Problem:
The 52 cards in a deck are numbered 1,2,…,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)≥21​ can be written as nm​, 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 1≤b,c≤a−1 or a+10≤b,c≤52 from the 50 cards from the deck excluding the cards numbered a and a+9.
Thus
p(a)=50⋅49(a−1)(a−2)+(43−a)(42−a)​=25⋅49a2−44a+904​.
Because p(a)≥21​, it follows that
p(a)=25⋅49a2−44a+904​≥21​
and thus
(a−22)2+420≥225⋅49​
Hence (a−22)2≥2385​. Because a is an integer, it follows that a−22≥14 or a−22≤−14; that is, a≥36 or a≤8. Thus the minimum possible value of p(a) is equal to
p(8)=p(36)=17588​,
and the requested sum is 263​ .
The problems on this page are the property of the MAA's American Mathematics Competitions