Help me in building the intution to solve these types of problems.

probability

A box contains 12 chocolates, 3 of which are white chocolate, 4 milk chocolate, and 5 dark chocolate.

I am sharing the box with three of my friends. The four of us take turns, each one drawing a chocolate at random from those available in the box. Yes, we're drawing without replacement. We intend to eat the chocolates, not put them back.

Politely, I let my friends draw before me.

For each of the subparts, provide your answer as an unsimplified fractional expression.

a) What is the chance that I draw a dark chocolate?

b) What is the chance that the friend who gets to select first draws a dark chocolate and I draw a dark chocolate too?

Best Answer

I get the following

a) $p=\frac{5}{12} \frac{4}{11} \frac{3}{10} \frac{2}{9}+3\frac{7}{12}\frac{5}{11}\frac{4}{10}\frac{3}{9}+3\frac{7}{12} \frac{6}{11} \frac{5}{10} \frac{4}{9} + \frac{7}{12} \frac{6}{11} \frac{5}{10} \frac{5}{9} \approx 0.42$

The first term accounts for three friends picking dark, the second for two of the three friends picking dark, the third for only one friend picking dark, and the last one for neither of them picking dark.

b) $p=\frac{5}{12}(\frac{4}{11}\frac{3}{10}\frac{2}{9}+2\frac{7}{11}\frac{4}{10}\frac{3}{9}+\frac{7}{11} \frac{6}{10}\frac{4}{9}) \approx 0.15$

Same principle, first term = both of the last two friends pick dark, second = one of the two pick dark, last = none of the two pick dark

Related Solutions

[Math] My last bars of chocolate

The fact that there are 6 people picking is irrelevant. The fact that you pick last is irrelevant. The fact that there are milk and white is irrelevant.

You get 4 chocolates, for them to be all dark you have to choose 4 dark from your 4 choices.

$$p=\frac{8}{24}\frac{7}{23}\frac{6}{22}\frac{5}{21}$$

How many draws to get one of each with replacement

This is a special case of the so-called "coupon collector's problem"

https://en.wikipedia.org/wiki/Coupon_collector

Calculating the expected number of draws required along with the variance is fairly easy using the linearity of expectation. From here, one can obtain probability bounds using Chebyshev's inequality. I don't know of a way to obtain the pmf directly, although you could try simulation.

Update: I played around with this in R. The Chebyshev bound says that 206 draws will be sufficient to have at least a 90% probability of drawing all 26. The exact tail probability from Ross, pointed out in @awkward's answer, gives $P(T \geq 206) \approx 0.0008$ and $P(T \geq 141) \approx 0.099$ for your example. Here's my R code if you're interested:

# This script calculates upper-tail probabilities for the coupon collector's
# problem in three different ways: 
#     1. A bound based on Markov's inequality
#     2. A bound based on Chebyshev's inequality
#     3. An explicit solution from Ross "A First Course in Probability"

# Compute expected number of trials, E[T], needed to obtain all n coupons
get_mean <- function(n) {
  n * sum(1 / (1:n))
}

# Compute an upper bound for P(T >= a) using Markov's inequality
get_markov_tail <- function(a, n) {
  stopifnot(a > 0)
  get_mean(n) / a
}

# Compute an upper bound for P(T >= a) using Chebyshev's inequality
get_chebyshev_tail <- function(a, n) {
  mean_T <- get_mean(n)
  stopifnot(a > mean_T)
  n^2 * pi^2 / (6 * (a - mean_T)^2)
}

# Exact formula for P(T >= a) from Section 4.1 of Ross
get_exact_tail <- function(a, n) {
  k_seq <- 1:(n-1)
  term1 <- exp(lchoose(n, k_seq) + a * (log(n - k_seq) - log(n)))
  term2 <- (-1)^(k_seq + 1)
  sum(term1 * term2)
}


# Using these functions we'll now try an experiment. Suppose there is a total of
# 26 coupons and we want to find a such that P(T >= a) < 0.1. In other words,
# we want to know how many trials we would need for there to be at least a 90%
# chance that we will obtain all 26 coupons. How do the two upper bounds compare
# to each other, and how do them compare to the exact formula?

# The Markov bound is terrible:
get_markov_tail(1002, 26)
get_markov_tail(1003, 26)

# The Chebyshev bound is a massive improvement:
get_chebyshev_tail(1002, 26)
get_chebyshev_tail(1003, 26)
get_chebyshev_tail(205, 26)
get_chebyshev_tail(206, 26)

# The exact tail probability is a substantial improvement over Chebyshev:
get_exact_tail(1002, 26)
get_exact_tail(1003, 26)
get_exact_tail(205, 26)
get_exact_tail(206, 26)
get_exact_tail(140, 26)
get_exact_tail(141, 26)

Best Answer

Related Solutions

[Math] My last bars of chocolate

How many draws to get one of each with replacement

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