# Math question

• Global
Joined: 08.08.2008
A jar containing 10 balls is placed in front of you. You draw 9 balls out of it without bias (randomly) and find that they are all green. You have no other information about the set. You are then asked if you'd like to make a wager on the last ball being green (the person offering the wager has no information about what's in the set). Of the options, what are the worst odds you'd take for this bet?

-------------------------

There are 11 different possible combinations for the jar:

G is green and N is not green.

NNNNNNNNNN
NNNNNNNNNG
NNNNNNNNGG
NNNNNNNGGG
NNNNNNGGGG
NNNNNGGGGG
NNNNGGGGGG
NNNGGGGGGG
NNGGGGGGGG
NGGGGGGGGG
GGGGGGGGGG

You select one ball - its green. So you can remove from the universe all jars with no green balls. Thats 1 impossible jar so there's 9 left. You select another ball - its green, so now you remove the jar with 9 non green balls. If we continue this method we eliminate all possibilities except:

GGGGGGGGGG
GGGGGGGGGN

so the answer should be 50:50

But this is in contradiction with the straightforward conditional probability problem ( If we select one of those 11 jars randomly and then randomly draw 9 balls from it which all happen to be green, then the probability that the 10th ball is green is 10/11) - if you have watched the movie 21 you should know what is this)

For easier understanding lets take 2 balls - then there are 3 possible jars:

NN
NG
GG

If you pick one of these jars randomly and draw a ball from it randomly and that ball happens to be green then the probability that the second ball is green is 2/3.

-----------------------------

I was told my answer is incorrect. I have no idea why it is incorrect and what is the right answer or how to find it.
• 57 replies
• Bronze
Joined: 23.02.2007
why would p(G) = p(N)?
• Global
Joined: 08.08.2008
Originally posted by swissmoumout
why would p(G) = p(N)?
without further information i think it is default to assume this
• Bronze
Joined: 07.06.2009
so the answer should be 50:50

only if there is 2 colors
if there's only RGB - 1/3
infinite number of colors - 0 chance of guessing last balls color
• Bronze
Joined: 04.12.2009
Originally posted by martins1337
only if there is 2 colors
if there's only RGB - 1/3
infinite number of colors - 0 chance of guessing last balls color
I think this is right, you wrote the remaining possibilities as being

NGGGGGGGGG
GGGGGGGGGG

but in fact it could be

YGGGGGGGGG
BGGGGGGGGG
RGGGGGGGGG
GGGGGGGGGG

etc.

The second point you make (from 21) is about variable change EDIT - I think this is what you mean - http://en.wikipedia.org/wiki/Bertrand's_box_paradox

In which case your 2nd answer is still only right if there are 2 colours involved
• Global
Joined: 07.05.2009
What are the chances of the last ball being not green if the previous 9 are green? What is the chance of 10 balls (9 green and 1 nogreen) in the same jar pulled out randomly that the first 9 balls pulled out are green? I think it is something like this

Chance of first being green 9/10 =0.9
Chance of second being green 8/9 =0.88
... =0.87
... =0.85
... =0.83
... =0.8
.. = 0.75
... =0.66
... =0.5

Chance of the last one being not green we multiply these ones (conditional probability i think its called) and we get 0.096

Not necessarily correct but that is how i would try to solve it
• Bronze
Joined: 24.05.2008
Originally posted by silent21
A jar containing 10 balls is placed in front of you. You draw 9 balls out of it without bias (randomly) and find that they are all green. You have no other information about the set. You are then asked if you'd like to make a wager on the last ball being green (the person offering the wager has no information about what's in the set). Of the options, what are the worst odds you'd take for this bet?
Without additional information, this is not a question that could be answered by math (probability) .. You need additional information to answer that, mostly what is the number of colours possible or what is the probability of a single ball being green

However, a question that could be easily answered is this:
You take out 9 balls, all are green, one is left in .. You return all 9 balls inside and pick one again .. What odds would you take for it to be green
• Global
Joined: 08.08.2008
Originally posted by martins1337
so the answer should be 50:50

only if there is 2 colors
if there's only RGB - 1/3
infinite number of colors - 0 chance of guessing last balls color
well i dont care how many are the possible colours of the last ball. for me it is either green or Non-green. i understand that i need to know what is the probability of the last ball being green, but without this information does this means there is no answer?

