1
$\begingroup$

Suppose you have 2 fair decks, let's call them deck A and deck B. Now take two cards from deck A and add them to the deck B, and shuffle; thus you now have a deck with 54 cards. Now draw 2 cards from deck B, what's the probability that you draw an ace?

EDIT: Let's say you draw exactly 1 ace.

This is kind of an arbitrary question I made up, I don't care so much about the specific answer but about the process. I'm really trying to understand conditional probability and I feel like if I can answer a question like this I'll be on my way.

My guess is that first we calculate the odds that one or both of the cards from deck A were aces. Then you would weigh the possibilities of two aces having been transferred, one ace having been transferred, or no aces having been transfered across drawing from deck B?

Can someone point me in the right direction?

Also as a complete aside, how do you calculate the odds of drawing exactly one ace when drawing two cards from a deck of 52? I've answered the question using complelemntary probability of drawing 2 aces, and no aces and then subtracting that from 1 but is there a "direct" way of doing it? i feel that I understand better when it's not through complements

$\endgroup$
  • $\begingroup$ The approach ("cases") that you sketched seems sensible. As to your second question, there are $\binom{4}{1}\binom{48}{1}$ "exactly one Ace" hands. Divide by $\binom{52}{2}$. $\endgroup$ – André Nicolas Apr 7 '14 at 5:42
  • $\begingroup$ Well I suppose I'm interested in both exactly or at least. I think exactly 1 ace is a more interesting proposition, can you provide a hint for that one and maybe I can try to figure out for at-least as well. And if I can ask is the rationale for your answer, that there are 4 choose 1 ways to pick an ace, and 48 choose 1 ways to pick the second card and then we just divide by all ways of choosing 2 cards from 52? $\endgroup$ – user141068 Apr 7 '14 at 5:46
  • $\begingroup$ The rationale you described is correct. $\endgroup$ – André Nicolas Apr 7 '14 at 5:48
0
$\begingroup$

We transferred $0$, $1$, or $2$ Aces.

The probability $p_0$ that we transferred $0$ is $\frac{\binom{4}{0}\binom{48}{2}}{\binom{52}{2}}$.

The probability $p_1$ that we transferred $1$ is $\frac{\binom{4}{1}\binom{48}{1}}{\binom{52}{2}}$.

The probability $p_2$ that we transferred $2$ is $\frac{\binom{4}{2}\binom{48}{0}}{\binom{52}{2}}$.

Given that we transferred $0$, the probability $a$ we draw exactly one Ace from the modified Deck B is given by $a=\frac{\binom{4}{1}\binom{50}{1}}{\binom{54}{2}}$.

Given that we transferred $1$, the probability $b$ we draw exactly one Ace from the modified Deck B is given by $b=\frac{\binom{5}{1}\binom{49}{1}}{\binom{54}{2}}$.

Given that we transferred $2$, the probability $c$ we draw exactly one Ace from the modified Deck B is given by $c=\frac{\binom{6}{1}\binom{48}{1}}{\binom{54}{2}}$.

Our required probability is $p_0 a+p_1b+p_2c$.

$\endgroup$
  • $\begingroup$ Thank you so much for this answer. It really makes a lot of sense, I really struggle sometimes knowing when to add probabilities but this was a terrific explanation. $\endgroup$ – user141068 Apr 7 '14 at 13:12
  • $\begingroup$ You are welcome. You had basically described the method. This a typical Cases argument. Maybe (i) happens. That has probability $p_{\text{i}}$. If (i) happens, the event we are interested in has (conditional) probability $a$. Maybe (ii) happens, $\dots$. This can be visualized by drawing a tree of possibilities, and calculating the probabilities of various paths. $\endgroup$ – André Nicolas Apr 7 '14 at 15:01
0
$\begingroup$

Compute as follows, letting $Y$ be number of aces drawn from deck B, and letting $X$ be the number of aces drawn from deck A: \begin{align} P[Y = 1] &= \sum_{i=0}^2P[Y=1 \mid X = i]P[X=i] \\ &= \sum_{i=0}^2 \frac{\binom{4+i}{1}\binom{50-i}{1}}{\binom{54}{2}} \frac{\binom{4}{i}\binom{48}{2-i}}{\binom{52}{2}} . \end{align} You can compute $P[Y=2]$ in a similar fashion.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.