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You would obviously need at least at least two draws but probably wouldnt expect seeing every card even after three draws. I took combinatorics but forget how to solve this.

RobPratt
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Troy Nemeth
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  • This is the coupon collector's problem with group drawings. The problem is discussed here. The paper cited in the answer is freely available on JSTOR. – saulspatz Feb 27 '21 at 00:00
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    See also https://math.stackexchange.com/questions/1375470/expected-number-of-times-a-set-of-10-integers-selected-from-1-100-is-selected – Mike Earnest Feb 27 '21 at 19:34
  • So no direct answer? I just want the number of times that you would exceed 50% probability that you would see all the cards. Or the formula for number of cards and number of cards drawn at a time. – Troy Nemeth Mar 03 '21 at 20:07

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It seems you’re referring to a $52$-card deck.

Your comment under the question contradicts the title of the question. Since the expected number of draws required to see all cards (asked for in the title) is easier to calculate than the number of draws required to have a $50\%$ chance to see all cards (asked for in the comment), I’ll compute the former. The formula for this is derived in the question Mike Earnest linked to, Expected number of times a set of 10 integers (selected from 1-100) is selected before all 100 are seen. In our case, $n=52$ and $m=26$, so the expected number of required draws is (Wolfram|Alpha computation)

$$ \sum_{j=1}^{52}(-1)^{j-1}\binom{52}j\frac1{1-\frac{\binom{52-j}{26}}{\binom{52}{26}}}\approx7.1\;. $$

joriki
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