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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

2031.0. "Cards, colour and probability" by AUSSIE::GARSON (achtentachtig kacheltjes) Sat Mar 16 1996 18:20

    From a normal deck of playing cards I choose at random n=4 cards in a
    manner such that I can see the cards and my colleague cannot. On noting
    the number of black cards and the number of red cards amongst those
    that I chose, I quickly compute that if my colleague chooses two cards
    at random from the cards that I chose, the probability that he chooses
    two red cards is exactly 0.5.
    
    I now repeat the procedure with a different value for n and find that
    the probability is again 0.5. What is this value for n?
    
    [dg: This problem comes from breakfast radio where listeners are
     invited to phone in as soon as they have a solution. There are
     obviously some people who are more alive at 7:30am than I am!
    
     P.S. There is a unique solution subject to the physical constraints of
     the scenario.]

T.R	Title	User	Personal Name	Date	Lines
2031.1		CSC32::D_DERAMO	Dan D'Eramo, Customer Support Center	`Sun Mar 17 1996 16:52`	5
	Spoiler... 21 Dan
2031.2	tough for brekkie time	TPOVC::BUCHANAN		`Mon Mar 18 1996 07:07`	8
	SPeller... And the next one would be 120, requiring three decks of cards, contrary to the problem statement. Andrew
2031.3		AUSSIE::GARSON	achtentachtig kacheltjes	`Mon Mar 18 1996 16:58`	4
	re .2 Wouldn't that be four packs of cards? 120 = 35->2 packs + 85->4 packs
2031.4	High quality radio	CHEFS::STRANGEWAYS	Andy Strangeways@REO DTN 830-3216	`Tue Mar 19 1996 05:34`	7
	So how long was it before the first calls came in? Does this program make a habit of throwing non-trivial math questions at breakfasting listeners? If so, they're to be highly commended for breaking with the "listeners/viewers should be always be treated as morons" code. We could do with more of this! Andy.
2031.5		AUSSIE::GARSON	achtentachtig kacheltjes	`Tue Mar 19 1996 16:37`	7
	re .4 With the caveat that I don't normally listen to it and wasn't listening all that closely...response time on this problem was not more than 10 minutes; and I gained the impression that some kind of mind-stretching "puzzle" is set each day. If I can catch another one, I'll post it so you have more of the flavour. This is on "state" radio.
2031.6		AUSS::GARSON	achtentachtig kacheltjes	`Wed Apr 03 1996 23:22`	4
	re .4 I caught one the other day...The question was "What is the probability of being dealt a royal flush?" [A royal flush is A,K,Q,J,10 of same suit.]
2031.7	Never in 5 cards, for me	EVMS::HALLYB	Fish have no concept of fire	`Thu Apr 04 1996 11:03`	6
	Standard 52-card deck, no wild cards? That's easy to calculate. But I can tell you from memory: 4 in 2,598,950 or ~0.00000154 If you're playing with a pinochle deck the odds are considerably lower. John
2031.8	misspent youth? (-:	AUSS::GARSON	achtentachtig kacheltjes	`Sat Apr 06 1996 05:16`	13
	re .7 Yes, standard deck. 4/C(52,5) is the easy way to see the answer immediately. The caller supplied solution was 20 4 3 2 1 -- * -- * -- * -- * -- 52 51 50 49 48 by looking at the number of cards that can be dealt at each stage that will lead to a royal flush and the number of cards remaining.
2031.9		AUSS::GARSON	achtentachtig kacheltjes	`Mon Apr 08 1996 20:05`	22
	re .7 So, how about the following... A standard deck of 52 cards is shuffled and then displayed to you i.e. all the cards and their order. Five cards are dealt to you. You then get the chance to discard from zero to five cards, replacing them with the same number of cards from the top of the deck. What is the probability of finishing with a royal flush (assuming 'sensible' decision making)? What if you get a second opportunity for discard and replace? [Presumably the probability is going up at each stage.] This of course corresponds to an ideal situation (of perfect information or possibly a bit of cheating). What if you have to specify your discard/ replace strategy in advance and can't see the cards that you could get as replacement? i.e. a more realistic scenario. What is the optimal strategy and what would the probability of finishing with a royal flush be? [Presumably the probability of finishing with a royal flush is between the answer in reply .7 and the answer above.]
2031.10		HANNAH::OSMAN	see HANNAH::IGLOO$:[OSMAN]ERIC.VT240	`Tue Apr 09 1996 16:41`	14
