You have a sixty-card deck. It has:

  • 3 Aa and 3 Ab cards (6 A cards)
  • 4 B cards
  • 4 C cards
  • 4 Da and 2 Db cards (6 D cards)
  • 14 Ea and 10 Eb cards (24 E cards)

At the begining of the game, you draw 7 cards. Then, at the begining of each turn, you draw one card.

What are the chances of having these cards by these turns?

  1. One E card
  2. Another E card, and a A card
  3. Another E card, and a B card
  4. Two C cards
  5. A D card

It doesn't matter if you drew your D card at the beginning of the game or on turn 5 - as long as you have on turn five, you satisfy the conditions. How would you calculate this?

What if you don't draw a card on your first turn?

  • $\begingroup$ Are you familiar with the rule of successive conditioning? If you can apply that, the rest is just a lot of tedious counting. $\endgroup$ – user17794 Jun 13 '12 at 22:53
  • $\begingroup$ @TimDuff: No, I'm not - care to explain? $\endgroup$ – Glycan Jun 13 '12 at 23:06

These questions should be pretty easily answered by standard counting arguments. For example, the answer to #1 is $$ 1 - \frac{\binom{60-24}8}{\binom{60}8} = \frac{176131}{178239} \approx 0.9882. $$ The numerator counts how many 8-card subsets of a 60-card deck avoid the 24 E cards; the denominator counts how many 8-card subsets the 60-card deck has in total. That ratio is the probability of failing to have an E card, hence the one-minus.

I agree that it gets tedious to do the inclusion-exclusion or conditioning to answer #2, ..., #5 in turn. I went ahead and programmed a simulation: in two million trials, the chances of your favorable draws were as follows.

  • #1: 98.8%
  • #1 and #2: 59.5%
  • #1-#3: 25.1%
  • #1-#4: 2.4%
  • #1-#5: 1.4%

The killer is the two C cards: just getting them in the first 11 cards, independent of anything else, is already just a 15% chance, and it goes down to 4.6% if we insist that the first 11 cards contain three Es and and A and a B in any order (much less by the right turns).

Of course these probabilities go down if you don't draw a card on the first turn: the chances of all five happening seem to be about 0.6% in that scenario.

  • $\begingroup$ Yes, M:tG. It's a combo deck, if you're interested. $\endgroup$ – Glycan Jun 14 '12 at 1:26
  • $\begingroup$ The slightly weird notation is hypergeometrical distribution, yes? $\endgroup$ – Glycan Jun 14 '12 at 1:27
  • $\begingroup$ Also, not quite - if you generalize it to "having {these} cards at turn N", that's more likely than "having {these} cards at turn N, and having {some subset of {these}} at turn N - 1", and having...". $\endgroup$ – Glycan Jun 14 '12 at 1:29
  • $\begingroup$ Finally, the expression reads as "one minus the number of unordered ways to draw 8 cards from a pool of 60 - 24 cards divided by the number of unordered ways to draw 8 cards from a pool of sixty cards"? Correct me if I'm wrong, please. $\endgroup$ – Glycan Jun 14 '12 at 1:31

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