rosenblg@csd2.UUCP (Gary J. Rosenblum) (09/06/85)
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? please reply either to me or the net. thanks!! gary rosenblum (ihnp4!cmcl2!rosenblg or rosenblg@csd2)
lambert@boring.UUCP (09/10/85)
> 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? 2404226057 My computations resulted in a probability of ---------- for "winning" 2710423730 (assuming that the shuffling leaves a completely random permutation of the deck), which is a tiny bit above 88.7% Can anyone corroborate this? Lambert Meertens ...!{seismo,okstate,garfield,decvax,philabs}!lambert@mcvax.UUCP CWI (Centre for Mathematics and Computer Science), Amsterdam -- Lambert Meertens ...!{seismo,okstate,garfield,decvax,philabs}!lambert@mcvax.UUCP CWI (Centre for Mathematics and Computer Science), Amsterdam
msp@warwick.UUCP (Mike Paterson) (09/20/85)
The probability needed can be viewed as the probability of getting two adjacent picture cards with different values (e.g. JK) in a random deal. The probability of failure seems to be the coefficient of x^38 in 41!40!/52!.e^x.(24+24x+12x^2+x^3)^3 My pocket calculator is smoking but the answer looks like 49220186227/447219915450 (about .110058129) This would give a winning percentage of a tiny bit below 89.0% compared with Lambert Meertens' 88.7%. Any other offers? Mike Paterson