1
$\begingroup$

You are playing a solitaire game in which you are dealt three cards without replacement from a simplified deck of $10$ cards (marked $1$ through $10$). You win if one of your cards is a $10$ or if all of your cards are odd. How many winning hands are there if different orders are different hands?

  1. $P(\text{1 card is a 10}) = 9 \times 8$ because if one of them is a 10, there are 9 options for position 2, and 8 for position 3. Multiply this by 3 because the order changing counts as another hand. (specified). $P(\text{All Odd}) = 5 \times 4 \times 3$ because there are $5$ and then there are $4$ and so on, multiplied by $3$ for the same reason.

  2. How many winning hands...$396$, adding those two together.

The next question is chance of winning. The total amount of possibilities would be $10!$ wouldn't it? So $396/10!$ ? It's been a while since I took combinatorics and this is a preliminary review/diagnostic for a CS course, and I feel like with a little review I know more than might be apparent without it.

  • 1
    It is not $10!$ total hands; that would be the total number of orderings of *ten-card* hands, not three-card hands. – 2017-02-01

2 Answers 2

1

Your first part produces $216$, which is correct. We can verify this by finding the total number of hands, and subtracting the number of hands that are drawn from just the cards $1$ through $9$:

$$ _{10}P_3\, - \,_9P_3 = \frac{10!}{7!} - \frac{9!}{6!} = 720-504 = 216 $$

The second part is not correct; it should be just the number of three-card permutations drawn from the set of five odd-numbered cards:

$$ _5P_3 = \frac{5!}{2!} = 60 $$

You can't reorder them (and multiply by $3$) because you've already counted them all in multiplying $5 \times 4 \times 3$: Each reordering corresponds to a hand already counted.

No hand can both contain a $10$ and contain only odd-numbered cards, so the total number of winning hands is just $216+60 = 276$. The total number of hands, as previously alluded to, is the number of three-card permutations drawn from the entire deck of ten cards:

$$ _{10}P_3 = \frac{10!}{3!} = 720 $$

So the desired probability is

$$ P(\text{win}) = \frac{276}{720} = \frac{23}{60} $$

0

I am led to believe that order does not matter. These kinds of problems are good to be dealt with in the following manner.

$(1):$ Calculate the number of possible outcomes.

This is simply the number of possible opening opening hands. That's $\binom{10}{3}=120$.

$(2):$ Calculate the number of success cases.

For a $10$ in hand, there are $\binom{9}{2}$ possibilities: $1$ of the cards is a $10$, the others are any two cards from the remaining nine.

For an all-odd hand, there are $\binom{5}{3}$ possibilities: any three of the five odd cards.

Since these success cases are mutually exclusive, we simply add them up, for a total of $\binom{9}{2}+\binom{5}{3}=36+10=46$ success cases.

$(3):$ Divide one by the other.

The probability is hence $\frac{46}{120}=\frac{23}{60}\simeq38.34\%$.

  • 0
    The order doesn't matter for the probability, but part of the OP's question asks for the total number of winning hands if order *does* matter. – 2017-02-01
  • 0
    .38 was correct for the chance of winning, but 46 is not correct for the first part of the question (success cases). Notice, it does say "How many winning hands are there *if different orders are different hands*?" so I tried multiplying 36 * 3 to get the diff. combinations for success 1 and 10 * 3 for success 2, added together. Not good either... – 2017-02-01
  • 0
    Oh, I didn't pay attention to that. I thought it was how you tackled the problem, rather than the question itself. Sorry. – 2017-02-01
  • 0
    I should note, however, that it is no coincidence that the answer ended up being correct. Since all cards are different, the number of outcomes and the number of success cases are each multiplied by $3!$ - the number of distinct ways to order $3$ distinct elements -, and so when you divide one by the other it is inconsequential. – 2017-02-01