Odds of never missing in a game of Battleship

There are possible hits. So, one the first draw, the odds of a hit are 17/100.

At that point, you have a 1/5 chance that it's the battleship, a 1/5 chance that it's a destroyer, a 2/5 chance that it's a sub or frigate, and 1/5 chance that it's a tugboat.

-If it's a tugboat, you have a 1/4 chance that your next shot will be the hit sinking the ship.

-If it's a submarine or frigate, you first have a 2/3 chance that the ship is on an edge and therefore a 1/4 chance that your next shot will be a hit. After the 2nd hit, you have a 1/2 chance that your next shot will sink the ship. On the other hand, if you are the 1/3 middle shop, you have a 50% shot that the next shot will be a hit and another 50% shot that you will sink the ship.

-For a destroyer, 1/2 chance that the ship hit on an end.
-For a battleship, 2/5 chance that the ship hit is on an end.

So for the first ship hit, 17% chance of hitting, (20% * 25%) + 40% * ((2/3 * 1/4 * 1/2) + (1/3 * 1/2 * 1/2)) + 20% * ((1/2 * 1/4 * 1/2 * 1/2) + (1/2 * 1/2 * 1/2)) + 20% * ((2/5 * 1/4 * 1/2 * 1/2 * 1/2) + (3/5 * 1/2 * 1/2 * 1/2 * 1/2))

This gives me a (17% * .157917)= .026846 chance of sinking my first ship without any misses. This is ignoring any boundary conditions (the fact that the ship is bordering an edge or another ship).

After the 1st ship is sunk I'll have a 20% chance of 15/98, 40% of 14/97, 20% of 13/96, and 20% of 12/95 left.
Add it all together and the odds of a 2nd hit on my second ship is .140691 * .026846 = .003777

After two ships, the odds of sinking two ships in a row is about .000596 (using .157917) as a factor.

So, after all of this I'm going to revise the odds to about 10,000,000 to 1.

Battleships where you can place ships adjacent is like 6:5 Blackjack. Sure it looks like the same game, but it just isn't :)

Quote: roony672

17 hits wins at Battleship. My mate and I were arguing which would be harder, never missing a single hit at Battleship (going 17 for 17) or winning the lottery. What are the odds of never missing your opponent at Battleship?

There is not enough information provided.
1. Will the placer of the batlleships be done at random? Or will the placer try and make it difficult?
2. Will the shooter have to announce all 17 shots at once, or shoot one at a time? If all 17 at once I am sure there is someone who can figure it out.
If not, then strategy in both placing the ships and shooting at them will make the answer unsolvable.

Quick math would say this is very similar to keno.

20 out of 20 out of 80 = 3.5 billion billion to 1.

I asked my unix box for the answer.

[2.20-/softdev/source] > bc
scale=99
a=(1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17)
a
355687428096000
b=(18*19*20*21*22*23*24*25*26*27*28*29*30*31*32*33*34*35*36*37*38*39*40)
c=(41*42*43*44*45*46*47*48*49*50*51*52*53*54*55*56*57*58*59*60)
d=(61*62*63*64*65*66*67*68*69*70*71*72*73*74*75*76*77*78*79*80)
e=(81*82*83*84*85*86*87*88*89*90*91*92*93*94*95*96*97*98*99*100)
f=b*c*d*e
f
26238266543048976361450768383434109693249889888956583299647758123055\
09774584920717887341946885961783168463991997208555683840000000000000\
00000000
g=f/a
g
73767764813906413749646930628844166943344564493281739236205759727834\
2511784732494558114871608611228852372373504000000000000000000.000000\
00000000000000000000000000000000000000000000000000000000000000000000\
0000000000000000000000000


and without division
scale=99

b=(2*19*2*21*2*23*2*25*2*27*2*29*2*31*2*33*2*35*6*37*38*39*10)
c=(41*42*43*44*45*46*47*48*49*10*51*52*53*54*55*56*57*58*59*10)
d=(61*62*63*64*65*66*67*68*69*10*71*72*73*74*75*76*77*78*79*10)
e=(81*82*83*84*85*86*87*88*89*90*91*92*93*94*95*96*97*98*99*100)
b
10834710698718466560000
c
339944897204613461881314754560000
d
153590673962987334080971304140800000
e
1303995018204712451095685346159820800000
f=b*c*d*e
f
73767764813906413749646930628844166943344564493281739236205759727834\
2511784732494558114871608611228852372373504000000000000000000

NEVER GOING TO HAPPEN.............