Maybe not exact number as answer, but some mathematical model or formula
• Bronze
Joined: 15.10.2008
Originally posted by sk345di
What are the chances of the last ball being not green if the previous 9 are green? What is the chance of 10 balls (9 green and 1 nogreen) in the same jar pulled out randomly that the first 9 balls pulled out are green? I think it is something like this

Chance of first being green 9/10 =0.9
Chance of second being green 8/9 =0.88
... =0.87
... =0.85
... =0.83
... =0.8
.. = 0.75
... =0.66
... =0.5

Chance of the last one being not green we multiply these ones (conditional probability i think its called) and we get 0.096

Not necessarily correct but that is how i would try to solve it

You can't hope more that 9,6% that 10th ball will be not green so the worst odds you could take are 1 : 10,42 (100/9,6).
• Bronze
Joined: 24.05.2008
[size=10][quote]Originally posted by ChoChikun[/size]
[size=10][quote]Originally posted by sk345di[/size]
What are the chances of the last ball being not green if the previous 9 are green? What is the chance of 10 balls (9 green and 1 nogreen) in the same jar pulled out randomly that the first 9 balls pulled out are green? I think it is something like this

Chance of first being green 9/10 =0.9
Chance of second being green 8/9 =0.88
... =0.87
... =0.85
... =0.83
... =0.8
.. = 0.75
... =0.66
... =0.5

Chance of the last one being not green we multiply these ones (conditional probability i think its called) and we get 0.096

Not necessarily correct but that is how i would try to solve it[/quote]

You can't hope more that 9,6% that 10th ball will be not green so the worst odds you could take are 1 : 10,42 (100/9,6).[/quote]
Not necessarily .. It assumes that the last ball is not green which is made up .. Unless you know that you cannot calculate it
• Global
Joined: 07.05.2009
Actually it is supposed to calculate the odds of the last color not being green it doesn't assume it. It's just the way that i phrased it in English that is not very clear i guess.
• Bronze
Joined: 22.11.2009
This is a game theory problem, not a question of probability. The general problem is, X puts N-n green balls and n red balls in a jar with probability c_n. Then Y draws N-1 balls from the jar, and is offered \$1 if he guesses correctly the colour of the remaining ball and loses \$P if he guesses wrong. Y chooses red y% of the time and green (1-y)% of the time (could make y dependent on the number of green balls he has seen so far if you want to make it even harder to solve!). What c_n should X choose and what y should Y choose to be unexploitable?

I'm trying to learn some game theory at the moment having just read 'The Mathematics of Poker', and I'm a maths professor, so I get paid to do this stuff!

Anyway, the simplest case is two balls (N=2). A quick calculation then shows that Y's profit is

c_0(1-y-Py) + c_2(y-P(1-y))+(1-c_0-c_2) (1-P)/2

= (c_2-c_0)(1+P) y + (1+c_0-c_2)/2 - P(1-c_0+c_2)/2

Therefore, to be unexploitable, X should choose c_0 = c_2, so that RR and GG are chosen equally often (no surprise), when his profit is (P-1)/2, so he should (obviously) make P as large as possible, and it doesn't matter what Y does. The game is obviously unfair if P>1.

Extending this to more balls looks a bit tricky. I may go and think about the 3 ball case. I think this is the right framework for talking about the problem though.

