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# 100 white marbles

## Question

There are 100 (quite small) marbles in a (quite large) jar.
Pull out one marble, look at it, put it back.
You have done it 100 times. All marbles were white.

Would you bet 5:1 that they are all white?

Bonus (and most interesting) question: Suppose you have 15-20 seconds to decide.

## Recommended Posts

• 1

If I had only seconds to decide I would bet those odds.

If I had the time to raise .99 to the 100th power,  (~.3734 = ~1/2.678) I would not.

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No, You can be deceived, if there is just 1 black,2 blacks or even 10 blacks marble there, and you are unlucky to find the black marble.

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@bmad: yes, "non-mathematical" equivalence of "the jar is quite large" and "the marbles are quite small" @jasen: In overwhelming majority of cases, I base my decision on probability, very exceptionally on luck.

@bonanova: The answer I expected. I would reason something like that: if there is one black marble, p(black)=1/100. Repeating 100 times, I would estimate p(all white) about 95% and would be very surprised if I were told it is less than 90%.

Knowing the solution, I recall the problem of n people having birthday the same day.

However, your formula is missing something. If you complete it, you will get another surprise. (Please hurry, I do not want to be faced with the problem of two best answers again.)

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There are 2 cases

case 1. Someone want to fool you, he just put one black
case 2. Nobody try to fool, maybe there are no black,1 or more than one black.

in case 1 :
you should not take the bet.
there is 35% probability that in 100 tries you cannot find the black one.

in case 2 :
I still confuse how to solve this case.

Edited by jasen
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fore case 2, I have run a computer simulation, which show that you should not bet.

The probability that all white is only .63

so the quite fair bet is 2:1  not  5:1

Edited by jasen
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@jasen: a computer simulation for such a problem is IMHO an overkill.

Bonanova's approach is correct, he just did not finish it.

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Event A: you only pulled out white marbles. Event B: there is at least one black marble in the jar. Event C: there is exactly one black marble in the jar. We want P(B|A). Unfortunately, while we can calculate P(A|B) (and bonanova calculated P(A|C) which is a partial result), we have no way of calculating P(B|A) unless we know the probability distribution with which the marbles in the jar have been selected. Did they uniformly choose a random number n from 1 through 100, then put n white marbles in the jar and 100-n black marbles? Did they flip a coin 100 times and put a white marble for each heads and a black marble for each tails? With the current information, the best we could do is propose a number of black marbles and see if the result (pulling only white marbles) is unlikely enough, based on the proposed number of black marbles, that we could reject the proposal with some level of statistical significance. Such an endeavour would not give us enough information to decide whether to refuse or accept the bet though.

TL;DR - we don't have enough information to properly solve the question.

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@Rainman:

In Europe, we slowly quit Dark Ages and advanced countries switch to Euro. Coins are identical on one side and national on the other side: French, German, Austrian, Greek... The coins travel quite well over the borders - there are studies on it, but that's another story.

I have three coins in my pocket.
I take one coin out and look at it before putting it back: it is French.
I do it three times and every time it is French.
What is p(all are French)?

Hell only knows how these coins got into my pocket.

Edited by harey
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@Rainman:

In Europe, we slowly quit Dark Ages and advanced countries switch to Euro. Coins are identical on one side and national on the other side: French, German, Austrian, Greek... The coins travel quite well over the borders - there are studies on it, but that's another story.

I have three coins in my pocket.
I take one coin out and look at it before putting it back: it is French.
I do it three times and every time it is French.
What is p(all are French)?

Hell only knows how these coins got into my pocket.

I probably shouldn't even be doing math now because I'm quite sick and feverish, but I maintain my stance that we don't have enough information to solve this problem either. We would have to know the probability distribution for French coins among all Euro coins. Imagine for example that French coins would be very rare, with only one Euro coin in a million being French. It is then much more likely that you only have one French coin and simply happened to pull it three times, than that all three coins are French which would be a one in 1018 anomaly. On the other hand, imagine that 90% of all Euro coins were French instead. Now the base probability that all three coins are French, even before you pull out any of them, is 0.93 = 0.729. Doing the test of pulling coins out and finding them French would only increase the probability that all are French.

On a side note, in my last post I claimed that we could calculate P(A|B), which upon further thinking seems to be false. We can't even do that without the probability distribution.

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i will take the bet . cause it only 20% will hit black . other than that is white no worry my friend.

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P(1 Black) = 0.99 100
P(2 Black) = 0.98 100
P(3 Black) = 0.97 100
P(4 Black) = 0.96 100
P(5 Black) = 0.95 100
.
.
P(99 Black) = 0.01 100

----------------------------------------------------- +
P(with 1 or some black) =  0.572....

P (all white) = 1
P (all white) : P(some black) = 1 : 0.572

Conclusion : do not take the bet !!

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Hidden Content

To reach this answer, you either have to assume a uniform distribution of the number of black marbles in the jar, or mistakenly reverse the conditional probabilities.

