Problem:
The 52 cards in a deck are numbered 1,2,β¦,52. Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked. The two persons with lower numbered cards form a team, and the two persons with higher numbered cards form another team. Let p(a) be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards a and a+9, and Dylan picks the other of these two cards. The minimum value of p(a) for which p(a)β₯21β can be written as nmβ, where m and n are relatively prime positive integers. Find m+n.
Solution:
Alex and Dylan are on the same team if Blair and Corey picked cards numbered b and c with either 1β€b,cβ€aβ1 or a+10β€b,cβ€52 from the 50 cards from the deck excluding the cards numbered a and a+9.
Thus
p(a)=50β
49(aβ1)(aβ2)+(43βa)(42βa)β=25β
49a2β44a+904β.
Because p(a)β₯21β, it follows that
p(a)=25β
49a2β44a+904ββ₯21β
and thus
(aβ22)2+420β₯225β
49β
Hence (aβ22)2β₯2385β. Because a is an integer, it follows that aβ22β₯14 or aβ22β€β14; that is, aβ₯36 or aβ€8. Thus the minimum possible value of p(a) is equal to
p(8)=p(36)=17588β,
and the requested sum is 263β .
The problems on this page are the property of the MAA's American Mathematics Competitions