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Context

I understand this proof and it is quite slick and throws up a counterintuitive result.

However, before looking at the solution, I tried my own ham-fisted approach of finding P(AoS immediately drawn after first ace), and wasn't sure where my mistake is.

Finding number of orderings s.t. AoS follows first ace.

$$\text{Let } A_1 = \text{first ace, which is not ace of spades}$$

$$\text{Let } A_s = \text{ace of spades} $$

Scenario 1: A_1 and A_s drawn first:

$$\text{A_1, A_s, ... = 50!*3 orderings (3 choices for A_1)}$$

Next scenarios:

$$\text{[X],} A_1, A_s, ... \text{ where [X] is a card other than an ace = 48*49!*3} \text{ orderings.}$$

$$\text{[X], [X]}, A_1, A_s ... = \frac{48!}{(48-2)!} * 48! * 3 \text{ orderings.}$$

...

Final scenario:

$$\text{[X], [X] ... [X],} A_1, A_s ... = \frac{48!}{(48-48)!}*2!*3$$

Sum these terms up. A general term can be written as:

$$\frac{48!}{(48-r)!} * (50 - r)! *3 = \frac{48!}{(48-r)!} * (48 - r)! * 49 * 50 *3 = 50! * 3 $$

Summing 49 of these terms would lead to an answer of 147*50!, which is not what the author had for the number of orderings that verify this event (51!).

What's gone wrong with my reasoning?

RobPratt
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threelinewhip
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  • Why other than an ace in the second line? You could even have all the aces together. And why go for a more cumbersome proof where there are more chances of error when a simple proof satisfies you ? Anyway, always remember the dictum "Cards don't have preference for positions." – true blue anil Mar 22 '21 at 15:32
  • Otherwise the AoS wouldn't be directly after the first ace? My idea was to write an exhaustive list of all possible orderings that verify the event 'AoS after first ace' just to test my understanding of combinatorics (ik the author's sol is v slick, but I tried this before reading the solution). – threelinewhip Mar 22 '21 at 17:55
  • And since I'm working out number of orderings that verify the event and divide this by total orderings (although I guess you can work this out with an unordered approach - nevertheless, an ordered approach can work as well), then cards would have a preference for position? – threelinewhip Mar 22 '21 at 17:55
  • Well, if you are counting, fine, I just meant that you should keep that dictum in mind for a possibly simpler approach for many problems. – true blue anil Mar 22 '21 at 18:14
  • Doesn't the author's slick solution make use of orderings? (51!) – threelinewhip Mar 22 '21 at 18:17
  • Yes, but from the dictum, it follows that all ordered pairs are equally likely, and we directly get the answer of 1/52. But yes, sometimes you can get misled without some counting – true blue anil Mar 22 '21 at 19:28
  • As a last comment, have a look at the solutions to this link. So many approaches have been given, you have to use whatever you are comfortable with. https://math.stackexchange.com/questions/1345413/is-there-an-alternative-intuition-for-solving-the-probability-of-having-one-ace/1345436#1345436 – true blue anil Mar 22 '21 at 19:39
  • Ok thanks - I was just checking for where I went wrong with brute force. As ugly as it is, brute force usually can work (if done correctly) and in a time-pressured situation, I might not be able to think of a slick sol. – threelinewhip Mar 23 '21 at 09:50

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