It is cyclic so: ...Jack, Q, K, A, 2... The first card you pick doesn't matter. You have a 100% chance of drawing a card that fits the requirements. There are now 51 cards in the deck. If the card was the n of some suit, the next card must be (n+1) or (n-1) of that same suit. Therefore there are two cards in the 51 that will satisfy the requirements. So, the P of drawing two consecutive cards of the same suit is 2/51 I wasn't given an answer, I was simply told I was wrong. Where is my error?
Want to improve this question? Add details and clarify the problem by editing this post. So, I am having a problem with this in that the method I use gives two completely separate answers.
So, for the first method I reason this. The first card picked has a $13/52$ chance of being in some suit. The second card picked has probability $12/51$ of being in the same suit. So... The probability should be $(13/52)(12/52) = 3/52$. The other method is by combinatorics. I have $52 \cdot 51$ one ways of creating a pair of cards. But I have $13 \cdot 12$ different ways of creating a pair of the same suit. Now to me, the logical thing to do is to multiply this number by $4$, because I would have to count each valid pair from each suit. This would give me $$\frac{4 \cdot 12 \cdot 13}{52 \cdot 51}$$ What's wrong with the reasoning on the second one? |