Puzzle time - who’s bullet?

[jon-nyc]

An easier one. Use a spoiler to give more folks a chance.

PUZZLE: Whose Bullet?

Two marksmen, one of whom (lets call him Jon) hits a certain small target 75% of the time and the other (lets call him Klaus) only 25%, aim simultaneously at that target. One bullet hits. What's the probability that it came from Jon?

[taiwan_girl]

||Thinking out loud here. Not sure if the correct answer is the easiest.

If each shoot 4 bullEts, 3 from Jon will hit and 1 from allays will hit, so 8 bullets. 3J and 1K will hit.

If scaled down to a total of 2 bullets shot, just divide by 4.

Chance would be 3/4 for Jon and 1/4 for Klaus

Seems too obvious so probably wrong. .||

[George K]

I hated statistics.

[jon-nyc]

Hint:

There are four possible outcomes when we both shoot:

1. we both hit,
2. we both miss,
3. Jon hits and Klaus misses,
4. Jon misses and Klaus hits

First step is to calculate the probabilities of each of those outcomes, knowing they must sum to 1.

[taiwan_girl]

@jon-nyc #1 and #2 cannot happen. In the riddle, one bullet hits you said. I think that leaves only 3 and 4

[Horace]

@jon-nyc said in Puzzle time - who’s bullet?:

Hint:

There are four possible outcomes when we both shoot:

1. we both hit,
2. we both miss,
3. Jon hits and Klaus misses,
4. Jon misses and Klaus hits

First step is to calculate the probabilities of each of those outcomes, knowing they must sum to 1.

maybe a simpler first step is to ignore all outcomes that don't fit the problem description and compare the chances of all that do.

[Horace]

@taiwan_girl said in Puzzle time - who’s bullet?:

@jon-nyc #1 and #2 cannot happen. In the riddle, one bullet hits you said. I think that leaves only 3 and 4

Yes good job TG. So regardless of what the absolute probabilities of 3 and 4 are, the puzzle is to determine the ratio between the two.

[taiwan_girl]

I will think some more. Should not be too difficult. Lol

[Doctor Phibes]

||spoiler||
Both hit - chance is .75 x .25 = .1875
Both miss - chance is .25 x .75 = .1875
J hit, K miss - chance is .75 x .75 = .5625
J miss, K hit - chance is .25 x .25 = .0625

If one bullet hits, either case 3 or 4 has taken place. Chance it was Jon is .5625/(.5625+.0625) = 90%.

[taiwan_girl]

@Doctor-Phibes your spoiler didn’t work. LOL

but what you did makes sense. I am going to claim that I would have done the same answer.

[Doctor Phibes]

Bugger, sorry.

[brenda]

@Doctor-Phibes said in Puzzle time - who’s bullet?:

Bugger, sorry.

Don't be sorry. You are t3h stats jock!

[jon-nyc]

@taiwan_girl said in Puzzle time - who’s bullet?:

@jon-nyc #1 and #2 cannot happen. In the riddle, one bullet hits you said. I think that leaves only 3 and 4

Well it’s an intermediate step which allows you to get to the conditional probability that the target is hit by only one bullet.

[jon-nyc]

To make the problem more realistic instead of shooting a target the puzzle should have been successfully playing a Rachmaninoff etude.

[taiwan_girl]

@jon-nyc

[Mik]

Neither one. Klaus can't shoot and you're drunk.

[Doctor Phibes]

What's the probability of somebody's family member shooting either Klaus or Jon as they attempt to play a Rachmaninoff etude?

[Klaus]

Exact Bayesian inference for the win.

normalize $ do
    k <- coin 0.25
    j <- coin 0.75
    if ((k && not j) || (j && not k)) then (if k then return "Klaus" else return "Jon") else fail

Result:

Dist [(0.9,"Jon"),(0.1,"Klaus")]

[Doctor Phibes]

Fucking IT people.

[Horace]

@Doctor-Phibes said in Puzzle time - who’s bullet?:

Fucking IT people.

Not in this lifetime. Nerds.