A football team enters a tournamet in which they play one ga

A football team enters a tournamet in which they play one game with each of three teams, but they get to choose the order in which they play their opponent teams. the team knows the probability of \"win\" against each of the three teams:

PA: is the probability to win when playing with team A

PB: is the probability to win when playing with team B

PC: is the probability to win when playing with team C

where pC<pB<pA

The team wins the tournament if they win 2 games in a row. The team wants to maximise the probability of winning. Show that the \"win\" probability is maximised if they play the weakest team second, and that the order of playing the other two teams doesnt matter.

Solution

The probability of winning tournament = either winning first 2 matches or winning second 2 matches

= [P1P2+(1-P1)P2P3]/2 where P1 is probability of winning first match and P2 second match and P3 for third match.

we know pA>pB>pC which means that descending order of product of probabilities will be

pApB, pApC, pBpC

P1P2 and P2P3 can be maximised if they are assigned the values of pApB and pApC in any order.

This means 2 represents A and 1 ,3 can be B,C in any order.

Hence proved that second game should be against A and other game opponents dont matter