	A standard deck of 52 cards is shuffled and then displayed to you i.e. all the cards and their order. Five cards are dealt to you. You then get the chance to discard from zero to five cards, replacing them with the same number of cards from the top of the deck. What is the probability of finishing with a royal flush (assuming 'sensible' decision making)? The answer is 0 or 1, depending on what the first 5 cards are on the top of the deck ! This is because you said "and then displayed to you". Hence we know for sure whether a replacement from zero to five cards will or won't produce a royal flush. /Eric
2031.11	nitpicking	FLOYD::YODER	MFY	`Wed Apr 10 1996 18:43`	5
	>The answer is 0 or 1, depending on what the first 5 cards are on the top of the deck ! Presumably you meant 10, not 5; and 15 for the case where you get to draw twice, etc.
2031.12		AUSS::GARSON	achtentachtig kacheltjes	`Wed Apr 10 1996 19:45`	13
	re .10 You have to specify the probability of finishing with a royal flush now before we even start the procedure - but perhaps the wording could have been better. Looked at another way, yes, each of the 52! shufflings of the pack leads to an outcome of 0 or 1 and the probability is the average of those outcomes. re .* Just in case people didn't notice, it is not the case that you finish with a royal flush if the first 10 (or 15) cards contain a royal flush. The order is important too.
2031.13	how to solve 1-draw case	FLOYD::YODER	MFY	`Thu Apr 11 1996 14:22`	26
	You can use the principle of inclusion and exclusion to simplify the problem when 1 draw is allowed. The answer is 4p - 6q where p is the probability that you can get some particular royal flush (say spades), and q is the probability that you can get either of two particular royal flushes (say spades and hearts). You can get two iff the top 5 cards are a royal flush, and the next 5 are also. This is because the 10th card is obtainable only if 5 cards are discarded, whence the 6th through 10th cards must be of the same suit. So q = 25!5!42!/52! (the factor of 2 is there since either spades or hearts can come first). p is the sum over 0 <= i <= 5 that i cards must be drawn to get the 10 through ace of spades; thus there must be 5-i of the needed cards in the first 5, and the next i cards must be exactly the remaining needed high spades. There are C(5,5-i)=C(5,i) ways that the 5-i high spades can appear among the top 5 cards; there are 5! ways to order the 5 high spades in the overall arrangement, and 47! ways to order the remaining cards, for any such set of positions, so the overall probability is [6(sum C(5,i))5!47! - 125!5!42!]/52! = (6325!47! - 125!5!42!)/52! since the sum of C(5,i) is 2^5 = 32.
2031.14		AUSS::GARSON	achtentachtig kacheltjes	`Wed Apr 24 1996 00:02`	23
	re .13 >[6(sum C(5,i))5!47! - 125!5!42!]/52! > > = (6325!47! - 125!5!42!)/52! It looks to me that both of those occurences of 6 should be a 4. Assuming that that is the case then this evaluates to 1090801 -------------------- 131749234744*43 which is slightly less than 32 times the probability of being dealt a royal flush. [The "32 times" is the "4p" and the "slightly less" is the "-6q".] Anyway the probability is about 1 in 20304. This was on the basis of full information. What happens if we are forced to decide how many to discard from the first five before seeing the next five? Obviously this means specifying some strategy. It seems to me (intuitively) that no strategy can equal the case of full information.
2031.15	I stand corrected	FLOYD::YODER	MFY	`Wed Apr 24 1996 10:23`	4
	"It looks to me that both of those occurrences of 6 should be a 4." You are right.
2031.16	strategy - the easy part	CSSE::NEILSEN	Wally Neilsen-Steinhardt	`Wed Apr 24 1996 13:20`	24
	Since we don't have to worry about bluffing or going for lower value hands, the strategy is pretty simple: Identify the cards you are holding which require the fewest cards to complete a royal flush. Discard the rest and draw. Two nits: If you have two equal ways to complete a royal flush (for example, holding Queen of Hearts, Jack of Diamonds and three low cards) then pick one way and discard the rest. If you have discarded once, remembering the discards might change your strategy, very occasionally. (For example, you are dealt King and Queen of Hearts, Jack of Diamonds, and two low cards. You hold the King and Queen, and you draw Ace, Queen and ten of Diamonds. In this case you discard the three diamonds and draw three.) I think this strategy is optimal, but I can't prove it. And I certainly can't compute the probability of ending with a royal flush using it. I agree that this strategy is inferior to perfect information, for some oderings of the deck. For example, the five cards to be drawn next may contain a royal flush. Empirical proof: marked decks exist.