = {11\cdot 0.5 \over 11\cdot 0.5 + 10\cdot 0.5} = 0.52381...$$

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