I'm working on a puzzle which describes a casino game where you draw 13 cards. You win if your first card is an ace, or your second is a two, or… all the way up to the 13th draw being a king. In order to analyse the problem, I started by simplifying to a two-card version. You draw two cards and win if the first is an ace or the second is a deuce.
The odds of drawing an ace on the first card is clearly 4/52 or 1/13.
My understanding, backed up by this question If you draw two cards, what is the probability that the second card is a queen?, suggests that the odds of drawing a 2 on the second card is also 1/13. What are the odds of drawing either? To my knowledge, you can't directly calculate that but instead calculate the inverse. The odds of not drawing a specific card should clearly be 12/13 and the odds of not drawing an ace followed by not drawing a two should be 12/13 * 12/13 = 144/169. The odds of drawing one or the other should therefore be 25/169.
But there's another approach to analysing the problem. There are 52 * 51 = 2652 different combinations for the first two cards. If A is an ace, T is a two and x is any other card, there are 5 combinations which give you a win:
AA
Ax
AT
xT
TT
Given that there are four aces, four twos and 44 other cards, the number of winning hands is:
4*3 + 4*44 + 4*4 + 44*4 + 4*3 = 392.
So the odds of winning should be 392 / 2652.
25/169 and 392/2652 are very close but they are NOT the same number. I believe the second number is accurate, but I can't see where the logic in the first method fails. I suspect that it has to do with the possibility of drawing both being double counted but I can't see how that should matter. It seems like you should be able to treat each draw as an independent event. Additionally, the second method doesn't scale well – it would be extremely tedious to directly calculate the number of winning hands for the full 13 card game.
edit: Clarified the rules for winning the game.
Best Answer
Let $A$ denote the event that the first draw will be an ace and let $B$ be the event that the second draw will be a two.
Then:$$\Pr(A\cup B)=$$$$\Pr(A)+\Pr(B)-\Pr(A\cap B)=$$$$\Pr(A)+\Pr(B)-\Pr(A)\Pr(B\mid A)=$$$$\frac4{52}+\frac4{52}-\frac{4}{52}\frac{4}{51}$$