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08-26-2002, 09:15 PM
two midget subs want to sink an enemy carrier

one sub carries the torp, the other is packed with radar-jammimg devices

they have to go through a narrow, guarded channel, one ahead of the other

the enemy can attack one sub only

the chance the first sub in the channel will survive an attack is 80% but for the second it is 60% as the enemy has more warning

if the sub carrying the torp gets through it is 100% certain to succeed in its mission

which sub should carry the torpedo and how likely are we to destroy the enemy aircraft carrier?

08-27-2002, 01:23 AM
If sub carrying torp is first, then carrier is sunk 80% of time by definition -- right?

If torp is in second sub, then 20% of time the carrier is sunk when the first sub is sunk; and (.8 * .6 = 0.48) and the carrier is sunk 48% of time when both subs get through the channel. 20% + 48% = 68% -- therefore I would send the torp with the first sub. It seems like there is really no need for the radar jamming sub -- it just makes things worse....

08-27-2002, 09:07 AM
I would send Torpedo first 2/3rds of time and would hope to have an 86.67% chance.

Please let me know if Im right, though Ive deliberately left out workings to let others try the problem.

If I am right, you have taught me something (or at least re-taught, I think I knew all this many years ago)

08-27-2002, 09:19 AM

08-27-2002, 03:02 PM
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!