larry@utcsstat.UUCP (larry) (09/11/85)
> i've come across a "magic trick" that i need help finding out why it > works. actually, i need the probability formula. here's the > description: > > pick any three cards from a deck of (52) playing cards, ignoring the > suit. if you get a duplicate, pick another one. find all the > combinations of pairs you can have. for example, if you pick the > king, seven and three, you can have > > k 7, 7 k, 3 7, 7 3, k 3, 3 k (clearly 6 choices). > > now, return the cards to the deck, shuffle to your heart's content, and > start dealing out the cards in any order. what will happen in most of the > cases is that 2 consecutive cards will be one of your choices. in other > words, if you have dealt out > > q a 2 3 1 k 4 7 3, > > you would "win" because the last 2 cards were 7 and 3, one of your choices. > things like "k k" don't count, but "k k 3" is fine. > > does anyone know how to figure out the probability of this problem? > > > gary rosenblum (ihnp4!cmcl2!rosenblg or rosenblg@csd2) Actually, as a kid I was taught the following "magic" trick. Ask someone to name any two cards(ignoring suit). Then shuffle the cards and most often you will find that these two cards occur beside each other. Given that this event occurs with high probability, it is not surprising that the aforementioned scheme does. I started to enumerate the probabilities but found the task tedious so I quit. I think a strategy is as follows. Suppose we want the probability that a 5 and 7 occur beside each other in the deck, say. Imagine 52 slots, in a row. There are 52 choose 4 ways of possible arrangements of 5's. The number of "available" or empty slots next to these cards ranges from 2 to 8. Given the number of empty slots, one can calculate the probability of any of four of the 7's falling in one of these slots. then let p(i)=Prob{ i available slots} x Prob{a 7 falls in one of the i empty slots}. The desire probability is then sum over i (from 2 to 8) of p(i). As I said, this is tedious and its early in the morning, so i won't attempt it. Perhaps there is a more succinct argument. If not, a monte carlo simulation may be the way to go. Larry Wasserman -- {allegra,ihnp4,linus,decvax}!utzoo!utcsstat!larry {ihnp4|decvax|utzoo|utcsrgv}!utcs!utzoo!utcsstat!larry
doyle@apple.UUCP (Ken Doyle) (10/03/85)
In trying to find the probability of a certain occurence, it is often easier to try to find the probability that it does NOT occur. So, in trying to find the probability that two card values X and Y occur consecutively in a deck, assume X is at position i in the deck. The probability that a Y does NOT occur at position i-1 is p1=47/51 (the number of non Y's over the number of cards left). The probability that a Y also does not occur at i+1 is p2=p1*46/50 (since we used up a non-Y at i-1). The probability that this happens for all 8 cards is p2**8 which equals about .267. Thus the probability that NO X and Y will be consecutive is about 26.7%. (There is probably a slight statistical variation at the beginning and end of the deck.) Ken Doyle