Tricky probability riddle

[edthayer]

You have a bag with three balls; 2 black, one white.

Your goal is to draw the white ball. First, you take one ball from the bag at random. Then, someone else removes a black ball from the bag. Now you have the option of taking the remaining ball in the bag or keeping your first choice. What's the correct move?

[gaming_mouse]

ed,

this is just a dressed up version of the monty hall problem, which has been posted many times.

it is always better switch. switching gives you a 2/3 prob of winning, while sticking gives you a 1/3.

gm

[edthayer]

Oh, I didn't know about that. Sorry if I'm being repetitive.

I understand the solution to some degree. Is there someone who can explain the thinking behind it?

[Mike Haven]

let's label the balls BA BB and WH

you pick one at random

if it's BA then WH is left in the bag

if it's BB then WH is left in the bag

if it's WH then BA or BB (B) is left in the bag

if you keep your first pick it's any one of BA BB or WH, a 1 in 3 chance of being WH

if you take the ball in the bag it's going to be WH WH or B, a 2 in 3 chance of being WH

[TomCollins]

Imagine you have 99 black balls, and 1 white ball. Now you choose one. Someone else removes every other ball but 1. If you have the white ball, it is of course black. But if you have a black ball, he guarantees it is white. It is pretty obvious that your chances of getting it right the first time was pretty slim, so switching will almost always result in the white ball.

[goldbach]

I think there is a more insightful way to show why you should switch. The person who pulls out the white ball is really saying "Hey, the black ball is still in the bag!" He's only telling the truth if you initially picked the white ball,which you did with 2/3's likeliehood. So believe him, and switch.

[SpaceAce]

[ QUOTE ]
Oh, I didn't know about that. Sorry if I'm being repetitive.

I understand the solution to some degree. Is there someone who can explain the thinking behind it?

[/ QUOTE ]

Simplification:

When you reach in and pick a ball the first time, there is a 2/3 chance that you selected a black ball. So, 2/3 of the time, when the other person removes a black ball from the bag, the remaining ball will be white and you will win by opting to trade for the ball still in the bag. Only 1/3 of the time will you have selected the white ball on your first try so only 1/3 of the time will you be making a mistake to switch the ball you selected for the ball still in the bag.

SpaceAce

[Buccaneer]

You have a bag with three balls; 2 black, one white.

Your goal is to draw the white ball. First, you take one ball from the bag at random. Then, someone else removes a black ball from the bag. Now you have the option of taking the remaining ball in the bag or keeping your first choice. What's the correct move?

This is a clasic probablility and sadistics question. The correct move is to worry about something else until someone writes the question so that it has only one answer. I suggest this:

You have a bag with three balls: one white ball and two black balls. The moment you possess a white ball you become ruler of the universe. You pick a ball at random and look at it. If it is white you win. You win right then and there, you look at the ball and if it is white you have won. If the ball is black you do not win or loose yet. Your friend then draws a ball at random and if it is white you loose because you can not possibly pick the white ball. If the ball is black then you win because you know that for your friend to pick a black ball it is dependent on two possible prior events. One you have not won because you did not pick the white ball leaving a white and black ball in the bag, the next event is that you did not loose because your friend picked the black ball and now we know where the white ball has to be. In the bag. You would then of course switch the obvious loosing ball, take possession of the white ball by switching your loosing black ball with the white ball in the bag and win the exercise in futility. You still become ruler of the universe. This then allows you to remove inside straights and this question from the universe forever. If you pick your nose while you pick your balls what is the probability that you will become ruler of the universe?

The problem with this brain cramp is that it is not very clear whether the events are dependent or independent of each other. In other words, can you see the first ball when it is picked? If you can it makes the rest of the games events dependent on the first event, if you are blind then the events are independent. If you are blind how do you know if the second ball is black or white. That is my not so humble oppinion, your oppinion may differ, I just always thought that this question had Hoover Dial A Matic written all over it!

If I may:
Professor French forgets to set his alarm with a probability of 0.3. If he sets the alarm to ring it rings with a probability or 0.8. If the alarm rings, it will wake him on time to make his first class with a probability of 0.9. If the alarm does not ring, he wakes in time for his first class with a probability of 0.2. What is the probability that Professor French wakes in time to make his first Class Tomorrow?

[invast]

Professor French wakes in time to make his first Class Tomorrow with a probability of .592

Am I right, or completely wrong?

[MickeyHoldem]

I may be completely wrong, but I get .544... but then again I skipped Prof. French's classes!

[Buccaneer]

You Sir are not right, you are completely right.

The problem is usually missed right here: If the alarm does not ring, he wakes in ............... What most people do is assume that the alarm will not ring only when the Professor remembers to set the clock. There is also another occurance that they forget about and that is of course that the alarm will fail to ring if he forgets to set the clock but there is still a chance that he may or may not wake up in time for class. The Prof will be ontime .06 if he forgets and late .24 if he forgets to set that alarm.

Just like the marbles in the bag problem you have to assume things from the information given but it is easier to define and filter the possible outcomes.

[invast]

Ya.. this is how I got the answer (all laid out)

1) sets alarm -> alarm rings -> makes it to class
2) sets alarm -> alarm rings not -> makes it to class
3) sets alarm not -> alarm rings not -> makes it to class

1) .7 * .8 * .9 = .504
2) .7 * .2 * .2 = .028
3) .3 * 1 * .2 = .06

.504 + .028 + .06 = .592

[MortalWombatDotCom]

[ QUOTE ]
You Sir are not right, you are completely right.

The problem is usually missed right here: If the alarm does not ring, he wakes in ............... What most people do is assume that the alarm will not ring only when the Professor remembers to set the clock. There is also another occurance that they forget about and that is of course that the alarm will fail to ring if he forgets to set the clock but there is still a chance that he may or may not wake up in time for class. The Prof will be ontime .06 if he forgets and late .24 if he forgets to set that alarm.

Just like the marbles in the bag problem you have to assume things from the information given but it is easier to define and filter the possible outcomes.

[/ QUOTE ]

your mode of speech is confusing and so i may actually be agreeing with you here. you do not state what the probability is that the alarm will ring if he does not set it. people seem to be assuming that it's zero, but that's just crazy.