What you call P(1 Black) is actually P(we only pulled white marbles|there is exactly one black marble in the jar). The "|" should be read as "given that" and is used to mark what is called a conditional probability. Conditional probability analyzes how one event having happened can influence the probability of another event. In the expression P(A|B), B is the event that is given to have happened, and A is the event whose probability we want to calculate, given the fact that B happened.

To reach the 0.99100 probability for exactly one black marble, you assume that there is exactly one black marble in the jar, and calculate the probability of only pulling white marbles based on that assumption. With only one black marble, there is a 0.99 chance of pulling a white marble, and so the probability for only pulling white marbles for 100 pulls is 0.99100. But the problem is you reverse the assumed event and the variable event. You want to calculate the probability of exactly one black marble in the jar, but instead you assume that there is exactly one black marble in the jar. You should assume that you only pulled white marbles, but instead you calculate the probability of only pulling white marbles. You calculate P(A|B) instead of P(B|A).

Bayes' theorem tells us that P(B|A) = P(A|B)*P(B)/P(A). In other words, to reach P(B|A), we must multiply P(A|B) by the factor P(B)/P(A). If and only if we have a uniform distribution of the number of black marbles, then P(B) will be constant throughout the 100 cases you have analyzed. Since P(A) is also constant we have a constant c = P(B)/P(A). So knowing that P(B|A) = c*P(A|B), we could use your calculations as follows:

P(1 Black) = c*0.99100, P(2 Black) = c*0.98100, ..., P(99 Black) = c*0.01100

P(some black) ~ c*0.572

P(all white) = c*1 = c

P(all white) : P(some black) ~ c/(c*0.572) = 1/0.572

This would lead us to the conclusion that the bet is bad for us, but it all depends on the assumption that the number of black marbles is uniformly distributed.

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I have three coins in my pocket.

I take one coin out and look at it before putting it back: it is French.
I do it three times and every time it is French.
What is p(all are French)?

There are 100 (quite small) marbles in a (quite large) jar.
Pull out one marble, look at it, put it back.
You have done it 100 times. All marbles were white.

Would you bet 5:1 that they are all white?

Bonus (and most interesting) question: Suppose you have 15-20 seconds to decide.

2nd one:

I'm assuming you tell us "there's an equal starting probability of having 0,1,2 or 3 French coins in my pocket", 'cos there are an infinite number of countries in the EU P(3 draws are French) = 1/27 if I had 1 French coin, 8/27 if I had 2 French coins, 100% if I had 3 French coins.
P(3 frogeaters in my pocket) = 1/3(1/27+8/27+1) = 1/3(36/27) = 12/27 ... I have the feeling I've fallen into an obvious trap (?)

1st one:

Worst case scenario, there's only one black among the 100 marbles. Chance of not picking it is 0.99^100. I'm no walking calculator, but I think that more than 20%, so I won't take the bet.

........ Windows tells me it's 36.6%, so even a 1:3 bet wasn't worth it.

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On 9/25/2015 at 5:18 AM, bonanova said:

If I had the time to raise .99 to the 100th power

Hint 0:

Spoiler

This problem dates from times when a pocket calculator with four basic operations cost 300\$. For 350\$, you could get one that could even calculate the square root.

Hint 1:

Spoiler

You can replace each term by an expression without powers, just a simple division.

Hint 1.1:

Spoiler

Bernoulli

Hint 2:

Spoiler

You get a series...

Hint 2.1:

Spoiler

There is a formula to sum of the first n terms.

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Spoiler

Let's think about it this way.

Say the bag has 99 white marbles and 1 marble that has a probability p of being white. If I'm giving 5:1 odds that all the marbles are white, I'm betting that p > 5/6. I'm basing the bet on a single event where the first 100 draws from the bag were white. That probability is Q = (.99 + p/100)100.

If we knew the value of Q, we could deduce p and then knowledgeably take the bet, or not. But we don't. And to be on the conservative side, we conclude, from a single success, only that Q > 0.5, which corresponds to p > 0.31. (The 100th root of .5 is .99309.) That means Q is favorable even when p is as low as 1/3, where we should be getting 2:1 odds rather than giving 5:1 odds. To be confident of p being > 5/6 we'd need to be confident that Q > (.99 + 5/600)100 = ~0.85. A single favorable outcome doesn't support that level of confidence.

Still not taking the bet.

@harey, some guidance about completing a formula would be appreciated. I'm interested in the "surprise."

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I agree with Rainman; we need a description of how the marbles were selected to be placed into the jar in the first place.

If you were to say that someone flipped a fair coin 100 times and put a white marble in the jar if he saw head and a black marble in the jar if he saw tails, then that would be enough of a description to take on the problem. And I bet the answer would be a lot different if someone picked a random number from 0 to 100 with equal probability and put N white marbles and 100-N black marbles in the jar.

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The way how the marbles were selected is not known, so you cannot do better then Bonanova.

However, instead of grabbing the calculator:

Spoiler

e**x=limit(1+x/n)**n and take x=-1;

n=100 is large enough for a quick estimation: 1/e

The exact value is about 58% (Euler-Mascheroni constant), but do not ask me for details.

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