I would have thought that the right answer to the original question is that you shouldn't play unless you're getting even money or better, and that you should choose red and green in equal proportions, just by symmetry.
• Bronze
Joined: 24.05.2010
Originally posted by ChoChikun
Originally posted by sk345di
What are the chances of the last ball being not green if the previous 9 are green? What is the chance of 10 balls (9 green and 1 nogreen) in the same jar pulled out randomly that the first 9 balls pulled out are green? I think it is something like this

Chance of first being green 9/10 =0.9
Chance of second being green 8/9 =0.88
... =0.87
... =0.85
... =0.83
... =0.8
.. = 0.75
... =0.66
... =0.5

Chance of the last one being not green we multiply these ones (conditional probability i think its called) and we get 0.096

Not necessarily correct but that is how i would try to solve it

You can't hope more that 9,6% that 10th ball will be not green so the worst odds you could take are 1 : 10,42 (100/9,6).
Actually, that is not the answer. But it's close. sk345di's thinking is correct, but the calculations are imprecise.
The odds of not extracting the non-green ball are indeed:
9/10 * 8/9 * 7/8 * ... * 1/2
This equals 1/10, which is 10%.
Therefore, I would wager anything that gives me better odds than 9:1 against.
That is, if I win \$1 if the last ball is green and lose \$9 if the last ball is not green, i'm even-money in the long run.
Any odds better than 9:1 against are profitable.

Of course, there are a few more things to take into consideration. Firstly, i would not call any bet from the guy who made the set of balls or has had a chance to see what's in it prior to the extraction.
This means i consider the set of balls pre-made by somebody unknown to both of us. (e.g. me and a friend are walking down the street and see a small sack on the ground; I lift it up, extract 9 balls out of it - all green, and then my friend stops me and proposes a bet for the last ball).

Another factor to take into consideration is: what are the odds that somebody placed 10 balls in a sack, 9 of which are green but the last one isn't? I would safely assume that 9/10 people who put 10 balls in a sack (out of which 9 are green) only put green balls in it.

This happens in poker also. If you have a combination of hole cards that beats 3/4 of the possible combinations of hole cards and you are facing a bet on the river offering you proper odds, that doesn't necessarily mean you should call. It is wrong to assume that he can hold any of the 1K+ combinations of cards with equal probabilities. He is very unlikely to hold one of the many very weak hands you can beat (he would have probably folded a fair share of them in the previous betting rounds).

Therefore, the odds I would accept are 9 times 9:1, that is 81:1 against.
• Black
Joined: 20.02.2008
'GGGGGGGGGG
GGGGGGGGGN

so the answer should be 50:50'

lol, wat?

