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Two cards are drawn successively from an ordinary deck of 52 well-shuffled cards. What is the probability that the cards are not of the same suits?

I did: enter image description here

where 13/52 is the probability to get a specific suit for the first draw and 39/52 is the probability to get not the same suit for the second draw. I multiplied by 2 because this can happen at the reverse order (first not Suit color and second Suit) I multiplied by 4 because there are 4 suits (Spades, Diamond, Heart, Club) but the result is obviously wrong.

I know the correct answer is 13/17 using another method but I don't know why my method is not working.

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  • $\begingroup$ I don't understand the factor of $2$. The factor of $4$ already allows for all possible first suits. $\endgroup$
    – lulu
    May 2, 2019 at 13:43

2 Answers 2

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What we want is the probability that our cards are of different suites.

The first card we pick...well it can be any card. There isn't anything we need to compare it against.

The second card isn't allowed to be of the same suite as the first. So as one of the suites have been chosen, that leaves the remaining $12$ cards of that suite as bad picks.

Our probability is then $$1\left(\frac{51-12}{51}\right) = \frac{39}{51} = \frac{13}{17}$$


Your method is almost correct; the multiplication by $2$ isn't needed. Let's say you draw a heart first then a spade. In the other order, you could draw a spade first then a heart. However your multiplication by $4$ already counted for this possibility.

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The multiplication with $2$ does not make sense. With this, you would be counting double. Think about why.

Moreover, instead of $39/52$, you should have $39/51=13/17$ because there are only $52$ cards remaining in the deck after you picked the first one.

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