Probability Puzzle #4 Russian Roulette

[Raikiribokken]

like zero said... Liar Game plagiarizing is bad...

but yeah:
in a vacuum environment, the initial odds are 2/6, increasing by 1/6 every shot

but since A & B dont live in a satelite, the spinning of the barrel is affected by gravity, making the shots more likely to appear in 3 positions depending on where the bullets were placed:

--1---2 if the shots were placed in opposing slots, you're screwed (1&6, 2&5, 3&4
3-------4 if the shots skip one slot (ex.1&4) then its likely the end result will be 3&6
---5---6 if the shots are next to each other, the result will be likely 3&5
either way... u were actually safer by going second

[Zabuza of the Mist]

I have 100% chance of winning. I load the gun, I shoot player B. Then I shoot him again with the second bullet, just to make sure he doesn't come back as a zombie.

P.S. Real men play with 6 bullets, 6 chambers, and go first. Are you a real man?

But seriously, chances increase by 1/6 every time someone lives, if the bullets are placed next to each other then you have a 3/5 chance of surviving, then you have a 2/4, or 1/2 chance of winning.

[animefanatic]

Noone has solved the problem yet. There is only one assumption you need to make for this problem. You need to assume that you want to win the game and live.

Raikiribokken, the gravity trick might have worked in Liar Game but it won't work here for two reasons.
1) Moment of Inertia is based on two factors, mass and distance. Inertia increases linearly with respect to mass but quadratically with respect to the distance from the center. If you go back and look at the cylinder in liar game, you should see how massive it was. However, the cylinder for a revolver that can hold 6 rounds is small and compact. Thus the center of gravity will not be offset significantly.
2) The biggest reason why the trick wouldn't work there is no guarantee that your opponent will let gravity stop the cylinder. Your opponent may stop the cylinder mid-spin.

[matthewlow]

Math problems about violence make me sad and not want to solve it.

I don't know much about guns in general, so I'm assuming that there are 6 slots and one at random has a bullet. So I'm going to instead think about this as someone having a deck of 6 cards, and the goal is to not draw the "bad" card.

1. Person B goes, and has a 1/6 chance of fail.
2. Assuming Person B succeeds, Person A has 1/5 chance of fail.
3. Assuming Person A succeeds, Person B has 1/4 chance of fail.
4. Assuming Person B succeeds, Person A has 1/3 chance of fail.
5. Assuming Person A succeeds, Person B has 1/2 chance of fail.
6. Assuming Person B succeeds, Person A will guarantee fail.

So Part A is 1/5, 20% fail -> 80% success, since 1) is already assumed to have happened.

Part B:
For Player A's failure rate
1. can't fail here
2. 5/6 * 1/5 = 1/6th chance fail on 2.
3. can't fail here
4. 5/6 * 4/5 * 3/4 * 1/3 = 1/6th chance fail on 4.
5. can't fail here
6. 5/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1 = 1/6th chance fail on 6.
-> 1/6 + 1/6 + 1/6 = 3/6 = 1/2 = 50% fail rate (or 50% success rate)

To explain:
1. 5/6 chance of passing to you
2. #1 must happen * 1/5 chance of failing = chances you'll fail here
3. 5/6 * 4/5 chance you'll get to this step * 3/4 to pass out of this step
4. #3 must happen * 1/3 chance to fail
5. 5/6th * 4/5 * 3/4 * 2/3 to get to this step * 1/2 to pass out of this step
6. #5 * 1 (guarantee fail since that's the only option left).

The idea is to calculate the chance of each of the three losing scenarios to come up, and add to find the losing %, then take a 100% minus the % you find for the winning %.

It's a 50%/50% chance.

Tom, you really stop posting these so I can actually do my prep work for teaching tomorrow instead of getting needlessly distracted by math puzzles.

EDIT: I just reread the problem and there are TWO ways to fail, not just one. There goes my entire calculation and about half an hour of my time -_-.

At least it's easier to do since I already know the algorithm.

1. Person B goes, and has a 2/6 chance of fail.
2. Assuming Person B succeeds, Person A has 2/5 chance of fail.
3. Assuming Person A succeeds, Person B has 2/4 chance of fail.
4. Assuming Person B succeeds, Person A has 2/3 chance of fail.
5. Assuming Person A succeeds, Person B will guarantee fail.

So Part A is 2/5, 40% fail -> 60% success, since 1) is already assumed to have happened.

