What Is the Probability of Specific Outcomes at a Blindfolded Easter Egg Hunt?

[moonman239]

Here's an interesting problem: At an Easter egg hunt, the host blindfolds all 20 children before sending them out to collect the Easter eggs. All 100 eggs are randomly placed throughout a 300-square-foot yard, whose boundaries are blocked off (except of course for the entrance).

Given that, what is the probability that:
1) Child #1 ends up with more eggs in his basket than child #2?
2) Child #1 ends up with the most eggs?
3) At least one child ends up with no eggs at all?

What I do know about this problem is that a hypergeometric distribution would not be the best fit, because removing one egg decreases the probability of another child finding an egg.

 

[SW VandeCarr]

Here's an interesting problem: At an Easter egg hunt, the host blindfolds all 20 children before sending them out to collect the Easter eggs. All 100 eggs are randomly placed throughout a 300-square-foot yard, whose boundaries are blocked off (except of course for the entrance).

Given that, what is the probability that:
1) Child #1 ends up with more eggs in his basket than child #2?
2) Child #1 ends up with the most eggs?
3) At least one child ends up with no eggs at all?

What I do know about this problem is that a hypergeometric distribution would not be the best fit, because removing one egg decreases the probability of another child finding an egg.

There are 100 eggs and 20 children, so one approach is to assume a uniform prior distribution since you've given no information on the individual child's skills. That would be that each child finds 5 eggs. A posterior distribution (based on the data) would give you information on the likelihood of of the uniform distribution vs the posterior distribution which probably would be best modeled by a Poisson distribution. You could then use the posterior distribution for a new experiment either on the same children (to test for consistency) or a new set of children (to test for representativeness). Alternatively, you could define a Poisson distribution with $$\lambda=5$$ and get some estimates from that.

I'm not sure what you mean by child # 1 and child # 2 out of 20 children or why it matters which child picks up an which egg at which time. It's a random stochastic process without more data.

 

[AlephZero]

Effectivel, the eggs are fouind one a a time (actually, it makes no difference that two could be found at the same instant, because they could both be found by the same child, or by two different children).

So this is the same as 100 repetitions of "throw a 20-sided die to decide which child gets the next egg".

Or looking at it anoother way, each child has 100 chances to find an egg, with the probability of each chance = 1/20 (because there are 20 children).

So the probability distribution of the number of eggs a child finds is binomial.

 

[SW VandeCarr]

Effectivel, the eggs are fouind one a a time (actually, it makes no difference that two could be found at the same instant, because they could both be found by the same child, or by two different children).

So this is the same as 100 repetitions of "throw a 20-sided die to decide which child gets the next egg".

Or looking at it anoother way, each child has 100 chances to find an egg, with the probability of each chance = 1/20 (because there are 20 children).

So the probability distribution of the number of eggs a child finds is binomial.

Why do you think it's binomial? The expectation is that a child finds five eggs. The most one child could find is 100, the least is zero. The value zero would have a non zero probability.

 

[moonman239]

I'm not sure what you mean by child # 1 and child # 2 out of 20 children.

Let's give them names: Child #1's name is Amy, Child #2's name is Robert.

None of the 20 children bear the same name.

 

[moonman239]

Alternatively, you could define a Poisson distribution with $$\lambda=5$$ and get some estimates from that.

How do I figure out the answer to questions #1 and #2 that way?

 

[SW VandeCarr]

How do I figure out the answer to questions #1 and #2 that way?

1. Think about your question. One child will have more than the other if they don't have the same number. What's the probability that two random children will have the same number of eggs?

Analytically you want $$P(x_j > x_i|x_i)$$ but x(i) can be located anywhere over the distribution with probability 1.

2. At least one child out of twenty will have the most eggs. However you will have to add the probabilities of ties from 2 to 20 if you want to include ties.

 

[SW VandeCarr]

Given that, what is the probability that:
1) Child #1 ends up with more eggs in his basket than child #2?

