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General*
Windows Bob - the best!
Sat 22nd Sep '07 1:20PM
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7th Apr '03
You're a cowboy, and get involved in a three way pistol duel with two other cowboys. You are a poor shot, with an accuracy of only 33%. The other two cowboys shoot with accuracies of 50% and 100%, respectively. The rules of the duel are one shot per cowboy per round. The shooting order is from worst shooter to best shooter, so you get to shoot first, the 50% guy goes second, and the 100% guy goes third, then repeat. If a cowboy is shot he's out for good, and his turn is skipped. Where or who should you shoot first?
    

Amanshu*
Giggity Giggity goo
Mon 24th Sep '07 9:58AM
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25th Aug '04
Ok, so for ease let's name the cowboys A, B and C. A is you, B is the 50% shot and C is the 100% shot. For ease of maths let us change those percentages to fractions of 1, so A has 1/3 chance of hitting, B has 1/2 chance of hitting and C has 1 chance of hitting. Let us also assume you have a choice of three things: shoot at B, shoot at C or deliberately miss.

[[

First consider the case that you deliberately miss. Now obviously if you're going to miss you're going to miss, so we can assume this is automatic. Cowboy B then shoots and the game continues as before. Let us say that your chance of winning in this situation is Y any your chance of losing is Z.

Now consider the case that you shoot B. If you hit B then C will shoot you and you'll lose. If you miss B then Cowboy B shoots and the game continues as above. In other words in the case that you shoot at B you have a 1/3 chance of hitting and if you do you'll lose. You also have a 2/3 chance of missing at which point the odds of you winning would be exactly the same as if you'd deliberately lost.
From this we can say the odds of you winning if you shoot at B are 2/3 x Y and the chances of losing are 1/3 + 2/3 x Z. So it should be obvious that shooting at B is just bad.

Now consider shooting at C. If you hit C then it descends into a two way fight between you and B. Let's say the odds of you winning this fight are S and the odds of you losing are T. If you miss then again it's as if you deliberately missed. So now the odds of you winning are (1/3 x S) + (2/3 x Y) and the odds of you losing are (1/3 x T) + (2/3 x Z). This will only be the better option if S > Y.

Unfortunately at this point logic fails me and I can't think of any reason why S or Y would be greater. Not without constructing the entire tree anyway. Hate that.

]]
   

Epicure_mammon
I'm not crazy cause I take the RIGHT pills :)
Mon 24th Sep '07 10:53AM
140 Posts
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12th Dec '06
How about this:
[[

Sticking to the ABC convention (I am A) and assuming that B and C will attempt to shoot the gunman who poses the most threat:

A deliberately misses:
B shoots C:
1: hits (1/2) -> A shoots C
1.1 lives (1/2) win.
1.2 dies (1/2) dead. (see below)
2: misses (1/2) -> C shoots B -> A shoots C
2.1 hits (1/3) win.
2.2 misses (2/3) -> C shoots A. dead.

=> dead: 1/2 * 1/2 + 1/2 * 2/3 = 1/4 + 2/6 = 21/36
win: 1/2 * 1/2 + 1/2 * 1/3 = 1/4 + 1/6 = 15/36

A shoots B:
1: hits (1/3) -> C shoots A -> Dead
2: misses (2/3) -> C shoots B -> A shoots C
2.1: hits (1/3) win
2.2: miss (2/3) dead.

=> dead: 1/3 + 2/3*2/3 = 1/3 + 4/9 = 7/9 = 28/36
win: 2/3 * 1/3 = 2/9 = 8/36

A shoots C
1: hits (1/3) -> B shoots A
1.1: hits (1/2) -> C shoots A dead.
1.2: misses (1/2)
1.2.1 dies (1/2) dead.
1.2.2 lives (1/2) win. (see below)

2: misses (2/3) -> B shoots C
2.1: hits (1/2) -> C shoots A dead.
2.2: misses (1/2) -> C shoots B -> A shoots C
2.2.1: hits (1/3) win.
2.2.2: misses (2/3) dead.

=> dead: 1/3*1/2 + 1/3*1/2*1/2 + 2/3*1/2 + 2/3*1/2*2/3 = 3/12 + 5/9 = 9/36 + 20/36 = 29/36
win: 1/3*1/2*1/2 + 2/3*1/2*1/3 = 1/12 + 1/9 = 3/36 + 4/36 = 7/36

So from this, rather surprisingly, you should deliberately miss

]]
  

Epicure_mammon
I'm not crazy cause I take the RIGHT pills :)
Mon 24th Sep '07 11:13AM
140 Posts
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12th Dec '06
I realised I missed out the "see below" bit:

[[
Above I have asserted that if A and B are in a head to head battle (with A shooting first) then there is a 50% chance of survival. This is why:

A shoots B:
1. Wins (1/3)
2. Misses (2/3) -> B shoots A
2.1 Hits (1/2) dead
2.2 misses - battle continues

So at this point look at the probabilities:
There's a 1/3 chance that A shoots B,
There's a 2/3 chance that B gets a go, and a 1/2 chance that B will kill A -> 1/3 chance that B hits A.

So the three scenarios in the 1st round ( A wins, B wins, neither win) are equal so in the limit there's an equal chance of A or B winning - isn't game theory great

]]
  

Epicure_mammon
I'm not crazy cause I take the RIGHT pills :)
Mon 24th Sep '07 11:17AM
140 Posts
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12th Dec '06
New puzzle:

There are two trains facing each other on a track. They are 100 miles apart and traveling at 20 mph. A fly flies from the front of the first train to the front of the second and turns round and flies to the front of the first train and turns round etc... Assuming that the fly is traveling at a constant speed of 50 mph and it can turn instantaneously how far will it have travelled when the trains collide?
  

Agentgonzo
There's no pee in catheter!
Mon 24th Sep '07 12:55PM
811 Posts
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Member Since
8th Aug '06
[[
The trains have a relative velocity of 40mph towards eachother. 100 miles at 40mph takes 2.5 hours. The fly flies for the same amount of time at a constant speed of 50mph. 2.5 hours at 50mph = 125 miles before it gets squished twixt the two trains.
]]
  

Amanshu*
Giggity Giggity goo
Mon 24th Sep '07 1:18PM
2708 Posts
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Member Since
25th Aug '04
An interesting variation on the cowboy one is if all three of them are deadshots (i.e. 100%) but they can also choose to shoot themselves in the foot. Each time they shoot themselves in the foot they'll reduce their accuracy by 10% due to pain and what not.
   

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