Homework Section

[Jubin]

I need some math challenges. :cool:

Three men, Adam, Bertie and Charlie, agree to take part in a lethal three-way duel. They each have a pistol and several rounds of ammo, and they draw lots to see who will shoot first and who second. Then they stand an equal distance apart, and they take turns to fire. When it is a player's turn, he can take one shot at any target of his choice, then the turn passes to the next player. The duel continues until only one player is left alive.

For simplicity, we can ignore the possibility of a player becoming injured and unable to shoot: a player is either alive or dead. We also assume that each player chooses to play in such a way as to give himself the best chance of winning.

Now Adam will always kill the person he shoots,Bertie hits 80% of the times and  Charlie is only a moderate shot and will only hit 50% of the times. All three men know this.

Of the three men, who has the best chance of winning the duel? Which are the surviving chances for the others. With the calculations please : )

[JDC]

You're not alone.
 
I saw this topic, and now I love this community even more.
 
Now, a simple trick question for all the smart minds here. Let's see if you can get it on the first try reading.
 
A bald lady lives in a yellow house. She wears yellow clothes, has yellow furniture, everything in the house is yellow. Later in the day, she rode a yellow taxi with a driver wearing yellow to the nearby mall, paying 3 kroner in fees. The mall is also colored yellow on the walls, the floor, the ceiling, and so is all their merchandise, and the lady shops for a day before riding home in another yellow cab. At this point, her eyes are already strained from yellow fatigue after using a yellow computer. When she looked at herself in the mirror after a tiring day, she was holding a green comb but she did not use it.
 
Why didn't she use it?

[happyday]

she is bald yo, unless she did use it -- but on her pubic hair?

[JDC]

It's always more confusing when told in person. Heh. Anyways, another one.
 
Imagine Gandalf is on vacation in Libya when the war erupts. When he went there, he wanted to build a sauna.
 
In a night of war, dozens of armies unleash legions of bombs upon the country, changing the night sky from black to yellow. The effect of the mass napalm shower makes itself felt in Libya, the light from the fires in the sky is bright enough to read a newspaper inside a forest at midnight. After this night, Gandalf does not need to build a sauna anymore.
 
Why does Gandalf no longer need to build a sauna?

[Daco]

Three men, Adam, Bertie and Charlie, agree to take part in a lethal three-way duel. They each have a pistol and several rounds of ammo, and they draw lots to see who will shoot first and who second. Then they stand an equal distance apart, and they take turns to fire. When it is a player's turn, he can take one shot at any target of his choice, then the turn passes to the next player. The duel continues until only one player is left alive.

For simplicity, we can ignore the possibility of a player becoming injured and unable to shoot: a player is either alive or dead. We also assume that each player chooses to play in such a way as to give himself the best chance of winning.

Now Adam will always kill the person he shoots,Bertie hits 80% of the times and  Charlie is only a moderate shot and will only hit 50% of the times. All three men know this.

Of the three men, who has the best chance of winning the duel? Which are the surviving chances for the others. With the calculations please : )

Already started doing it but it's late... *sleepy* I'll finish it tomorrow.

[MrFrancis]

It's always more confusing when told in person. Heh. Anyways, another one.
 
Imagine Gandalf is on vacation in Libya when the war erupts. When he went there, he wanted to build a sauna.
 
In a night of war, dozens of armies unleash legions of bombs upon the country, changing the night sky from black to yellow. The effect of the mass napalm shower makes itself felt in Libya, the light from the fires in the sky is bright enough to read a newspaper inside a forest at midnight. After this night, Gandalf does not need to build a sauna anymore.
 
Why does Gandalf no longer need to build a sauna?

Coz it has an 'atmospheric' sauna already in Libya

[Call_me_Dad]

Three men, Adam, Bertie and Charlie, agree to take part in a lethal three-way duel. They each have a pistol and several rounds of ammo, and they draw lots to see who will shoot first and who second. Then they stand an equal distance apart, and they take turns to fire. When it is a player's turn, he can take one shot at any target of his choice, then the turn passes to the next player. The duel continues until only one player is left alive.

For simplicity, we can ignore the possibility of a player becoming injured and unable to shoot: a player is either alive or dead. We also assume that each player chooses to play in such a way as to give himself the best chance of winning.

