# Old Maid

##### New Member
Hi everyone! So, my wife and I are playing an old children’s version of Old Maid which has a 37 card deck which contains 18 pairs of 2 identical characters and one odd card out, the Old Maid. She just shuffled and split the deck and we each got one of the 18 cards so that neither of us could make a pair (I got the old maid). I’m wondering, what is the probability of shuffling and dealing the deck evenly like that? Thanks!

#### fed2

##### Active Member
very unlikely id say.

Basically you got '37 choose 19' possible deals, assuming you get the more cards. For each possible pair you can get one of the possible pairs, so there are 2^18 deals that give no pairs. Something like that....

So ur looking at 2^18 / choose(37, 18 )... or like, 0 or 1 in a million about. (==wife is cheating?)

Heres some r code for your enjoyments:

C-like:
NCards = 27;

myDeck = 1:NCards;

runSim = function(i){

myDeal = sample( myDeck );

spouse1 = myDeal[  1: floor(NCards/2) ]

spouse2 = myDeal[  ceiling(NCards/2):NCards ]

pairs2 = intersect( spouse2[ spouse2 %% 2 == 0 ], spouse2[ spouse2 %% 2 == 1 ] + 1);

noPairs = length(pairs2) == 0;

length(pairs2)

}

results = unlist(as.vector(  lapply(1:900,runSim)   ))
print(   1 -   (  sum(  results >= 1 )/900  )  )

print( 2^floor( NCards/2) / choose(NCards, floor(NCards/2) )  )