Part B:
For Player A's failure rate
1. can't fail here
2. 4/6 * 2/5 = 8/30 = 4/15 chance fail on 2.
3. can't fail here
4. 4/6 * 3/5 * 2/4 * 2/3 = 2/15th chance fail on 4.
5. can't fail here
-> 4/15th + 2/15th = 6/15 = 40% fail rate (or 60% success rate)

To explain:
1. 4/6 chance of passing to you
2. #1 must happen * 2/5 chance of failing = chances you'll fail here
3. 4/6 * 3/5 chance you'll get to this step * 2/4 to pass out of this step
4. #3 must happen * 2/3 chance to fail
5. 4/6 * 3/5 * 2/4 * 1/3 to get to this step * 1 (guarantee fail)

Now I have a headache. -_-

[Kakashi_Hatake777]

Reading is tech..... again.

[animefanatic]

Heh it seems like Matt Low missed the one important factor. I drew a picture to help explain this better

http://img.photobucket.com/albums/v8...anRoulette.png

1-4 are empty slots. And everyone is right that there is a 2/6 chance of dying for player B on the first shot. But everyone misses the mark for the second shot.

Consider the configuration above. Given that the first shot is empty, the cylinder will rotate 1 round clockwise (or counterclockwise, depending on how you look at it). Given that the first round is empty, there are only 4 options for the second round, which I labeled as A,B,C and D.

That's all for my hint.

[Kakashi_Hatake777]

If the barrel is rotating one round per shot and the first shot is empty, we can basically tell that the bullets are in the other 5, making it a 2/5 chance and if the next shot happens to be empty as well, we now know that the chambers 1 and 2 are empty, leaving the bullets on 2 of the 4 left and so on so forth.

1st question was what were the odds of you surviving after Player B survives. Since you rotated one barrel and it was empty, the chances are 3/5 or 60% of you surviving.

The overall chance of you winning would be something like:

The odds of winning on the first try is 2/6 or 33.33333333%
Then you must shoot yourself 2/5 40%
Then the other one shoots 2/4 or 1/2, 50%
Then you shoot 2/3 or 66.666666%
Then the 2/2 shot done by the opponent which is 100%.

Now adding the overall chance of you winning and subtracting the overall chance of you loosing.

End calculation: 77.77777777777%

[matthewlow]

As I said, I have no idea how a gun like this works. My assumption is that it just moves to the next slot. From my understanding of your picture, after slot 1 is cleared, only 4 of the 5 other options can be reached next. However, I don't think that matters in the least bit, since all five slots are unknown, thereby leaving the randomization factor intact. Each of the five slots would have a 2/5th chance of being the "fail", with a 4/5 chance of being a possible choice, 1/4th being the actual choice, reverting back to a 1/5 chance of being chosen.

I can't see how my calculation is wrong unless my understanding of the how the gun works is completely off base and not mathematically withstanding, which is something I wouldn't be able to factor into my calculation anyway.

[animefanatic]

Originally Posted by matthewlow

As I said, I have no idea how a gun like this works. My assumption is that it just moves to the next slot. From my understanding of your picture, after slot 1 is cleared, only 4 of the 5 other options can be reached next. However, I don't think that matters in the least bit, since all five slots are unknown, thereby leaving the randomization factor intact. Each of the five slots would have a 2/5th chance of being the "fail", with a 4/5 chance of being a possible choice, 1/4th being the actual choice, reverting back to a 1/5 chance of being chosen.

I can't see how my calculation is wrong unless my understanding of the how the gun works is completely off base and not mathematically withstanding, which is something I wouldn't be able to factor into my calculation anyway.

You missed something key in my problem statement. And that is that YOU load the two bullets in yourself. I also said you should assume that you want to win. Given these conditions, you want to load the bullet to your advantage.

Edit: Just so you know, you're understanding of how the gun works is correct.

[matthewlow]

Should have said that "you choose" what slots to put them in... since I thought they go in at random. I wouldn't know that you get to choose where they go in a gun. I figured that you dropped the bullets in, and then they found their place in the gun at complete random.

This narrows it down to either placing the bullets next to each other, opposite each other, or skipping a slot.

Next to each other or opposite each other means that both players have danger of getting hit by one of the two. Skipping a slot means either both can hit you or neither can.

You'd have to calculate both scenarios and see which one works out best. I'll consider doing this later.