Hint: This distribution does not have an infinite tail. There are 101 possible numbers. There's 0.5 probability density that child two will have 0-5 balls and 0.5 probability density that child two will have 6-100 balls. However, if child two has over 50 balls, child one cannot have the same number. The probability of child two getting over 50 balls is infinitesimal and can be ignored. Now since there are fifty numbers to consider, each number has a non zero probability of being the location of child two's number of eggs which can be computed under the distribution since they are discrete fractions of the whole distribution. Now what is the probability that a number will be chosen twice, by both child one and two? Sum over these probabilities for all numbers up to fifty. (You'll be very close if you just do the first 10-20). I'm pretty sure this will be a close estimate of the probability that child one and child two have the same number of eggs.

I was looking for a short cut but I didn't trust them. This is the brute force computational way.

 

[moonman239]

1. Think about your question. One child will have more than the other if they don't have the same number. What's the probability that two random children will have the same number of eggs?

Analytically you want $$P(x_j > x_i|x_i)$$ but x(i) can be located anywhere over the distribution with probability 1.

2. At least one child out of twenty will have the most eggs. However you will have to add the probabilities of ties from 2 to 20 if you want to include ties.

Running a simulation of the distribution of the eggs shows that the probability of Amy having more eggs than Robert is about 43.7%. Is this correct?

 

[SW VandeCarr]

Running a simulation of the distribution of the eggs shows that the probability of Amy having more eggs than Robert is about 43.7%. Is this correct?

How did you do it? I just suggested summing over the the squares of the probabilities of the individual numbers (at least the first 10. Twenty would be more than adequate). That gives the probability of the two children having the same number of eggs. Subtract that from 1 and take half of that for one particular child over the other. So 1-(0.437)(2) = 0.126 probability of them having the same number. It sounds reasonable but I can't give you a number without doing the computation and I'm just using my laptop where I am right now.

 

[AlephZero]

Why do you think it's binomial? The expectation is that a child finds five eggs. The most one child could find is 100, the least is zero.

The probaibilties of a child finding 0, 1, 2, ... 100 eggs are given by the terms in the expansion of
(0.95 + 0.05)^100.

That's what I meant by "binomial". Apologies if that isn't the correct term for this probability distribution. (Maybe Bernouilli trials would be better?)

If you want to approximate this by a Poisson distribution, that's fine.

The value zero would have a non zero probability.

Sure, it is 0.95^100 = 0.00592.
Or using the Poisson approximation, exp(-5) = 0.00674.

 

[AlephZero]

Running a simulation of the distribution of the eggs shows that the probability of Amy having more eggs than Robert is about 43.7%. Is this correct?

I agree with that. For any two children A and B, either A has more eggs than B, B has more than A, or they have the same number.

By symmetry, p(A has more than B) = p(B has more than A).

From the Binomial distribution (or whatever other people want to call it) the probabilities for the number of eggs per child are

 N   P(N)
 0  0.00592
 1  0.03116
 2  0.08118
 3  0.13957
 4  0.17814
 5  0.18001
 6  0.15001
 7  0.10602
 8  0.06487
 9  0.03490
10  0.01671
11  0.00719
12  0.00280
13  0.00100
14  0.00032
15  0.00010
16  0.00003
>16 to small to worry about.

The probability that A and B get have the same number of eggs is the sum of the squares of P(N) = 0.131
So the probability that A gets more than B is (1 - 0.131)/2 = 0.4345.

 

[moonman239]

How did you do it?

I ran 500,000 iterations each for two variables (the number of eggs in Alex's basket, and the number of eggs in Robert's basket.) I then subtracted the results of each iteration in variable 1 from the results of each iteration in variable 2 to obtain the third variable, the difference.

I found that approximately 43.69...% of the observed values in the third variable were greater than 0.

