 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Number Cards Before Face Cards (Posted on 2019-05-02) If you turn over a well shuffled deck of cards one card at a time, what is the probability that you'll see at least one of each of the denominations of the numeric cards, ace through ten (1 - 10) before seeing any face card (J, Q, K)?

 See The Solution Submitted by Charlie Rating: 3.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Full solution: part 1 formatting correction | Comment 3 of 6 | The formatting genii swallowed my last attempt. I'll have another go and save up one of my three wishes for improving this 1990 type user interface for something in the direction of LaTeX.
Som here goes nutin'.

I have the full solution, but it is rather too long (and tedious!) to fit in the box. I will put down part now, maybe adding to it later.

Consider the case where we have already drawn N cards, which include M distinct number cards and no face cards. The remaining deck consists in

• 12 face cards
• 4M - N “familiar” number cards
• (52-N) - 12 - (4M-N) = 40 - 4M “new” number cards.

The next card to be drawn will lead to one of three events:
• Event A - With probability PA = 12/(52 - N) Draw a face card, ending the game
• Event B - With probability PB = (4M - N)/(52 - N) Draw a familiar number, leaving M unchanged
• Event C - With probability PC = (40 - 4M)/(52 - N) Draw a new number, increasing M by 1

I will consider the probabilities of meeting the requirement (=winning) with exactly 10, 11, and 12 draws, each time generalising on previous results, finally leading to a formula for the general case.

In order to win, you would need 10 type C events, and any number between 0 and 27 of type B events to occur with no type A event. This allows us to calculate probabilities for various scenarios.

For instance, the possibility of winning after only 10 draws is.
W(10) = (Product from M=0 to M=9) (40-4M)/(52-M) = 4^10 10! 42! / 52! , or about 6.6e-5

Now consider wins with one or more type B events. Note that type B events cannot occur on first draw, since there are no familiar numbers at this point. Neither can they occur on the final draw, since then the game would have been won beforehand.

For instance, a game won after exactly 11 draws includes exactly one type B event, which may occur on draws 2, …, 10 and the probability of a win with a single event B event is Probability of one B event * Probability of ten C events. Thus the win probability with a single type B event on the second draw is
U(2) = 40/52 * (3/51) * 36/50 * 32/49 * 28/48 * .. * 4/42 = (3/42) W(10)

Here, the term in parentheses corresponds to the single B event
Similarly, the win probability with a single type B event on the third draw is
U(3) = 40/52 * 36/51 * (6/50) * 32/50 * 28/48 * .. * 4/42 = (6/42) W(10)

Continuing in this fashion, and finally summing, we get the probability of winning on exactly 11 draws:
W(11)  = W(10) (3 + 6 + .. + 27) / 42 = W(10) 45/14, or about 2.1e-4

From this line of thought, it should be clear that
W(12) = W(10)  Sum (3p-3)(3q-3)  / (42 * 41 )

where p,q correspond to the two B events occuring on draws p and q, so the sum is over 12 > p > q > 1
But, sum (3p-3)(3q-3) = 9 Sum k^2 from k = 1 to k = 9 which gives 5130, so
W(12) = W(10) 5130 / (42 * 41 ) = W(10) 855/287, or about 2.0 e-4; pretty much the same as W(11)

Following this reasoning, we see:
W(10 + R) = W(10) ( (42 - R)! / 42! ) 3^R Sum a1*a2*..aR
Where the sum is carried out over all integer combinations 10 + R > aR > … > a2 > a1

The probability of a win regardless of the number of draws sums this expression for W(10+R) from R = 0 to R = 27.

 Posted by FrankM on 2019-05-05 11:35:30 Please log in:

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