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From an ordinary deck of 52 cards, cards are randomly drawn one by one and without replacement. What is the probability that an ace will appear before any of the cards 2 through 10?

Consider the sample space Ω to be the set of all possible sequences of 52 cards drawn from the deck (without replacement). So we do not remove the Jacks, Queens and Kings from the deck.

I am struggling... the correct answer is 0.1 but I can't reach the right solution.. please help me

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    $\begingroup$ Hint: the face cards have no meaning here, just the $40$ cards that you are watching for, $4$ of which are Aces. $\endgroup$ – lulu Nov 20 '17 at 1:46
  • $\begingroup$ If you truly insist on using the sample space as all possible sequences of all fifty two cards... consider counting by breaking into cases based on the position of the first ace. If the first ace is at the front of the deck, choose which ace it is in $4$ ways and then order the remaining $51$ cards, giving $4\cdot 51!$ arrangements. If the first ace is second from the top, pick which face card appears and then which ace giving $12\cdot 4\cdot 50!$ arrangements, and then $12\cdot 11\cdot 4\cdot 49!$, then $12\cdot 11\cdot 10\cdot 4\cdot 48!$ and so on... $\endgroup$ – JMoravitz Nov 20 '17 at 1:57
  • $\begingroup$ An important observation however that the other answers are trying to point out is that this is a very poor choice for sample space and there are many more convenient choices that we can make instead while still keeping the elements in the sample space equiprobable. $\endgroup$ – JMoravitz Nov 20 '17 at 1:58
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Let's look at the first non-face card that appears. There are 40 possible cards, and 4 of these cards are aces. The probability that this card is an ace is 4/40.

(the statement "The first non-face card that appears is an ace" is equivalent to "an ace appears before any of the cards 2 through 10")

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  • $\begingroup$ Can you please explain a bit more in detail... I think your solution is shortcut which is better but do you also know another way to Use the following events: • Ai be the event ‘ith card is an ace’. • Bi be the event ‘All cards before the ith card are Jacks, Queens or Kings’. $\endgroup$ – Kellen Choi Nov 20 '17 at 2:16
  • $\begingroup$ See my answer for more details, then accept Air Conditioners. $\endgroup$ – Graham Kemp Nov 20 '17 at 2:19
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You seek the probability that one from the four aces is the first from the $40$ non-royal cards encountered.

There will be $0$ to $12$ royal cards in the deck before the first of the $40$ remaining cards is encountered.   Four of these $40$ cards is an ace, so the probability that an ace is encountered first among the non-royal cards is clearly $1/10$.

Consider the sample space Ω to be the set of all possible sequences of 52 cards drawn from the deck (without replacement). So we do not remove the Jacks, Queens and Kings from the deck.

We do not remove them, but we may safely ignore them.

Still, if we insist on not doing so, there are $52!/40!$ ways to select places in the deck for the 12 royal cards and arrange them.   For each of these arrangements, there are $40!$ ways to arrange the remaining cards.   Among these arrangements there are $4\cdot 39!$ in which one from the four aces are placed before the other thirty nine cards.

Thus there is a total of $52!$ ways to arrange the cards in the deck, among which there are $52!\cdot 4/40$ ways to arrange the cards so an ace occurs before the first non-royal.

Hence the probability that an ace is encountered before any other non-royal is simply $1/10$.

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  • $\begingroup$ Thank you guys!!! very clear now!! :) $\endgroup$ – Kellen Choi Nov 20 '17 at 11:44

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