Western Shoot Out Puzzle

i listen to Car Talk, so I'll disqualify myself.

I bet Andy chose to use a Mac10 auto. 60 rounds a min.

Intentionlly miss Bill. He has a one-in-three chance to Kill Bill and it could happen on the first shot. Then, no chance to kill Charley and Andy's a dead man!

Regards,

Mike

Andy shoot himself first, in the foot. Charley and Bill will laugh so much they'll miss. Giving him his second shot. Charley will shoot Bill because odds are he will shoot him being the best shot, and thus the bigger threat. Andy shoots Bill as well, his second shot. He now stands a fifty fifty chance of success.

Think of a good tune.

Does it involve a certain type of watch?

Interesting puzzle. Skipping Andy, if given the choice Bill would always shoot Charlie first because Bill is twice as dangerous as Andy, so Charlie if given the choice would always gun down Bill.

In a match between Andy and Bill, if Andy gets to shoot first, he has a 50% chance of eventually winning. But if Bill gets to shoot first, he has an 86% chance of winning.

What I get is that if Andy shoots at Bill first, he has about a 30% chance of survival. If he shoots at Charlie first, he has about a 38% chance of survival. But I could be wrong.

"Andy scores a kill in 1 out of 3 shots"

Andy MUST shoot at Charley first. IF Andy kills Charley there is a 66% chance Bill will kill Andy. This is Andy's best chance for survival. if Andy were to shoot at Bill first (33%) and get lucky, Charley kills Andy on the next shot. However, if Andy's first shot at charley misses, Bill MUST shoot at Charley. This is Bill's only chance for survival. If Bill kills charley, Andy must then shoot at Bill. Andy's new odds for killing Bill are up to 50%. If Bill does NOT kill Charley, Andy MUST shot at Charley again. Again Andy's odds of killing Charley are 50%. So on and so fourth. . .

My basic assumption is that Andy WILL kill in one out of three shots as directly stated in the question. It's like pulling buttons out of a bag. 1 bag, 3 buttons, 2 black 1 red. there is a 33% chance the first button is red. there is a 50% chance the second button is red, and 100% chance the third button is red (provided the first 2 buttons were black).

-A-

The method I used was to first figure out what would happen if only Andy and Bill were left. This is a useful subset to use for calculating the overall odds.

I errored in my calculation previously. If Andy shoots first he has about a 43% chance of winning. Bill still has the approximately 86% chance of winning if he shoots first.

The logic is:

• Andy shoots first, 33% of time he wins on first shot.
• 67% of the time Bill will then get a shot and he will win 67% of this 67% or 44%
• Then about 23% of the time Andy will get a second shot and 33% of this is about 7.5% chance he will win on his second shot.
• About 15.5 % of the time Bill will get a second shot. Odds of wining on this shot 9.9% etc.
• The number quickly dwindle toward 0 and it sums up to about 43% that Andy wins if he gets to shoot first. Similarly if Bill gets to shoot first, it comes out to about 86% that Bill wins.

Plugging these probabilities into my spreadsheet, I still get that Andy should shoot at Charlie first. His odds of winning are 33%. If he shoots at Bill first, his odds of winning drop to 27%.

He should buy a lotto ticket. With his winnings buy the saloon and supply free drinks to Bill and Charley. A drunk cowboy cannot aim or farm. They will all live happily ever after.

Get someone to stand in for him

or use a shot gun so he wont miss

tALLY HO

What I really want to know is...if Andy deliberately misses on his first shot, has he hit what he was aiming at ( anything but Bill and Charlie) and therefore used his 1 in 3 chance of hitting his target ???????