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Find the probability that in a hand of $5$ cards from an ordinary deck of $52$ cards, some suit appears on $2$ cards in the hand and each of $3$ other suits appears on $1$ card in the hand.

Okay, so here's what I did so far:

We know there are 4 suits in the deck of $52$ cards. For $2$ cards, we want $1$ suit. So that means we can pick $1$ suit out of $4$ by $\binom{4}{1}$, and for each of these ways we want to have $2$ cards that have this same suit. Since there are $13$ cards in the deck that have the same suit, and we want $2$ cards out of $13$ with the same suit, $\binom{13}{2}$.

$$\binom{4}{1} \binom{13}{2}$$

Now we need $3$ other cards, and these $3$ cards are cards of the other $3$ remaining suits, and they're all different.

So we have $\binom{3}{1}$ way to pick $1$ suit out of $3$ left, and for each of these ways, we want to pick $1$ card out of that $1$ suit, so $\binom{13}{1}$. Likewise, we have two more cards left, and two suits left to pick from, $\binom{2}{1}$ to pick $1$ suit out of $2$, and for each of these ways, we have $\binom{13}{1}$ ways to pick $1$ card out of that suit of $13$ cards. Finally, we have $1$ card left, $\binom{1}{1}$ way to pick $1$ card out of $1$ suit, and for each of these ways, we have $\binom{13}{1}$ to pick $1$ card out of the last suit.

So altogether, we can add these up:

$$\binom{4}{1}\binom{13}{2} + \binom{3}{1}\binom{13}{1} + \binom{2}{1}\binom{13}{1} + \binom{1}{1}\binom{13}{1}$$

And divide it by the denominator, which is $$\binom{52}{5}$$

Is this correct?

Thanks

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    $\begingroup$ Many errors . the order in which the 3 remaining suits' cards are chosen does not matter, so the terms $\binom{3}{1}$ and $\binom{2}{1}$ should not be included. Also, all of the terms in the numerator should be multiplied together. Whenever you are saying things like "for each ..." it is usually an indication that multiplication is the way to go. $\endgroup$ – Zubin Mukerjee Feb 14 at 18:54
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Approach by rule of product by breaking apart via the following steps.

  • Pick which suit appears twice in your hand: $4$ options

  • For the suit which appears twice, pick which two numbers appear: $\binom{13}{2}$ options

  • From the unused suits remaining, pick which number appears for the earliest suit in terms of alphabetical order: $13$ options

  • From the unused suits remaining, pick which number appears for the earliest suit in terms of alphabetical order: $13$ options

  • From the unused suits remaining, pick which number appears for the earliest suit in terms of alphabetical order: $13$ options

Multiplying these together we get a total number of hands with all suits appearing at least once each (exactly one of which appearing twice) as being:

$$4\times \binom{13}{2}\times 13^3$$

Note: in the third, fourth, and fifth steps, we do not need to bother ourselves with choosing which suit that is. The choice is forced. In your attempt (after replacing your addition with multiplication) you are incorrectly making the decision to treat the suit as being able to be freely selected but in doing so you are treating multiple identical hands as being "different" according to your count. For example the sequence of choices $\clubsuit$ for the suit which appears twice, $2,3$ as the numbers for $\clubsuit$, then $\heartsuit$-$Q$ as the next card followed by $\diamondsuit$-$K$ as the next card and $\spadesuit$-$A$ as the final card, this is the same result as the different sequence of choices: $\clubsuit$ for the suit which appears twice, $2,3$ as the numbers for $\clubsuit$, then $\diamondsuit$-$K$ as the next card followed by $\heartsuit$-$Q$ as the next card and $\spadesuit$-$A$.

Dividing the count we obtained by $\binom{52}{5}$ gives the probability of such an occurrence from a well-shuffled fair deck.


As for addition versus multiplication... remember you add together two collections of outcomes if you treat each set of outcomes as being final outcomes, being of equal importance/validity, and there being no overlap. For example, a kid has four different shirts and three different pairs of pants. In how many ways can he choose a single article of clothing. There are $4+3$ ways.

On the other hand, multiplication is used when you have separate parts of an outcome that when combined give a full outcome. For example, a kid has four different shirts and three different pairs of pants. In how many ways can he select one shirt and one pair of pants to make an outfit? $4\times 3=12$.

In the above problem, this is analogous to the situation we are making an outfit out of multiple pieces of clothing and so use multiplication, not the situation of choosing a single piece of clothing where addition would have been used.

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  • $\begingroup$ Okay, so in the third, fourth and fifth step, the reason we dont do $\binom{3}{1}$, etc. is because when we pick 1 card out of 13, we are implying we are picking 1 card out of 13 cards that all have the same suit? And the same goes for the other 2 suits, is that why? $\endgroup$ – Stuy Feb 14 at 19:19
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    $\begingroup$ We use the trick of picking the earliest suit remaining from the list Clubs, Diamonds, Hearts, Spades so as to avoid overcounting. When applying the rule of product, we want to make sure that each outcome is counted only once, but if we were to "choose" which suit comes next after having picked the suit that is duplicated, this incorrectly introduces information into our calculation that is irrelevant since there is no way to know after mixing our hand which the "next" suit was. $\endgroup$ – JMoravitz Feb 14 at 19:24
  • $\begingroup$ So because earlier we stated there are 4 ways to pick 1 suit, then it forces the rest of the choices to have the remaining 3, 2, 1 suit? And picking the suit again would have duplicates? $\endgroup$ – Stuy Feb 14 at 19:31
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    $\begingroup$ Yes, @Stuy . The order of the suits does not matter, you just need to select which one is special. So you just need to count the ways to: select $2$ from $13$ faces for $1$ from $4$ suits, and then $1$ from $13$ faces for each of the remaining $3$ suits. $$\binom{13}2\binom 41\binom {13}1^3$$ $\endgroup$ – Graham Kemp Feb 14 at 23:15

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