In R, my console looks like this (after running another simulation of the data):
>amy = rpois(500000,5)
>robert = rpois(500000,5)
>difference = amy - robert
>table (difference > 0)

FALSE TRUE
282033 217967
>217967 / 500000
[1] 0.435934

 

[moonman239]

I just suggested summing over the the squares of the probabilities of the individual numbers (at least the first 10. Twenty would be more than adequate).

Please explain what you mean.

 

[SW VandeCarr]

Please explain what you mean.

AlephZero calculated it above using the binomial distribution. The Poisson is the limiting case for the binomial when the expected value is small. You want the probability value for each number under the distribution. You get the probability mass by summing over the distribution between two values. In this case the number of eggs could be anywhere from 0 to 100 for either child, but with an expected value of only 5, the probabilities for any value over 16 are negligible by AlephZero's calculations. Since we a allow the number of eggs k for the child B to be anywhere between 0 and 16, we take the total probability mass of this interval, which is close to 1, to be 1 for all practical purposes.

Now we want to know the probability that the child A will also have those values at the same time as child B; that is, the probability that both children have the same number of eggs at the end of the game. We get this by squaring the probability. If it were three children we would cube the probability. We then sum over all squared values to get the probability of the two children having the same number of eggs k over the interval 0 to 16. We then subtract this from one to get the probability that the two children have different values. Taking half that value gives the probability that child A will have more eggs then B (or that B will have less eggs then A.)

 

[moonman239]

AlephZero calculated it above using the binomial distribution. The Poisson is the limiting case for the binomial when the expected value is small. You want the probability value for each number under the distribution. You get the probability mass by summing over the distribution between two values. In this case the number of eggs could be anywhere from 0 to 100 for either child, but with an expected value of only 5, the probabilities for any value over 16 are negligible by AlephZero's calculations. Since we a allow the number of eggs k for the child B to be anywhere between 0 and 16, we take the total probability mass of this interval, which is close to 1, to be 1 for all practical purposes.

I think I understand. In other words, we find the probability of no eggs, probability of 1 egg, probability of 2 eggs, then 3, and so on and so forth, up to 16.

Now we want to know the probability that the child A will also have those values at the same time as child B; that is, the probability that both children have the same number of eggs at the end of the game. We get this by squaring the probability.

Squaring what probability? 1/20?

We then sum over all squared values to get the probability of the two children having the same number of eggs k over the interval 0 to 16.

Huh, what?

 

[SW VandeCarr]

I think I understand. In other words, we find the probability of no eggs, probability of 1 egg, probability of 2 eggs, then 3, and so on and so forth, up to 16.

Squaring what probability? 1/20?

No. It's the probability of each number 0-16 under the Poisson (or the binomial) distribution which AlephZero lists in his post, just as you yourself acknowledged above. (As I said before but will say again, the Poisson is the limiting case of the binomial distribution for small expected values.)

Huh, what?

From AlephZero's post of calculations (post 12):

"The probability that A and B get have the same number of eggs is the sum of the squares of P(N) = 0.131
So the probability that A gets more than B is (1 - 0.131)/2 = 0.4345."

I've explained this several times. Given the probability that one child A will have p(An(x)) of a number n eggs under the distribution, the probability of two children p(An(x)) and p(Bn(x)) having the same number of eggs at the end of the game is p(An(x))p(Bn)(x)). This is a basic law of multiplying probabilities for two randomly concurrent events. We sum this over all numbers 0-16 to get the probabilities for the relevant distribution because the location of both children in the distribution is random. I'm surprised you don't understand this.

I also told you that my approach was to calculate the probabilities of the two children having the same random number of eggs, From that we calculate probabilities of the inequalities. This is absolutely clear from my quote of AlephZero's response above.

You don't seem to be paying attention to the people that are trying to help you.

 

[awkward]

No one has mentioned so far that if we define $$X_i$$ to be the number of eggs found by child i, and we assume that each child is equally likely to find any given egg and the finds are independent, then the vector
$$(X_1, X_2, X_3, \dots , X_{20})$$
has a Multinomial distribution.