Now Adam will always kill the person he shoots,Bertie hits 80% of the times and  Charlie is only a moderate shot and will only hit 50% of the times. All three men know this.

Of the three men, who has the best chance of winning the duel? Which are the surviving chances for the others. With the calculations please : )

I tried this:

There are 6 permutations for the draws: (representing the order of who shoots after who):
1 ABC
2 ACB
3 BAC
4 BCA
5 CAB
6 CBA

Each draw has a probability of 1/6=0.16

Lets calculate the probability of A winning the battle=
Sum of probabilities of A winning the battle in all six cases=
Case 1: ABC
P(event) denotes the probability of occurrence of 'event'

Case1 && A shoots B && C misses A = P(Case1) x P(A shoots B) x P(C misses A)
=0.16 x 1 x 0.5
=0.08

Case 2: ACB
Case2 && A shoots B && C misses A = P(Case1) x P(A shoots B) x P(C misses A)
=0.08

Case 3: BAC
Case3 && B misses A && A shoots B && C misses A = P(Case3) x P(B misses A) x P(C misses A)
=0.16 x 0.2 x 0.5
=0.016

Case 4: BCA
Case4 && B misses A && C misses A && C misses A again = P(Case4) x P(B misses A) x P(C misses A) x P(C misses A)
=0.16 x 0.2 x 0.5 x 0.5
=0.008

Case 5: CAB
Case5 && C misses A && C misses A again = P(case5) x P(C misses A) x P(C misses A)
=0.16 x 0.5 x 0.5
=0.04

Case 6: CBA
P(Case6) x P(C misses A) x (B misses A) x P(C misses A)
=0.16 x 0.5 x 0.2 x 0.5
=0.008

So probability of A winning the threesome=0.08 + 0.08 + 0.016 + 0.008 + 0.04 + 0.008= 0.232

Some1 else calculate plssss....ima sleep now

[Jubin]

I have to say  that Call_Me_Dad already has mistakes in his train of thoughts.

[Ayden]

From the way I see it, Charlie has the best chance at winning if he has knowledge to win even if he is the worst shooter. He will just shoot and aim at pure air in order to prevent the two others to shoot at him. So Charlie has a 2/3 chance to win, while the two others have just 1/3 chance to win as they will shoot at each other if they get drawn.
I figured this out since you said yourself that they all knew who were shooting the best and who weren't
Not sure on the calculation part though.
Correct me if my theory is wrong

[Altair_Carter]

A trick question:

What colour would chameleon be if you put him in a mirror room?

[JDC]

Can it become transparent?

[Daco]

I will refer to Adam: A, Bertie: B and Charles: C.
A does 100% damage and the target dies.
B does 80% damage.
C does 50% damage.

Now, who will begin shooting first, then who will begin shooting second, so each player's turn will be clear? Since they'll play lottery to determine it, this is where probabilities come in.
I will take each case alone: If A starts shooting, or B starts shooting, or C starts shooting.

Since it has been 3 years since I studied probabilities in school, I'll use l'arbre de choix or what we call selection tree in English, or close to it.

1^st scenario: If A starts shooting:



The lines linking a letter to another is the trajectory of the bullet from the left to the right. Equal signs move on to the next action in the same scenario.

2^nd scenario: If B starts shooting:



3^rd scenario: If C starts shooting:



Now we count the chances.

It's obvious that A has the most winning chances, now let's count B's chance and C.

B's chances are 7 wins out of total 24 wins. So his chance his 7/24 of winning, which is approximately 29 %.

Same goes for B.

Cheers :girl:

[Drastic]

fucking amazed

[Daco]

f**king amazed

 :girl:


also guys, maths or gtfo

[Jubin]

Daco - For simplicity, we can ignore the possibility of a player becoming injured and unable to shoot: a player is either alive or dead.
So no 50% alive option.

I am not amazed at all.

From the way I see it, Charlie has the best chance at winning if he has knowledge to win even if he is the worst shooter. He will just shoot and aim at pure air in order to prevent the two others to shoot at him. So Charlie has a 2/3 chance to win, while the two others have just 1/3 chance to win as they will shoot at each other if they get drawn.
I figured this out since you said yourself that they all knew who were shooting the best and who weren't
Not sure on the calculation part though.
Correct me if my theory is wrong

Ayden you do realize, that shooting into the air does not prevent others from shooting him. Because once A or B is dead. they will shoot C.