Probability of winning a game of battleships without a single error?

Question

I mean every 'guess' at the ships locations is a success and a hit.

My cousin pointed out to me that once u make a hit the probability of a 2nd hit is greatly increased until the ship is sunk then its a complete guess again.

Answer 1

really small....

Answer 2

It seems to me that this is a non-trivial calculation to make because:

*When you start the game, you just have to randomly guess... Simple so far

*When you get a hit, though, you now have additional information making the 4 squares directly adjacent much more likely to have a ship than you would otherwise think from just assuming every square has an equal chance with a battle ship.

So, to really account for that additional information, you probably are getting into some fairly complicated case-analysis.

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On the other hand, if you just wanted to be naive about it and choose all of the spots you were going to attack before the game began, then the probability is easy to calculate:

Battleship is played on an 10x10 grid with 17 spaces occupied by ships. The number of different ways to chose a set of 17 spaces from a total grid of 100 is given by

C(100,17) = 100! / ( 17! (100-17)! )

Which is

6,560,134,872,937,201,800

And since only 1 of the choices for a set of 17 squares is completely correct the probability is 1 in 6,560,134,872,937,201,800

But keep in mind that's only if you don't take into account the fact that for every hit you get new information about the location of other spaces to hit.

For information on the rules of the game, see the Wikipedia article below. For information about the math, see the wikipedia article on combinations.

Answer 3

In any game of battleship, there are 100 spaces (a 10x10 grid) to place your ships. Each player has five ships; a destroyer (2 spaces), a frigate (3 spaces), a submarine (3 spaces), a battleship (4 spaces), and an aircraft carrier (5 spaces). Combined that gives us 17 spaces that count as hits. The probability would be calculated as thus:

17/100 * 16/99 * 15/98 * 14/97 * 13/96 * 12/95 * 11/94 * 10/93 * 9/92 * 8/91 * 7/90 * 6/89 * 5/88 * 4/87 * 3/86 * 2/85 * 1/84
= 1.50x10^(-19)

Basically, one quadrillionth of a percent. Not too high.

Hope this helps.

Answer 4

I have a different method af addressing this problem.

The odds other people calculated were based on the odds of 17 random guesses all being hits. However, that is not a reasonable estimate. LIke others pointed out, the odds get better as you see the order the ships are hit, and where the ships are. Ships on the borders or in the corners have less random options than ships in the center. But most importantly, you can eliminate impossible combinations before guessing the 17 shots. You will not, for example, pick a combination of shots that would make the possibility of hitting all of the ships impossible - For example, you wouldn't launch a spacing the shots so that none are next to each other, or that at least one combination is not at least 5 in a row to sink the aircraft carrier. We can use this to simplify the math.

I'm going on a mathmatical limb here, but I think the odds are closer to this. (Actually, your odds are better than even this, since your first sunk ship will eliminate possible combos on later ships, but lets keep it simple):

Pre-plan your 17 shots in groups of 5, 4, 3, 3 and 2, either vertically or horizontally.

Lets assume you launch the 5 shot combo first. Since there are 120 possible 5-shot combos, your odds of sinkng the Aircraft carrier is 1 in 120, or .00833 (rounded off)
-Then launch your 4-shot combo, that has 140 possible combos, or a .0071 possiblity.
-The 3 shot combos has a 1 in 160, or .0065 possiblity (two times)
-The 2 shot combo has a 1-in 180, or .0055 possiblity.
To combine these odds, multiply by each other:
.00833 x .0071 x .0065 x .0065 x .0055 = .0000000000137, or .00000000137% chance. Way more favorable odds than those estimated by random selection of shots.

There are other variables, such as selecting the actual order that you hit the ships, but I think this calculation of odds is closer (and millions of times better) than the odds provided in other answers.

Answer 5

None of the answers given so far seem to be correct. They are not taking into account the fact that you are getting a hit each time and therefore the probably of getting the second, third, fourth hit on a particular ship is increased greatly. To do the math it would depend on the order in which you sink each ship and whether ships are touching each other. The probably cannot be determined with the information given.

Answer 6

I have it at 1 in 2,700,000,000,000,000. Can someone more experienced with stats check my work? Here s how I got there...

Probably going to find the biggest ships before smaller ships.

First guess: 1/100
Next guess: 1/8 from surrounding squares
Hit that one and you re down to a 1/2 guess on the linear shape of the boat (carrier in this case)
Guess four: 1/2
Guess five: 1/2 . . . And it s sunk!

Continuing on, 95 squares left to choose from for next boat. 1/95.
Repeat, 8, 2, 2

Next boat, 91, 8, 2
Next boat, 88, 8, 2
Last boat, 85, 8

From there, I multiplied all the denominators and got the number listed above. I m not sure this is right, or what to do with the chance of hitting a boat on the edge of the board. Sheesh, eh? :)

Answer 7

HOW MANY SHIPS ARE THERE FOR EX. 5 IN YOUR SIDE AND 10 ON THE OTHER SIDE THEN THE PROBABILITY OF WINNING IS 5/10 OR 1/2 OR 50% SUCESS IN HITTING WILL BE NUMBER OF SHIPS / THE WHOLE AREA IN CHEATING TO WIN USE A SUBMARINE

Answer 8

Let x be the number of squares on the board and y be the number of squares that all the battleships occupy. Then the chance is:

y!(x-y)!/x!

Answer 9

Probably not

Answer 10

100% if you get the mirrors set up just right......