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Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.


Puzzle ID:#144
Fun:*** (2.3)
Difficulty:*** (2.13)
Submitted By:duckrocket**
Corrected By:sugarnspice4u7




Peter throws two darts at a dartboard, aiming for the center. The second dart lands farther from the center than the first. If Peter now throws another dart at the board, aiming for the center, what is the probability that this third throw is also worse (i.e., farther from the center) than his first? Assume Peter`s skillfulness is constant.

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(user deleted)
Dec 31, 2001

I could agree with the solution if all 3 darts were thrown at the same time, or if any results were unknown until all three darts had been thrown independently (or for that matter, any scenario where the question being asked is "Three darts are thrown at a dartboard, what are the chances that any particular dart will not be the closest to the center?". However, since you are running 2 independent tests, you are actually asking a different question twice (will the dart currently being thrown be closer than the 1st dart thrown?) the odds will be 50-50 for every dart thrown. What if you threw 100 darts, or 1000? Would your odds really increase with every dart? What if you removed the 2nd dart from the board after it was thrown? Would it decrease your odds of being farther from the center?
Mar 06, 2002

The question and explanation are a little out of synch. The answer is correct only before any darts are thrown.

It is true that if we are about to throw N darts independently, then each dart has a future probability 1/N of being closest to the center, and a (N-1)/N probability of not being closest. But this is only true before we throw any darts.

However, in the question, exactly as it is phrased, we already know the relative positions of the first two darts. The question states that the second dart is NOT the closest. Therefore the probability that the second dart is the closest is clearly zero, not 1/3 as given in the explanation.

So I suggest that the correct answer to the question, as stated, is 1/2 not 1/3. Because there are only two darts that could potentially end up being closest – the 1st or the 3rd.

This holds true for any number of darts if we wait until just before we throw the last dart to ask our question. Suppose N=1000 and we have already thrown 999 of them, then we already know that 998 of the thrown darts now have zero probability of being closest. With this prior knowledge, we can state that the final dart has a 50% chance of being closest, since there are now only two possible outcomes – either the last dart will hit closer than the current closest, or it won’t.
Mar 08, 2002

I like vega's thinking, but there is not enough info. 2/3 is definitely wrong though, 50/50 is closer
Apr 10, 2002

I disagree with the solution. There is no way to predict if a shot will be better than the first two.
Apr 17, 2002

If you don't know "how good" the other ones were all you can say is 50/50 the last being better
If you know how good the others are, eg bulls eye, 0 chance of improving.
If the other two went miles away =~ 1 of improving
Jun 12, 2002

Duck is 100% correct. The information that the second dart is farther from the bullseye than the first is irrelevent. The question could be rephrased: what's the probablity the third dart won't be the closest yet? The answer to this question is (N-1)/N for any number, N, of darts thrown.
Jul 23, 2002

Duckrocket's solution is correct, provided that there is no chance
of a tie between the different darts. If there is a chance of a tie,
that will reduce the probability a bit, since the question asks for
the probability that the third dart is further than the first.
The really neat thing about this puzzle is that it works for any probability
distribution function where ties have probability zero (most continuous density
functions without singularities should satisfy this), no matter what the
shape of the underlying probability function is.

The information that the second dart is further than the first *does*
change the probability distribution of the first dart. To see this,
suppose that we were rolling fair dice instead of throwing darts.
If I tell you that the second die was higher than the first, the
probability that the first die is now showing a "1" is five times
that of the first die showing a "5" (you are looking at
five combinations, 1-2, 1-3, 1-4, 1-5, and 1-6, vs. 5-6),
and the probability of the first die showing a "6" is now zero.
Before I told you that additional information, all sides
were equally probable for the first die.
If you do this problem with dice instead of darts, you won't get
exactly 2/3rds, because of the chance of a tie, but as you
increase the sides of each die towards infinity, the limit
of the probability should be 2/3rds.
Jul 29, 2002

The information that the second dart is farther off the mark than the first is irrelevant, as long a the question is: 'what's the probability that the third dart won't be the closest yet?'. The probability would be unchanged if instead the second dart was closer than the first, and question was to find the probability that the third dart is farther off the mark than the second. Of course the statistics of the location of the first dart are influenced by the information that it's closer than the second dart. But the question does not ask about the location of the first dart. As for the probability of a tie, that is clearly zero. The distance of a dart from the center of the target is a continuous variable, not discrete like the faces of a die. For more discussion of this problem, see the probability section at, where Duck probably got it.
Jul 31, 2002

The probability of a tie doesn't have to be zero, despite it being a (mostly) continuous variable.
For example, you could have a model where the probability of being immediately next to a boundary stripe
is non-zero, because any hit on the boundary stripe is deflected to the closest point of the target next to the stripe.
As for the relevance of the second dart, it becomes irrelevant once you paraphrase the question as you suggest.
But it is clearly relevant to the original question, which asked for the probability that the third dart is
farther than the first. Just disregarding the second dart without accounting for its affect on the first dart's probability distribution
leads to confusion such as shown in vvega's comment. You can't disregard the second dart without either explicitly or implicitly acknowledging that
the first dart's distribution has changed with the additional information.
Nov 21, 2002