why.....?
• Bronze
Joined: 05.12.2008
According to problem description, we are not interested in the first nine ball came out from the box. They were taken out and they are green and this is separate event from the last one which is actually question about. If the number of Green and not green balls are unknown. We can't assume anything about the future outcomes for 1 event. Not knowing the number of balls make the last event (taking one of the last 2 balls) an independent event like a coin flip - 50:50. For this event to be an independent from the previous ones, both sides shouldn't know about the number of green or not green colored balls.
• Black
Joined: 20.02.2008
one thing i'm sure about is that the probability of it being green is higher than iof the others had been other colours
• Bronze
Joined: 26.04.2010
I'm not really good at maths, but here's my guess. Take note that we win if the ball we put out is green.
Theory 1:
After extracting 9 balls we are confronted with the two possible situations:
1) The 10th extracted ball is green, so it is certain that we had 0% chances of extracting a non-green ball, and so 100% chance of winning.
2) The 10th extracted ball is not green, so now we know that jar contained 9 green balls and a non-green one. By replaying the game we have 1:10 odds of selecting the one non-green ball out of the 10 possible balls.
You said that you win if the last ball is green. So the worst chance of selecting a non-green ball is 1:10, meaning we have 90% chances of winning.
Theory 2: To simplify the problem, we must pick from a jar which contains either a green ball or a non-green ball. The odds here would be 1:1. But I think this reasoning is wrong because seeing 9 green balls taken from the jar, we would have to think that the frequency of green balls is way bigger than that of the non-green balls.
Theory 3: I belive "the person offering the wager has no information about what's in the set" is crucial to knowing the right answer. Because this is mentioned, we could abstractly assume that the ball color is generated as we pick, just as cards are generated in poker , and not modify the result of the problem.
If the first ball we pick is green, and the next one is not green, we can assume that there is a 1:1 green ball-non green ball generation ratio so we would have 50% chances of the next ball picked being green.
If the first ball we pick is green,the 2nd ball is green, and the 3rd ball is not green we can assume that the generation ratio is 2:3. This is correct "on the long run" since I'm guessing somehow that we would have a chance of (0.33)^9 ( I hope this is correct) chance to hit 9 green balls with these odds.
If the first 9 balls are green and we don't know anything about the last ball, we must come back to the two cases: 50% chance that we won 100% and 50% chance we have a 9:10 chance of winning. From here I would conclude we have 95% chance of winning.
T3 seems like a deeper explanation of T1, so the odds of 95%( or 19:20) being the correct answer are 2:1 D.
I notice the problem is asking for "worst" odds so I'm wondering if my answer is that they want.
PS: The problem does not include the option of having 0 chance of winning.
As silent21 said, we care only if the ball is green or not. That's two options, even if the non-green could mean 16million colors.
• Bronze
Joined: 26.04.2010
Originally posted by roswellx
According to problem description, we are not interested in the first nine ball came out from the box. They were taken out and they are green and this is separate event from the last one which is actually question about. If the number of Green and not green balls are unknown. We can't assume anything about the future outcomes for 1 event. Not knowing the number of balls make the last event (taking one of the last 2 balls) an independent event like a coin flip - 50:50. For this event to be an independent from the previous ones, both sides shouldn't know about the number of green or not green colored balls.
I think it's actually the other way arround. For the second event to be unrelated to the first event, someone SHOULD know the color of the last ball or even all of the balls. If no one had any clue about the colors of the balls we could assume the colors were distributed through a mathematical algorithm and so it will be possible to calculate the odds of the last ball being a certain color using the information we have from the first event.
I noticed two interesting things here:
1) Someone KNOWING something, but not influencing it in ANY way affects something. I think from what we both written, it would result we both agree to this. This is totaly strange. It's like the moon turning white only when someone is looking at it, and be..pink.. when no one's watching it.
2) What is the mathematical algorithm that generates green balls 9 out of 10 times? Who created it? Why does it exist?

Well that was my moment of madness for today. We assumed that the two events are linked in case no one knows about the ball colors. What we actualy did is we assumed that no one actually ARRANGED the colors because if they did, the colors of the balls might have been the result of a chaotic sistem, so the color of the last ball would be totaly unpredictable and unrelated to the first event and have the value assigned by the equivalent of a 50:50 coinflip( best guess).
If this was not the case, the only remaining option is that the colors of the ball have been assigned by a standard non-chaotic algorithm, which is predictable through odds and chances, so the first event IS related to the final ball selection through the frequency of green balls.
I hope at least one person can make some sense of what I wrote here and I hope I am not using wrong judgement
• Bronze
Joined: 05.12.2008

Well that was my moment of madness for today. We assumed that the two events are linked in case no one knows about the ball colors. What we actualy did is we assumed that no one actually ARRANGED the colors because if they did, the colors of the balls might have been the result of a chaotic sistem, so the color of the last ball would be totaly unpredictable and unrelated to the first event and have the value assigned by the equivalent of a 50:50 coinflip( best guess).
If this was not the case, the only remaining option is that the colors of the ball have been assigned by a standard non-chaotic algorithm, which is predictable through odds and chances, so the first event IS related to the final ball selection through the frequency of green balls.
I hope at least one person can make some sense of what I wrote here and I hope I am not using wrong judgment
You wrote the same thing as me... It all depends if the guy who has offering a bet knows about the number of green and not green balls. If he doesn't know then it's 50:50, if he knows then it's still 50:50 for us but he will know exactly what odds are we getting and can manipulate us to take a bet with wrong odds.
• Bronze
Joined: 17.02.2010
Originally posted by silent21
If we continue this method we eliminate all possibilities except:

GGGGGGGGGG
GGGGGGGGGN

so the answer should be 50:50