[unreality]

yes I made these up, so don't be too harsh on me

Warmup Problem

You have a bag with 20 smooth, spherical stones. Each stone has a 1/2 chance to be either white or black. You draw 19 black stones in a row out of the bag... what is the probability that the final stone in the bag is black?

Problem One

You have two bags... one with 9 black stones and 1 white stones, and another with 1 black stone and 9 white stones. You arrange the two bags into a random order, and then draw from the first one: you pull out a black stone. What is the probability that you will draw a white stone from the second bag?

Problem Two

You have three bags this time... one with 9 black and 1 white, one with 1 black and 9 white, with the third bag containing 5 black stones and 5 white stones. You arrange the bags into a random order of 1-2-3 and then draw from bag 1: you remove a shiny black stone from the bag. You then draw from the second bag... what is the probability that you will draw a white?

Problem Three

You have 3 bags again- the same 3 bags from problem two. Put them in a random order of 1-2-3, and draw randomly from the first bag. Place this stone into the second bag, shake the second bag, and then draw randomly from the second bag. Put the stone you just drew into the third bag, shake the third bag, and then draw from the third bag. What are your chances of drawing a white from the third bag?

Problem Four

You have one bag, with 3 white stones and 1 black stone. You and your friend decide to play a little game... you have two stones (white and black) in your pocket, and he has two stones (white and black) in his pocket. You put one of your stones into your hand as a prediction of what will be drawn out of the bag. Your friend puts one of his stones into his hand as a prediction of what color stone will be in YOUR hand. Then both of you reveal your stones at the same time. If your friend predicted right, he wins. If he didn't, and you guessed correctly, you win. If you NOR your friend guessed correctly, nobody wins. What color stone should you put in your hand?

[HoustonHokie]

50/50 - Your statement that each stone has half chance to be black or white and not how many white or black stones are in the bag means that each is independent of the other. So even though you've drawn 19 in a row, the 20th is independent of the others.

That being said, it doesn't seem like your random stone generator did a very good job of adhering to your statistical specifications. It would take a large pool of otherwise half-chance black/white stones somewhere else (i.e., not in your bag) to validate the 50/50 chance for the stones that are in your bag.

[HoustonHokie]

Your selection of a black stone from the first bag did not alter the composition of stones in the second bag. So if the second bag is the 9W1B bag, your chances are 90%. If the second bag is the 9B1W bag, your chances are 10%. However, the selection of a black stone from the first bag did indicate the probability that the second bag is either the 9W1B or 9B1W bag. Because you drew a black stone, you have a 90% chance that the second bag is the 9W1B bag and a 10% chance that the second bag is the 9B1W bag. Multiplying the probability of the bag by the probability of picking the correct stone and then adding the two products together gives 0.81 + 0.01 = 0.82, or 82% chance of picking a white stone from the second bag.

[unreality]

HH: good job

I agree that the warmup is 1/2... the probability that all 20 are black is 1/(2²⁰), however that probability doubles each time a black is drawn because you have more information. Each stone is independent, so yeah, 50-50 ;D

for problem 1, spot on

[HoustonHokie]

ur - thanks for kudos! BTW, good problems...

The logic here is similar to that for problem one. Although the selection of a particular stone from bag one doesn't change the composition of stones in the other bags, it does indicate the probability that the first bag is either of the three bags available. By selecting a black stone from the first bag, the probabilities of the first bag are: 9B1W - 60%, 5B5W - 33.3%, 9W1B - 6.7% (these are proportional to the probability of drawing a black stone from each bag, summing to 100%). The second bag could be either of the two remaining with equal probability (50/50), so you take the probability of drawing a white stone from each of the remaining bags, multiply by 0.5, and add together to get the probability of drawing a white stone from the second bag. Then multiply the probability of drawing the black stone from the first bag by the probability of drawing a white stone from the second bag, and add it all together. I get 0.6*0.7 + 0.333*0.5 + 0.067*0.3 = 0.607, or 60.7% chance of getting a white stone from the second bag.

Okay, I've done the easy ones - I'll let someone else get the rest

Edited August 2, 2008 by HoustonHokie

[unreality]

HH: right once again, though your methods are different than mine

the answer is 91/150, which is 60.6666666% if you're using percentages

edit: I'll add how I do #2:

the bags (and their chances of drawing white) arranged in increasing order are:

1/10

5/10

9/10

those bags' chances of being picked first (ie, of a black being picked from them first) are:

9/15

5/15

1/15

* If the first bag was bag #1 (9/15 of this happening), the average of the other two is 7/10

* If the second bag was bag #1 (5/15 of this happening), the average of the other two is 5/10

* If the third bag was bag #1 (1/15 of this happening), the average of the other two is 3/10

9/15 for 7/10 = 63/150

5/15 for 5/10 = 25/150

1/15 for 3/10 = 3/150

63+25+3 = 91

91/150

Edited August 2, 2008 by unreality

[unreality]

Okay, I've done the easy ones - I'll let someone else get the rest

hehehe

#3 may be easier than you think

[HoustonHokie]

Okay, I'll give this one a go. It seems like you should have even odds of picking a white stone from the third bag. It would take a fair amount of math to show it, but I'm pretty sure I'm right. Basically, you have even odds of moving either color from bag one to bag two, and slightly greater than even odds of moving the same color stone from two to three. Because all three bags have an even chance of being the third bag, and you have even odds of moving a black or white stone into that third bag, I'd say that you have even odds of picking a white stone from the third bag. Am I on the right track?

[unreality]

HH: yes, though it can be done with much simpler work:

The bags are even for white and black (1/10, 5/10, 9/10), and no color is mentioned during the transfers. There is no reason to suggest that either color would be more or less likely than the other. They have 50-50 chances ;D

[Guest]

Black. Your friend is only right if he guesses the stone and your guess, so he has to play the odds and guess white. You can only win if he guesses wrong, so you have to pick the unlikelier choice.

[unreality]

Spaceman Spiff: But what if he's expecting you to do that? DUN DUN DUN!

[HoustonHokie]

Spaceman Spiff: But what if he's expecting you to do that? DUN DUN DUN!

Yeah. That's where I finally laid down my pencil and gave up. You're now out of statistics and into rational behavior... Or, in my case, irrational!

[unreality]

yeah ;D check this out

[HoustonHokie]

yeah ;D check this out

Thnx for the link - I would never have arrived at that answer, or even known how to approach it. But, in case I ever find myself playing that game, at least I know what to do...

[Guest]

I havent looked at the answers yet... after I post we'll see how far off i am.

Warmup through #3 are 50/50.

And the answer to #4 is Black. If you know that you have a 75% chance of drawing white, then so does the other guy. Theres no reason to give him the win...

[unreality]

Telamon: you're wrong on #1 and #2, and #4 has no real answer, it depends on how well you know your friend