It says, "Assume Peter`s skilfulness is constant." Then wouldn't the 3rd dart be the worst?
Nov 22, 2002

I read "Assume Peter's skillfulness is constant"
to mean "Assume that Peter throws each dart
with the same probability distribution function."
So if he has (say) a 10% probability of hitting the
bullseye with the first dart, he would also have
a 10% probability of hitting the bullseye with the third
dart. Likewise for any other area on the dart board.
Mar 14, 2003

I think there are some before and after misinterpretations. Before he throws the darts, if you ask the question which of the three throws is going to produce the closest dart, the 1st, 2nd or 3rd throw, then they are all equally likely. After he has thown one dart, you are now in possession of new information. If he is an expert dart player who rarely misses and his first dart barely makes the board, then clearly the next dart is not equally likely to be futher away. I'd like to be at a tournament where the world darts champion has just thrown the worst shot of his career and have some of these theoreticians tell me the next dart is 50/50 or 33% etc likely to be worse. I'll take $100 000 bet at these odds if you are willing to put up the money. Unless you know something about the skill of the player and the average distribution of shots, you cannot make such predictions using this kind of Mathematical probability which is based on counting the number of equally likely cases!
Apr 25, 2005

Dec 09, 2005

I think you people have to much time on your hands.
Jan 09, 2006

I also think that the 2/3 solution can't be right. After all, you do need more information. For example, say the first dart landed right by the bull's eye, with just enough room for one additional dart to fit on the exact center of the bull's eye. In order to fit the dart on the bulls-eye, you would need perfect accuracy to hit that tiny target. Since the first dart did not hit, and the second dart was worse, the statement that the thrower throws with a constant degree of accuracy demonstrates that the thrower does not have perfect accuracy. The third dart might land anywhere within a circle (or more likely an elllipse) defined by the throwers degree of accuracy. Just looking at the surface area available to hit would say that if the thrower is not perfectly accurate, the odds of the third dart landing in the tiny space inside the first dart is extremely low, while the odds of it landing outside is much larger. The probability of making a better throw would then be the area of the circle of probability defined by his level of accuracy divided into the area of the circle within the location of the first dart. In any event, it is not a simple or straightforward as presented, and more information is needed to answer the question properly.
Feb 23, 2006

ILL miss YOUR dartboard
Mar 24, 2006

Nice problem and beautifully explained by dewtell.
Can't understand how the hint is supposed to help in this case?
Mar 31, 2006

dunt undrstand even a lil bit!!
Apr 25, 2006

i got it,, but i dont get the HINT.
May 13, 2006

There is no correct answer.

Say the first dart was a bullseye. Then the third dart will automatically be farther from the center. But if the second dart hits the very edge of the board, and the first dart was just barely closer, then the third dart will have a great chance to be the closest. So the answer really depends on where the first dart is specifically.
Aug 15, 2006

? i just did not get it, it was a tough one
Oct 25, 2006

The answer to this teaser is clearly wrong, why is it allowed to remain?
Apr 06, 2007

btw, how did you make "wrong" bold?
Aug 12, 2008


Throwing three darts you can have these possible outcomes.
(the numbers are which dart that comes closest to center, in the middle and furthest away. So for example 213 means second dart was best and third dart was worst)

We know that the second dart lands farther away then the first so we only have these left:

Out of these there is a 2/3 chance third dart will worse than the first one. QED

The information that the second dart is farther from the bullseye than the first is relevent contrary to what manyof you think. Imagine we had 1000000 darts. First dart is better than the next 999998 darts how likely is it then that the last dart will be better than the first? Obviously in this case the first one was really good, probable bulls eye, so its very veryunlikely we can beat it with the last one.

I have also runned a computer simulation on this and it confirm the answer 2/3.
Feb 20, 2009

You know you've got a great probability problem when more than half the people posting claim you have the wrong answer.

BTW, as Pixit explained quite clearly, the author is correct. Everyone who disagrees with them is full of baloney.

With these sorts of puzzles, if you start with a bunch of cases that have equal probability, then you have some additional information that eliminates some of the cases, then the remaining cases still have equal probability. (Of course, you do have to define your cases correctly so that the new information doesn't re-weight them. But that isn't hard, usually.)
(user deleted)
Jun 30, 2011

The solution is correct.

Let the darts be called A, B, and C. Let us throw A, B, and C in that order. We have two possibilities (where the order is how close the dart was to the center):


What most of you don't understand is that, given that dart A is closer than dart B, it is more likely that dart A is the best throw.
Jul 29, 2013

Let's suppose that dart A landed exactly in the centre of the bullseye and dart B landed on the edge of the dartboard. Are you seriously trying to say that the events ABC, ACB and CAB are all equally likely?

The answer depends upon the proximity of dart A, which is known at the time of throwing dart C. Assuming that Peter does actually hit the dartboard with his last dart, the answer is 1-d/r, where d is the distance of dart A from the centre and r the radius of the board. If we allow for the possibility that Peter misses the dartboard with equal probability, then we have a problem, as r becomes infinite.

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