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Problem 1 - You are taking out candies one by one from a jar that has 10 red candies, 20 blue candies, and 30 green candies. What is the probability that there are at least 1 blue candy and 1 green candy left in the jar when you have taken out all the red candies?

Strategy 1

The desired probability is $P($last candy is green and last non-green candy is blue$) + P($last candy is blue and last non-blue candy is green$) = P($last candy is green$)\cdot{P(}$last non-green candy is blue | last candy is green$) + P($last candy is blue$)\cdot{P(}$last non-blue candy is green | last candy is blue$)$

Problem 2 - You deal cards from a well-shuffled pack one-by-one until the Two of Hearts appears. What's the probability that you see exactly one king, one jack, and one queen before the Two of Hearts?

My questions are:

(i) I'm trying to create a broad list of probability strategies such as spotting symmetry and conditioning on well-chosen events. What does category would you say strategy 1 falls into?

(ii) Is there a similar strategy for the second problem?

(iii) Are there other problems like both of these, and what are some examples?

Thank you!

Abhay Agarwal
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  • And a doubt... candies of same color are distinct or identical? – Blue Cat Blues May 09 '24 at 22:23
  • Please edit to address the ambiguity in the phrasing of problem $2$: I say that $9\spadesuit, 3\clubsuit, K\heartsuit, 5\diamondsuit, Q\spadesuit, J\spadesuit, 2\clubsuit, 2\heartsuit$ is a winning sequence, but someone else has posted a solution which reads the problem as saying it wouldn't be (as it involves unrelated draws). – lulu May 10 '24 at 10:05
  • Also worth noting: for the first problem, this question is effectively a duplicate. I do understand that you are asking about methodology and not explicit solutions to these specific problems. – lulu May 10 '24 at 10:07

2 Answers2

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A point of clarification: For problem $2$, I am reading the problem as ignoring all but $13$ cards (the four Kings, Queens, Jacks, and the $2\heartsuit$). Thus, I'd say that the path $9\spadesuit, 3\clubsuit, K\heartsuit, 5\diamondsuit, Q\spadesuit, J\spadesuit, 2\clubsuit, 2\heartsuit$ is a winner as you see exactly one each of $K,Q,J$ before the $2\heartsuit$. As the phrasing of the problem is (at least slightly) ambiguous, perhaps the OP should clarify the post.

I would say that the two problems are very different. For the second, once you remove all the irrelevant cards from the deck, there are only $13$ cards left and you can easily enumerate the winning paths (they are $KQJ2, KJQ2$ and so on. The exact list doesn't matter, there are $6$ of them and they are all equally likely).

For the first, it is difficult to enumerate the paths, and it isn't necessary. Instead, you can use the fact that, in any collection or sub-collection of the candies, each of them has an equal chance of being last.

Phrased differently (if a bit tersely), the first problem concerns Final Cards and ignores specific paths, while the second concerns Initial Cards and centers on specific paths.

lulu
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For the second problem

$P(\operatorname{desired})=\frac{4}{52}×\frac{4}{51}×\frac{4}{50}×6×\frac{1}{49}$

Blue Cat Blues
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  • It doesn't seem to me that you answered OP's asks of (i), (ii), (iii). They don't seem concerned with obtaining the numerical value. – Calvin Lin May 09 '24 at 23:38
  • @CalvinLin I'll edit the answer if I come up with something that answers OP's actual questions – Blue Cat Blues May 10 '24 at 06:22
  • And as a matter of interpretation: I take question $2$ to ignore all but $13$ cards. That is to say, drawing the $9\spadesuit$ is irrelevant, the problem says nothing about $9's$. I agree the phrasing is somewhat ambiguous, perhaps the OP could clarify) – lulu May 10 '24 at 10:00
  • @lulu I get what you're saying...I solved for the case where only these three cards can appear before $2$ but it may happen that other cards can also appear but not two kings, two queens or two jacks – Blue Cat Blues May 10 '24 at 11:26
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    Thanks, that means my comments are no longer needed, so I deleted them. – David K May 11 '24 at 18:47