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From a standard 52 card deck, we draw one card at a time without replacement. After having drawn the first ace, the game continues until either

  1. We draw another ace, or
  2. We draw a 2,

and the game then stop. Which events of the two is more likely, and what are the corresponding probabilities?

Edit: It is not very hard to get the probability = 0.5 with calculations. but can we come up with some more intuitive explanations?

wanderlust
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    What have you tried? Please post some of your thoughts on the problem along with the problem statement – user1001001 Jul 14 '20 at 14:39
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    Hint: You can ignore all cards but the Aces and deuces. – saulspatz Jul 14 '20 at 14:48
  • @saulspatz thanks for your input, it simplify the problem a lot, I will try – wanderlust Jul 14 '20 at 15:01
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    The probability that it is an Ace turns out to be $\frac12$, which I found very surprising; I felt sure a deuce would be more likely. Can anyone give an intuitive explanation why the two events are equiprobable? – saulspatz Jul 14 '20 at 16:07
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    @saulspatz I also got 0.5... what I did is break things down into five scenarios, assuming there will be 0 - 4 "2" appearing before the first ace, with probability 1/2, 2/7, 1/7, 2/35 and 1/70, with each scenario, we have a "2" before the second ace with probability 4/7, 1/2, 2/5, 1/4 and 0 respectively. it is not hard to have 0.5 after that. – wanderlust Jul 14 '20 at 18:00
  • @wanderlust Yes, that's the right way to calculate the probability. What I was asking for is an intuitive way to see that the answer is $\frac12$, without calculation. – saulspatz Jul 14 '20 at 18:47
  • @saulspatz I am trying to think up something. The best I have is that you have numbers one through eight. Choose four of them at random. What is the probability that the smallest two are a distance of 1 from each other? This is essentially the same problem. But, it does not seem any more intuitive at first glance. – SlipEternal Jul 14 '20 at 18:49
  • @InterstellarProbe I agree; it's the same problem, but I don't see how it helps, except that it's easier to visualize, somehow. – saulspatz Jul 14 '20 at 18:56
  • @saulspatz Again, this does not seem overly intuitive, but I thought about expanding the problem. Suppose there are $n+1$ aces and $n+1$ twos. Then, given the same premise, the probability becomes: $$\sum_{k=0}^{n+1}\dfrac{n \dbinom{n + k}{k}}{(n+k)\dbinom{2n+2}{n+1}} = \dfrac{1}{2}$$ I thought perhaps we could relate this to another combinatorial problem, but I am not seeing it. Do you know how this might be shown combinatorially? If not, it still seems like an interesting combinatorial identity. – SlipEternal Jul 15 '20 at 21:01
  • @InterstellarProbe I don't recognize anything off the top of my head. I'll have to think about it. – saulspatz Jul 15 '20 at 21:09
  • @wanderlust If you make a new question (maybe generalized), basically restating the problem then pointing to this question, explaing shortly that standard methods yield $p=1/2$ which is mildely supring and that you would like to find a bijection or similar proof that does not need to go into calculating probabilities excessively and rather finds an inherent symmetry, I'd put a bounty on it. I'd do it for this question, but it's closed, so no answers here. – Ingix Jul 16 '20 at 09:10
  • @Ingix sure will do – wanderlust Jul 16 '20 at 12:49

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