Emprworm's Riddle #17: Mess of Tokens

[emprworm]

hmmmm well the only thing i can say is that no one has it right so far.

the best way to solve it is to figure out the odds of each person, with each variation.

[emprworm]

no one has yet accurately stated the odds of robot #1, let alone any of the others.

[SurlyPIG]

...I knew I should have taken statistics...

Ironic, I think I know how to solve it but I don't know how to calculate the final odds...

[Khan]

[hide] If robot 1 shot at no-one robot two and robot 3 would try and destroy each other, then robot 1 could try and kill the victor. I'm not right am I[/hide]

[guusr5644]

[hide]robot 1 shoot robot 2. robot 2 tryes to shoot miss, robot3 shoots robot2 robot2 down, robot 1 shoots and hit robot 1 win... [/hide]

[guusr5644]

i tought about it again:

[hide]robot1 shoots robot2 but misses him, robot2 gets angry and shoot robot1 misses tho the shot of robot 2 hits robot 3 and kills it, robot1 shoots robot2 and hit score is for robot1 am i right?.

and btw emprworm a silly q: is there are diffrences of the armors? [/hide]

[Anathema]

Grmbl I hate math >:(

[emprworm]

no differences in armors goldeagle. we can assume that any shot that hits will destroy the robot. And surprisingly, the math is not that complex. I really think most of you can compute this. Its just that no one has really thought this thing through well enough. heh, like i said, this riddle is a bit harder than the others. maybe too hard eh? well, we shall see...

if you all give up, i'll post the answer in a few hours.

[guusr5644]

can you IM me the answer and then ill edit this post and say the answer and take credit about it? ::) :P

[emprworm]

no.

:P

[Edric O]

Someone wrote an entire book without the letter E. Was it 'The Butterfly and the Diving bell', or am I getting confused?

It was a novel in french.

the butterfly = le papillon

There's an E in the "le". ;D

[Edric O]

Now for the riddle:

[hide]Let's assume that R2 and R3 choose their target randomly, so they each have 1/2 chance of choosing R1 as their target. (assuming they both live) Due to lack of informaton, this is the best assumption we can make.

Now let's see what R1 can do:

1. Shoot R3.

1.a. HIT (1/3 chance) -> R3 is dead, and then R1 is left with 1/2 chances to survive (as R2's only target). 1/2 * 1/3 = 1/6 chance of survival

1.b. MISS (2/3 chance) -> R2 then has a 1/2 chance to choose R1 as a target and 1/2 chance to hit him, so R1 is left with a 3/4 chance of survival. But R3 has a 1/2 chance to choose him and 1/1 chance to hit him => 1/2 chances of survival. 3/4 * 1/2 = 3/8 chances of survival. But since this possibility has a likelihood of 2/3 -> 3/8 * 2/3 = 1/4 total chances of survival.

TOTAL chances of survival for shooting R3 = 1/4 + 1/6 = 10/24 = 5/12

so TOTAL for shooting R3 = 5/12

2. Shoot R2.

2.a. HIT (1/3 chance) -> That was stupid! Now R3 is sure to kill you. 0 chances of survival. 0 * 1/3 = you are the weakest link. Goodbye. :)

2.b. MISS (2/3 chance) -> same as 1.b. -> 1/4 chances of survival

so TOTAL for shooting R2 = 1/4

Therefore, with 5/12 chances of survival, it's better to shoot R3. And whatever you do, DO NOT kill R2. ;)[/hide]

[emprworm]

edric has the best computations so far...

but I am sad to say, no one has solved this riddle yet.

hmmmpfft

i guess i'll make my next one a bit easier.

i'll post the solution in a couple hours.

[Edric O]

Hmmm, yes, I see that my computations were a bit incomplete...

[hide]I did not take into account all the things that might happen if you miss. R2 might or might not kill you, but he also might or might not kill R3. So let's see what we can add to your 3/8 chances of survival...

R2 has a 1/4 chance of killing R3 (1/2 for decision * 1/2 for accuracy). That means you survive, since there is no more R3 to shoot you. So if you miss, you get an extra 1/4 chances of survival... 1/4 + 3/8 = 5/8

5/8!!!! That's more than the 1/2 if you hit and kill R2!

