Three guys are fighting over TimeLady and they agree to decide with a "truel" (thats a duel of three ). All three get a gun and last man standing wins.
Man 1 shoots pretty bad. 1 in 3 kill
Man 2 shoots better. 2 in 3 kill
Man 3 shoots perfect - 3 in 3 kill = 100%
They agree to shoot one at a time. First the worst shooter (Man1) followed by Man2 (if alive), followed by Man3 (if alive), after which the cycle repeats.
We assume that a hit = death and a miss = complete miss.
You are Man 1. What do you do.
For the mathematicians: What are the probabilities of Man1, 2 and 3 to win this truel?
Spoiler:
Man 1 has to shoot at Man 3 because if he kills Man 2, Man 3 shoots him and it's game over.
It's actually ideal for him to miss, in which case Man 2 kills Man 3 or vice versa and he gets a guaranteed shot at the win. If he kills Man 3, he has to dodge Man 2's 2/3 shot
Man 1 31.2%
Man 2 54.0%
Man 3 14.8%
Optimal IRL strategy is to call a truce and all shoot TL
Ok so I actually wrote an algorithm to have a look at how many iterations various size decks required and the pattern is crazy. I have no clue how to go about finding a formula for this lol.
I havent thought about it at all, but if you have a sequence of numbers you cant explain, try entering them into the online integer sequence database. It is a pretty neat tool.
I'll give you a hint based off of what you have posted so far. Use it if you want or ignore it.
Spoiler:
You don't want to go about this by looking for a pattern as n increases. It is a good idea, but won't work for this problem as you have noticed.
Here is another hint to get you in the right direction. Again feel free to ignore.
Spoiler:
You can actually calculate the total number of iterations by doing a single iteration of the shuffle procedure and then doing some calculations based off of the result of that single iteration.
I'll give you a hint based off of what you have posted so far. Use it if you want or ignore it.
Spoiler:
You don't want to go about this by looking for a pattern as n increases. It is a good idea, but won't work for this problem as you have noticed.
Here is another hint to get you in the right direction. Again feel free to ignore.
Spoiler:
You can actually calculate the total number of iterations by doing a single iteration of the shuffle procedure and then doing some calculations based off of the result of that single iteration.
Spoiler:
Ok that's interesting. Obviously the last half of the order after the first iteration will be the odd positions in reverse order (so for a 4 card deck it's xx31, for a 7 card deck xxx7531 etc). That means the calculation must be based off the positions of the even cards. I'll have a look through and see if I can come up with anything.
Edit: On first impressions it seems as though the decks with a number of cards that is a power of two are the key to solving the problem. These always require far fewer iterations than those around them and there is a definite pattern to the order of the cards.
OMG SOME BODY ANSWERS MY PROBLEM DON'T JUST IGNORE IT IT'S RUDE!!!!1
Well, this is not really a brainteaser. I mean 11 times /12 h so 22 times/day and it is kind of relevant whether you include both the 00:00 and the 24:00 or only one of them. If you include both the answer is 23.