we have to make our table to show the percentage probability the carrier is destroyed



our only two possible strategies are to put the torpedo in the first or the second sub



the enemy's choices are to attack the first or the second sub



<PRE> attack on first attack on second

torp in first 80 100

torp in second 100 60 </PRE>

using our algorithm (the mathematical word for a recipe that leads to the answer), as described in the previous problem, on the two columns, the differences are 80-100 = -20 and 100-60 = 40 which translates to proportions 40 and 20



therefore, we have to put our torpedos in the first submarine with the probability of 40/60 = 2/3 and in the second with probability 20/60 = 1/3



this gives us the maximum chance of success of 80 x (2/3) + 100 x (1/3) = 86.67%



there is NO enemy strategy that can reduce our chance of success below 86.67%



for example, if they always attacked the first sub, then we would succeed 80 x 2/3 + 100 x 1/3 = 86.67%



if they always attacked the second sub we would succeed 100 x 2/3 + 60 x 1/3 = 86.67%



if they attacked the first sub 1/2 times and the second 1/2 times we would succeed 1/2 x 2/3 x 80 + 1/2 x 2/3 x 100 + 1/2 x 1/3 x 60 + 1/2 x 1/3 x 100 = 86.67%



any naval officer who concludes that he should automatically put the torpedo in the first submarine because it has an 80% chance of success should be demoted - if not worse!