For the probability that everyone finds at least one egg, apply the Inclusion / Exclusion method. Let's say event $$E_i$$ is the event that child i finds no eggs. There are $$\binom{20}{1}$$ ways to pick the child. Given a child, the probability that he finds no eggs is $$(19/20)^{100}$$. So
$$S_1 = \sum P(E_i) = \binom{20}{1} (19/20)^{100}$$

There are $$\binom{20}{2}$$ ways to pick two children. Given two children, the probability that they both finds no eggs is $$(18/20)^{100}$$. So
$$S_2 = \sum P(E_i \cap E_j)= \binom{20}{2} (18/20)^{100}$$

And so on.

So the probability that at everyone gets at least one egg is
$$1 - S_1 + S_2 - S_3 + \dots - S_{19} = 1 - \sum_{i=1}^{19} \binom{20}{i} \left( \frac{20-i}{20} \right) ^{100}$$

 

[SW VandeCarr]

No one has mentioned so far that if we define $$X_i$$ to be the number of eggs found by child i, and we assume that each child is equally likely to find any given egg and the finds are independent, then the vector
$$(X_1, X_2, X_3, \dots , X_{20})$$
has a Multinomial distribution.

The OP got a probability of 0.4359 with a simulation (500,000 iterations). Using the binomial distribution (given the Poisson $$\lambda=5$$) and the approach I suggested, AlephZero got P=0.4345. This is pretty good correspondence for two different approaches. I'm curious what you would get using a multinomial distribution to answer the OP's question: What is the probability that Child A gets more eggs than Child B?

 

[moonman239]

No. It's the probability of each number 0-16 under the Poisson (or the binomial) distribution which AlephZero lists in his post, just as you yourself acknowledged above. (As I said before but will say again, the Poisson is the limiting case of the binomial distribution for small expected values.)




From AlephZero's post of calculations (post 12):

"The probability that A and B get have the same number of eggs is the sum of the squares of P(N) = 0.131
So the probability that A gets more than B is (1 - 0.131)/2 = 0.4345."

I've explained this several times. Given the probability that one child A will have p(An(x)) of a number n eggs under the distribution, the probability of two children p(An(x)) and p(Bn(x)) having the same number of eggs at the end of the game is p(An(x))p(Bn)(x)). This is a basic law of multiplying probabilities for two randomly concurrent events. We sum this over all numbers 0-16 to get the probabilities for the relevant distribution because the location of both children in the distribution is random. I'm surprised you don't understand this.

I also told you that my approach was to calculate the probabilities of the two children having the same random number of eggs, From that we calculate probabilities of the inequalities. This is absolutely clear from my quote of AlephZero's response above.

You don't seem to be paying attention to the people that are trying to help you.

So (just to get it in my head) it's like this:

P(A and B having the same number of eggs) = P(A and B having 0 eggs) + P(A and B having 1 egg) + P(A and B having 2 eggs) + P(A and B having 3 eggs)...+ P(A and B having 16 eggs)

 

[moonman239]

Thanks, sorry! I wasn't thinking enough.

 

[SW VandeCarr]

So (just to get it in my head) it's like this:

P(A and B having the same number of eggs) = P(A and B having 0 eggs) + P(A and B having 1 egg) + P(A and B having 2 eggs) + P(A and B having 3 eggs)...+ P(A and B having 16 eggs)

Yes, because anyone these concurrent events is possible (including more than 16 up to 50, but these probabilities are negligible given an expectation of 5 eggs).

 

[awkward]

The OP got a probability of 0.4359 with a simulation (500,000 iterations). Using the binomial distribution (given the Poisson $$\lambda=5$$) and the approach I suggested, AlephZero got P=0.4345. This is pretty good correspondence for two different approaches. I'm curious what you would get using a multinomial distribution to answer the OP's question: What is the probability that Child A gets more eggs than Child B?