That's it! Your best chance is to actually MISS! Don't shoot any of them![/hide]

[nemafakei]

[hide]Written before reading other answers:

R1 does not shoot. R2 then should take out the most powerful opponent, R3. If he succeeds, R1 has a 30% chance of killing R2 in the first shot. If he fails, R3 would kill R2, R1 would have a 30% chance to hit, and otherwise die. Chances are almost balanced, because R1 gets the first shot.

If R1 were to shoot R3 (and win), then R2 would have a 50% chance of winning next shot, and so on. Chances are, R2 will win.

So R2 must refrain from shooting[/hide]

I've now read the other answers, and I reckon the ones who've posted first have it too, but the probabilities aren't quite right.

I can't be bothered to work out the sum to infinity of the probablilities, but approximations are that the best method gives you about a 50% chance.

[Edric O]

Well? Am I right? ???

[guusr5644]

mm nema's answer seemed to me like a right answer..

[Edric O]

Err, yes, but I gave the same answer before he did...

[emprworm]

SOLUTION TO THE ROBOT DUEL

(contains spoiler unhidden, so please skip if you dont want to know the answer)

Ok, here is the solution to the riddle. IT WAS A HARD ONE! whew.

I am not giving credit to anyone. Exact were odds not needed, I was just looking for the ballpark. I almost gave Nema credit, but in his answer he said that R2 must refrain, and this is incorrect. And even though Edric was the first person to actually state Robot 1's best move, his odds were not close enough.

Before I explain the odds, I need to show you some basic math probabilities. Assume two people are in a contest flipping a coin. First player to flip a heads wins. Player A flips first, followed by player B, repeat.

What are the odds that player A will win? Well since he flips first, we know it will be more than 50/50. Here is a diagram of the probabilities:

LET H = heads, and T = tails. Whoever lands HEADS first wins and the game stops. In a series of flips, here are your odds that the last flip will be HEADS:

String Probability. Player A is in red, player B is in Green.

H 1/2

TH 1/4

TTH 1/8

TTTH 1/16

TTTTH 1/32

TTTTTH 1/64

TTTTTTH 1/128

To prove this, imagine flipping a coin 3 times. What are the possible combinations of flips?

HHH

[zamboe]

Emprworm your answer does not seem mathematically correct.

Unless you prove all the chances (odds) give a grand total of 1 or 100% the whole solution is incorrect.

I've done it with your results and it's not 100%. As you know, all odds in the bottom line should be 100%.

I keep my answers.

[emprworm]

zamboe, the odds are precisely correct.

how do you figure they are wrong?

try using my utility with 1 million iterations and you will see it adds up precicely.

added:

ahh yes i see a typo i made in my solution. i fixed it.

[zamboe]

zamboe, the odds are precisely correct.

how do you figure they are wrong?

try using my utility with 1 million iterations and you will see it adds up precicely.

You didn't answer my Q, how much is total add (sum) of those odds in your perspective ?

If it is not 100%, then is wrong.

My calculator, that can use the very same principles as any utility program, gives me a different result, and I tell you the your numbers are wrong, at this precise moment I double checked it, try using exact percentages like 0.3 (30%) and 0.7 (complementary odd), 0.5 (50%) and so on.

:)

[emprworm]

yes, zam, i had a typo in my solution and i fixed it. sorry bout that.

[zamboe]

yes, zam, i had a typo in my solution and i fixed it. sorry bout that.

Well as I said in my answer-post (when i posted the solution i think it's the correct one) we agree in two things :

1. R1 should miss the first shoot.

2. If R1 chooses R2 or R3 for the first shoot, that is not important.

But the odds, I just don't agree with you, to make it short :

Better chance, one case :

Odds = Chance of R1 missing a shoot * Chance of R2 hitting R3 * Chance of R1 shooting R2 = 0.7*0.5*0.3= 10.5%

This is a max value, and it repeats it self in a few combination more.

[emprworm]

If R1 chooses R2 or R3 for the first shoot, that is not important.

But the odds, I just don't agree with you,

what do you mean its not important what R1 chooses? Of course it is. If R1 chooses R2 that would be the dumbest thing he could do.

I'm not following you here Zamboe.