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Consider a game with 24 cards with 4 suits and 6 ranks: 2,3,4,5,6,7 and lets say 5 cards are dealt to 4 people (jim, joe, mary, fred) also the other four cards are unused.

I am trying to find out the probabilty of the following:

(a) jim is dealt at least one "5"

(b)joe is not dealt any 3's

(c) What is the chance that mary is dealt neither any 5s nor any 3s?

(d)that fred is dealt at least one 5 and at least one 3?

(e)jim is dealt at least one 5, given that you know she was dealt at least one 3?

Attemped solutions:

(a) $$ ({4 \choose 1} * {20 \choose 4})/{24 \choose 5} $$ (b)$$ 1 - {20 \choose 5}/ {24 \choose 5} <--complement$$ (c)$$ 1 - 2*{20 \choose 5}/ {24 \choose 5} <--complement$$ (d) $$ ({4 \choose 1}*{4 \choose 1} * {16 \choose 3})/{24 \choose 5} ?? $$ (e) No clue?

I feel like I'm off, any clarification would help?

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Complement, not compliment. –  Yuval Filmus Sep 23 '12 at 0:03
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1 Answer 1

up vote 3 down vote accepted

(a) Your calculation overcounts considerably. Consider the hand $\{\clubsuit 5,\heartsuit 5,\clubsuit 3,\spadesuit 6\}$: you count it once with $\clubsuit 5$ as the $5$ and $\heartsuit 5,\clubsuit 3$, and $\spadesuit 6$ as the three other cards and once again with $\heartsuit 5$ as the $5$ and $\clubsuit 5,\clubsuit 3$, and $\spadesuit 6$ as the other three cards. You count the hand $\{\heartsuit 5,\clubsuit 5,\diamondsuit 5,\spadesuit 5\}$ four times.

When you see at least one, you should usually calculate the number of things that don’t have at least one of whatever it is and then take the complement. Here there are $\binom{20}5$ hands that contain no $5$ at all, so there are $\binom{24}5-\binom{20}5$ hands that contain at least one $5$.

(b) You’ve got it just backwards: you’ve counted the hands that have at least one $3$. In order to get no $3$’s, you must get all $5$ of your cards from the $20$ non-$3$’s, which you can do in $\binom{20}5$ ways.

(c) If she is to get no $5$’s and no $3$’s, all of Mary’s cards must come from the $16$ cards that are not $5$’s and not $3$’s; how many ways are there to choose $5$ cards from those $16$?

(d) This one’s a bit harder. Once again we’ll count the complementary set of hands that we don’t want. We’ve seen that there are $\binom{20}5$ hands that contain no $5$ and $\binom{20}5$ hands that contain no $3$, so you might at first think that there are $2\binom{20}5$ hands that we don’t want. However, some of those unwanted hands have no $5$ and no $3$, so we counted them twice. We need to subtract the number of hands that we’ve counted twice. Call that number $N$; the number of unwanted hands is $2\binom{20}5-N$. But isn’t $N$ exactly the number that you were supposed to calculate in (c)? So once you have the right answer to (c), you’re also practically done with (d).

(e) This is the hardest of the bunch. I’m going to do it in the most straightforward way, with no attempt to be clever. If we know that Jim has at least one $3$, there are four possibililities.

  1. Jim has four $3$’s. There are $20$ cards that aren’t $3$’s, so there are $20$ possible hands of this type.

  2. Jim has three $3$’s. There are $\binom43=4$ ways to pick three of the $3$’s, and there are $\binom{20}2=190$ ways to pick the two non-$3$’s, for a total of $4\cdot190=760$ hands of this type.

  3. Jim has two $3$’s. There are $\binom42$ ways to pick two of the $3$’s, and there are $\binom{20}3$ ways to pick the three non-$3$’s; in this case and the last I’ll leave the arithmetic to you.

  4. Jim has just one $3$. There are $\binom41$ ways to pick the $3$, and there are $\binom{20}4$ ways to pick the four non-$3$’s.

Now go through each of the four types and work out how many of the hands have at least one $5$. I’ll do the first two:

  1. There are four $5$’s, and he has to get one of them in order to have at least one $5$, so just $4$ of the $20$ hands in this case meet the requirements.

  2. The only way not to get at least one $5$ is to get $2$ non-$5$’s for the two cards that aren’t $3$’s. $16$ of the $20$ non-$3$’s are non-$5$’s, so there are $\binom{16}2$ ways to choose two non-$5$’s and therefore $\binom{20}2-\binom{16}2$ ways to choose two non-$3$’s that include at least one $5$.

The last two are similar to (2) above. When you’re all done, add up the number of hands in all four categories that meet the requirements; call this total $R$. Then add up the number of hands that are in the four categories altogether; call this total $H$. Given that Jim has at least one $3$, each of these $H$ hands is equally likely, so the probability that Jim has a hand that meets the requirements is $\frac{R}H$.

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just to make sure for event 3 for quesion e) would be (20 choose 3) - (16 choose 3) and event 4 would be the same except we choosing 4 ? –  Thatdude1 Sep 23 '12 at 16:53
    
@Beginnernato: Yes, that’s right. –  Brian M. Scott Sep 23 '12 at 19:49
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