# Which is the better deck of cards?

You have two decks of cards: a 52 card deck (26 black , 26 red) and a 26 card deck (13 black, 13 red). You randomly draw two cards and win if both are the same color. Which deck would you prefer? What if the 26 card deck was randomly drawn from the 52 card deck? Which deck would you prefer then?

The first part seemed easy since $\dfrac{25}{51} > \dfrac{12}{25}$. But the second part seems more complicated. Then the probabilities for drawing both reds or blacks would be different so how would we compare them?

• The first question: You prefer the bigger deck, because it is "closer" to infinite - that is, the second draw is less skewed towards the other suit. The second is no difference - it is the same as picking two cards at random from the 52-card deck. – Thomas Andrews Nov 27 '15 at 23:56
• For the skewed deck, let there be $20$ cards that are red and $6$ cards that are black. The probability here is $\dfrac{20}{26} \cdot \dfrac{19}{25}+\dfrac{6}{26} \cdot \dfrac{5}{25} > \dfrac{25}{51}$. This is greater than the probability of a greater deck so how is there no difference? Also, when we take the cards out of the $52$ card deck we are assuming we replace them after, correct? – user19405892 Nov 28 '15 at 0:03
• Because the process is equivalent to shuffling the deck and then taking the top 26 cards, then taking the top two cards. That is no different from just taking the top two cards. – Thomas Andrews Nov 28 '15 at 0:44

The answer is the same as the first part: the bigger deck, because no new information is revealed to you when a bunch of the cards are thrown out of the big deck.

Perhaps it is simpler to see if you think about what is happening when you randomly choose two cards. It's kind of like you are throwing out 50 of the original 52.

It doesn't make a difference if you take a break after throwing out the first 26 cards, then throwing out another 24 to 'choose two.' The distribution is the same as it would be if you were choosing from 52.

expanding to something more formal...

Let the set of cards $C=\{R_1,\dots, R_{26}, B_1,\dots, B_{26}\}$ where reds are $R=\{R_1,\dots,R_{26}\}$ and blacks are $B=\{B_1,\dots,B_{26} \}$. Let $\mathcal{T}_i$ be the set of $i$-card decks for $i\geq 2$: $$\mathcal{T}_i=\{T \subseteq C: \ |T|=i\}.$$

Let $\mathcal{D}_S$ be the set of 2-card draws from $S$, so that $$\mathcal{D}_T=\{D \subset T:\ |D|=2\}.$$

Now we are interested in the space:

$$\Omega_i = \underset{T\in\mathcal{T}_i}{\prod} T \times D_T$$

Everything is chosen randomly, so we know if $T\in \mathcal{T}_i$ then $P(T)=1/{|C| \choose i}.$ Also, if $D\in \mathcal{D}_T$ then $P(D|T)=1/{i \choose 2}.$ It is easy to see that this induces a unique probability measure on the entire space. In particular, $P(T,D)=1/\left({|C| \choose i}{i \choose 2}\right)$ if $D\in \mathcal{D}_T$ and 0 otherwise.

Let $X_i$ be a RV on $\Omega_i$ where if $(t,d)\in \Omega_i$ then $X_i(t,d)=1$ if $d\subseteq R$ or if $d\subseteq B$, and 0 otherwise.

You've already determined that the probability that two guys are the same color choosing from the specific deck of 52, $S=\{R_1,\dots, R_{13},B_1,\dots, B_{13}\}$ is $25/51$.

We are also interested in the probability that two guys are the same color choosing from an arbitrary deck of size 26. Or: $P(X_{26}=1).$ This can be rewritten as:

\begin{array}{rl} & P\left(\cup_{T\in\mathcal{T}_{26}}\left\lbrace D \subseteq \mathcal{D}_T\left| \ D \subseteq R \lor D \subseteq B\right. \right\rbrace\right) \\ &= \sum_{T\in \mathcal{T}_{26}}P\left(\left\lbrace D \subseteq \mathcal{D}_T\left| \ D \subseteq R\lor D \subseteq B\right. \right\rbrace\right) \\ &\overset{\text{(a)}}{=} 2\sum_{T\in \mathcal{T}_{26}}P\left(\left\lbrace D \subseteq \mathcal{D}_T\left| \ D \subseteq R\right. \right\rbrace\right) \\ &\overset{\text{(b)}}{=} 2\sum_{T\in \mathcal{T}_{26}: |T\cap R|\geq 2}P\left(\left\lbrace D \subseteq \mathcal{D}_T\left| \ D \subseteq R\right. \right\rbrace\right) \\ &\overset{\text{(c)}}{=} 2\sum_{i=2}^{26}{26 \choose i}{26 \choose 26-i }P\left(\left\lbrace D \subseteq \cup_{T: |T\cap R|=i}\mathcal{D}_T\left| \ D \subseteq R\right. \right\rbrace\right) \\ &\overset{\text{(d)}}{=} 2\sum_{i=2}^{26}{26 \choose i}{26\choose 26-i }{i \choose 2}/\left({52 \choose 26}{26 \choose 2}\right)\\ &\overset{\text{(e)}}{=} 25/51. \end{array}

(a) is due to symmetry in $R$ and $B$ and their disjointness. The probability we get two blacks is the same as the probability that we get two reds.

(b) is due to our inability to get two reds if there are one or less in the deck we are choosing from. This makes those terms have probability 0 and not contribute to the sum.

(c) Fixing $i,$ we enumerate all possible ways of choosing $i$ reds and filling the rest of our small deck with blacks. ${26 \choose i}{26 \choose 26-i}$

(d) By the way we rewrote the sum in our previous step, we know each deck has ${i}$ reds, so the number of 2-draws of red we could make is ${i \choose 2}.$ Further we know the probability of each of these $D$ by previous investigation of our model.

(e) Rewriting the summation terms, we see that it's the same.

• So when we take out the cards from the large deck in the second part, we aren't replacing them, correct? – user19405892 Nov 28 '15 at 0:08
• In my answer I am assuming no. I'm typing out the formal version right now. – enthdegree Nov 28 '15 at 0:10
• Yes, because otherwise as I commented above the probability would be greater sometimes. – user19405892 Nov 28 '15 at 0:11
• Greater sometimes, less other. Probability is about averages. @user19405892 The randomness of the sub-deck selection is part of the randomness total. And it is not true that the bigger deck is better - the second process is exactly the same as picking two cards from the first deck at random. – Thomas Andrews Nov 28 '15 at 0:47
• By greater sometimes, less others, you mean if we were replacing them in the second part? – user19405892 Nov 28 '15 at 3:17