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Suppose we shuffle a deck of 10 cards, each bearing a distinct number from 1 to 10, to mix the cards thoroughly. We then remove three cards, one at a time, from the deck. What is the probability that we select the three cards in sorted (increasing) order?

user2232305
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3 Answers3

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Let's first consider that we draw 3 cards without ordering them, then we order our 3 drawn cards. This is an equivalent problem and regardless of the cards we drew, there are 6 equiprobable orderings, only one of which yields an increasing order.

The probability is therefore 1/6.

TerranDrop
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  • I don't understand your reasoning. First we need to draw a card that should be between 1 and 8 inclusive, then two others with probability 1/9 and the last one with probability 1/8: So the probability is ${8/10 * 1/9 * 1/8}$ NO ?? – user2232305 Jan 23 '14 at 10:16
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    It's right the first card we need to draw must be strictly less than 9, but then the other "allowed" cards depend on the first drawn card. Computing the probability in this way becomes a combinatorial nightmare, so I think it is better to consider we just pick three cards at random, then order them randomly, which is equivalent but easier to compute. – TerranDrop Jan 23 '14 at 11:04
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Let E = event that we draw 3-cards that are in increasing order
Then P(E) = |E|/|S|
Note we can use P(E)= |E|/|S| because each outcome is of equal probability.

First, we find the size of the sample space |S| as the total permutations of 3 cards out of 10 cards: n!/(n-k)! = 10!/7! = 720. That's the easy part.

Second, what are the total # of 3-card orderings that are in increasing order? i.e. |E|
This is simply the total combinations of 3 cards out of 10 cards:
n!/[n!(n-k)!] = 10!/(3!7!) = 120
Why can we use the combinations formula?
Well, when counting combinations, order doesn't matter,
so {1,2,3} = {2,1,3} = {2,3,1} = {3,2,1} = {3,1,2}
In any set of the same numbers, there is only one ordering where the numbers are increasing. Therefore, by counting the combinations, its the same as counting the only sorted order and disregard the non-sorted orders.

P(E) = |E|/|S| = 120/720 = 1/6

Please feel free to let me know if I made some logical shortcuts in the second part.

s.ka
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  • The second part seems a bit confusing to me. You can make it different by merely saying that the set of strictly increasing sequences with length three over ${1,\ldots,10}$ is equipotent to the set of all subsets of ${1,\ldots,10}$ with cardinality $3$. The latter has $\binom{10}{3}$ elements. – Gabriel Romon Sep 01 '14 at 17:18
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I used the approach of counting. Since we are drawing one card at a time: $S = 10 * 9 * 8$ (10 ways to pick first card, then 9, then 8) and $$E = \sum_{i = 1}^8 \sum_{j = i+1}^9 \sum_{k = j+1}^ {10}i.j.k$$

As somebody already pointed out, a combinatorial nightmare. Wondering how this shall turn out to be $10C3$

Sloan
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Skartik
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