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I recently came across the question of how many cards we expect to draw before drawing an ace. I found a nice solution with indicator variables, and it seemed natural to me to ask this question next:

Suppose we draw cards one by one from a standard deck without replacement. How many cards do we expect to draw before our first consecutive pair, e.g. two 7s in a row?

My previous method doesn't work for this question, and I haven't found it tackled anywhere. What are some different proofs for it? I would like as many as possible! Is the problem even tractable, or is the solution always going to be ugly (and if so, why?).

And, if anyone is able to answer a slightly broader question: where might I find a resource that has many of these kinds of problems (with dice, cards and such) and maybe general methods for solving them? I am awful at probabilities but would like to get really good at this.

This is my first post, so please let me know if I have made any fatal errors (I have really tried not to!) I apologise if anything is too general, and I'm really grateful for any comments :)

TheOppositeOfEuler
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  • might be nice if you added the combinatorics tag it may help with views and people in the know. –  Sep 29 '17 at 16:17
  • Thank you for the tip! :) – TheOppositeOfEuler Sep 29 '17 at 16:25
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    there are a lot of orders without a pair on consecutive pulls as well though. more than 12^8*(11!)^4 at last check. –  Sep 29 '17 at 16:44
  • If you have no replacement, surely drawing the same card twice is impossible? – Joffan Sep 29 '17 at 18:38
  • Could be in a different suits - I just mean drawing two 7s in a row, or two aces etc. Thanks for helping me clarify! – TheOppositeOfEuler Sep 29 '17 at 20:43
  • Are cards drawn one by one, or two by two ? – true blue anil Sep 30 '17 at 05:13
  • They're drawn one by one, but do you have an answer to the two-by-two question? I feel like it should be easier, but I don't immediately see a method – TheOppositeOfEuler Sep 30 '17 at 06:29
  • "And, if anyone is able to answer a slightly broader question" This is a different question. I believe it deserves being asked separately. – zhoraster Sep 30 '17 at 07:13
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    What is the value of the variable in the case where you go through the whole deck without getting a consecutive pair? – bof Sep 30 '17 at 09:43
  • @TheOppositeOfEuler Would you be content with a simulation ? I think it is pretty difficult to find the exact value. – Peter Sep 30 '17 at 12:07
  • @bof I think 52 makes the most sense, though I realise I left it ambiguous – TheOppositeOfEuler Oct 01 '17 at 01:38
  • @Peter a simulation would be nice, though I'd be more content with a good approximation not requiring computers. Even knowing that the problem seems very difficult is interesting to me - I'm mainly trying to ascertain what questions are and aren't asked in the field, so it's informative to see that a question which appears superficially similar to "easy" questions (e.g. # of cards before first ace or # of dice rolls before k heads) is in fact hard! – TheOppositeOfEuler Oct 01 '17 at 01:41
  • The closely related problem of the expected number of draws from a deck without replacement until some value repeats for the first time recently appeared at this MSE link. – Marko Riedel Oct 02 '17 at 22:34

1 Answers1

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On the first draw you have, of course, no chance for a match. On each subsequent draw, the probability of matching the previous card is $\frac1{17}.$ Treating these matches as independent events (which of course they are not) we get the crude approximation: $$1+17\left[1-\left(\frac{16}{17}\right)^{51}\right]\approx17.2$$ Namely, let $X$ be the number of cards drawn, and let $X_k$ be the indicator variable whose value is $1$ if at least $k$ cards are drawn, $0$ otherwise. Then $X=\sum_{k=1}^{52}X_k,$ so $$E(X)=\sum_{k=1}^{52}E(X_i)=\sum_{k=1}^{52}P(X_k=1).$$ Now $P(X_1=1)=P(X_2=1)=1,\ P(X_3=1)=1-\frac3{51}=\frac{16}{17},\ P(X_4=1)\approx\left(\frac{16}{17}\right)^2,$ and $P(X_k=1)\approx\left(\frac{16}{17}\right)^{k-2}$ for $k\ge2,$ so $$E(X)\approx1+\sum_{j=0}^{50}\left(\frac{16}{17}\right)^j=1+\frac{1-\left(\frac{16}{17}\right)^{51}}{1-\frac{16}{17}}=1+17\left[1-\left(\frac{16}{17}\right)^{51}\right].$$

bof
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  • Thanks! It looks to me like [1-(16/17)^51] is the probability of a series of 51 draws yielding at least one consecutive pair. I'm curious why multiplying that by 17 gives the expected number of draws before a pair? – TheOppositeOfEuler Oct 01 '17 at 03:28
  • @TheOppositeOfEuler I added an explanation to my answer. – bof Oct 01 '17 at 03:56
  • Nice, thanks! This looks like a good approximation, and I had not seen this method before. Gave you an upvote, but it doesn't show since I have no rep :/ – TheOppositeOfEuler Oct 01 '17 at 04:11