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I'm trying to reconstruct the blackjack basic strategy chart. I am particularly interested in the question right now, when dealing with two decks of 52 cards each, how many ways can we choose 2 cards? A common response online is $\binom{104}{2}$ but I feel like it should be $\binom{52}{2}$ regardless of how many decks there are. My reasoning is that since the second deck contains indistinguishable cards from the first deck, there are only $\binom{52}{2}$ ways regardless of which deck a card came from.

Is my intution correct? If so, how could we calculate it from $\binom{104}{2}$? If not, why?

Thanks

Leo
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  • I think it should be ${52\choose 2} + 52$ as it is also possible that the two cards are the same but from different decks. – Shreya Mundhada Jan 11 '24 at 10:15
  • To be clear... if you are wishing to ask a question about the probability of events surrounding this scenario... even if the decks are otherwise identical you should treat them as distinct. There may be $\binom{52}{2}+52$ ways if we could not tell the difference between $A\clubsuit J\heartsuit$ where the ace was "from the first deck" and jack from the second vs $A\clubsuit J\heartsuit$ where the ace was from the second deck and the jack also from the second deck... but that does not help us answer probability questions since outcomes counted in this way are not equally likely to occur. – JMoravitz Jan 11 '24 at 13:06
  • For that reason... when discussing strategy and likelihood of outcomes... $\binom{104}{2}$ is the preferable count so we can use counting techniques to answer questions about probabilities of events, regardless of whether or not we could have "condensed the sample space into fewer possibilities." If all you are after is tabulating (but not using for purposes of calculations) outcomes, then the suits of the cards should also have been irrelevant... condensing it down to $\binom{13}{2}+13$ or even $\binom{10}{2}+10$ outcomes where all of a rank (maybe even all 10-K) are treated as same. – JMoravitz Jan 11 '24 at 13:09
  • @JMoravitz Could you explain a little bit further the pros and cons of each options? Also, how was $\binom{52}{2} + 52$ derived? – Leo Jan 11 '24 at 23:37
  • As @ShreyaMundhada already correctly stated... it is possible that the two cards you draw are the same. $\binom{52}{2}$ counts how many ways you draw two different cards like $2\heartsuit 3\clubsuit$. There are $52$ ways to draw two identical cards like $J\clubsuit J\clubsuit$. You could if you insist count this instead with stars-and-bars but given that the earlier explanation is so incredibly basic and the numbers are so small there should be no reason to. – JMoravitz Jan 12 '24 at 00:04
  • Pros of only writing down the $\binom{10}{2}+10$ outcomes: You might actually be able to fit them all on one page. Reading information for each if you insist on writing each out will be much easier. You don't even need to mention suits which are irrelevant for the game of blackjack. Cons: These are not equiprobable and do not lend themselves to easy calculations for probability purposes. Each case would need to be treated separately depending on their relative probability of occurrence. Pros of $\binom{52}{2}+52$: all distinguishable outcomes are listed... – JMoravitz Jan 12 '24 at 00:07
  • ...Cons: still not equiprobable. Again, each case will need different calculations depending, but at least fewer cases than the proposed $\binom{10}{2}+10$. Pros of $\binom{104}{2}$: All cases are equally likely to occur. You can actually use counting techniques for probability. Recall $\Pr(A)=\dfrac{|A|}{|S|}$ is guaranteed true only when outcomes are equally likely to occur. Cons: some "indistinguishable cases" become distinguishable adding some redundancy in the list, but that is a very small cost to pay when considering we can now calculate probabilities easily. – JMoravitz Jan 12 '24 at 00:10
  • Related: Why is flipping a head then a tail a different outcome than flipping a tail then a head? – JMoravitz Jan 12 '24 at 00:16

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