Let $$X_i$$ be the number of eggs found by child i. The marginal distribution of $$X_1$$ is Binomial(n=100, p=1/20), and the conditional distribution of $$X_2 | X_1=r$$ is Binomial(n=100-r, p=1/19); so

$$P(X_1 = X_2) = \sum_{r=0}^{50} P(X_1=r) P(X_2=r | X_1=r)$$
$$= \sum_{r=0}^{50} \binom{100}{r} \left( \frac{1}{20} \right) ^r \left( \frac{19}{20} \right) ^{100-r} \binom{100-r}{r} \left( \frac{1}{19} \right) ^r \left( \frac{18}{19} \right) ^{100-2r}$$
$$\approx 0.1273289$$

Using $$P(X_1 > X_2) = (1/2) [1 - P(X_1 = X_2)]$$,
we find the probability that child 1 finds more eggs than child 2 is approximately 0.4363355.

 

[SW VandeCarr]

It looks like we all nailed it! (At least close enough for government work.)

 

[moonman239]

For the reference we can also employ the normal approximation to solve these problems, because it satisfies the condition that both n * p and n * (1 - p) >= 5.

For question 1, the parameters to use for the normal approximation are:
mean = 5 - 5 = 0 (since we can expect both children to obtain 5 eggs each, on average)
sd = sqrt(Var(Amy) + Var(Robert)) = 2.738613 (to 6 decimal places)

Then, the approximate answer is: P(Amy having more eggs than Robert) = Norm(>0.5,0,2.738613) = 42.7661%.

To answer question #2:

mean = -90
sd = 8.215838

resulting in a probability of 0.

 

[SW VandeCarr]

Here's an interesting problem: At an Easter egg hunt, the host blindfolds all 20 children before sending them out to collect the Easter eggs. All 100 eggs are randomly placed throughout a 300-square-foot yard, whose boundaries are blocked off (except of course for the entrance).

Given that, what is the probability that:
1) Child #1 ends up with more eggs in his basket than child #2?
2) Child #1 ends up with the most eggs?
3) At least one child ends up with no eggs at all?

Then, the approximate answer is: P(Amy having more eggs than Robert) = Norm(>0.5,0,2.738613) = 42.7661%.

To answer question #2:

mean = -90
sd = 8.215838

resulting in a probability of 0.

How do you get P=0? Question 2 is the probability that child 1 (Amy) has the most eggs. How do you conclude that Amy has zero probability of having the most eggs (of all the children)? Since all the children are random, Amy has about a 1/20 chance of having the most eggs as does Robert or any other child (unadjusted for ties).

 

[moonman239]

How do you get P=0? Question 2 is the probability that child 1 (Amy) has the most eggs. How do you conclude that Amy has zero probability of having the most eggs (of all the children)? Since all the children are random, Amy has about a 1/20 chance of having the most eggs as does Robert or any other child (unadjusted for ties).

Sorry for the late response, but what I did was run the calculation with the knowledge that the the sum of the eggs collected by the other 19 children would equal 95, on the average.

 

[moonman239]

For the reference we can also employ the normal approximation to solve these problems, because it satisfies the condition that both n * p and n * (1 - p) >= 5.

For question 1, the parameters to use for the normal approximation are:
mean = 5 - 5 = 0 (since we can expect both children to obtain 5 eggs each, on average)
sd = sqrt(Var(Amy) + Var(Robert)) = 2.738613 (to 6 decimal places)

Then, the approximate answer is: P(Amy having more eggs than Robert) = Norm(>0.5,0,2.738613) = 42.7661%.

I wonder how I arrived at 2.738613. Maybe I rounded some place. On my graphing calculator I came up with about 3.08227 as the standard deviation. The variance of the binomial distribution = n*p*(1-p).

Is there any way I can use the normal distribution to estimate the answer to question #2?