Yet another coin flipping problem

Question

Problem

Starting with a set of 10 coins at the start where all coins are tails up, and given n number of integers \$x_1, x_2, x_3... x_n\$ representing n rounds of coin flipping.

At each round, we randomly flip \$x_i\$ number of coins at random. i.e Coins that were heads become tails, and vice versa. Within each round, every coin can be flipped at most once, i.e no repeats.

Objective Write the shortest function that takes as input a list of integers, and calculates the expected number of heads at the end of all rounds.

Assume that the inputs will always correct, i.e every element is between 0 to 10.

Example 1:

# 3 coins chosen at random were flipped over one round, hence E(Heads) = 3
Input = [3]
Output = 3

Example 2:

# 5 coins chosen at random were flipped in the first round
# At the second round, only 1 was random flipped with 50% chance of 
# picking a head/tail. E(Heads) = 0.5*6 + 0.5*4

# Hence E(Heads) = 5
Input = [5, 1]
Output = 5

Answer 1

Python 2, 40 bytes

r=0
for c in input():r+=c-r*c/5.
print r

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Let \$Z_i, 1 \le i \le 10 \$ be random variables indicating the state of the \$i\$-th coin after all flips, \$1\$ for heads and \$0\$ for tails. The number we are interested in is:

$$ E \left[ \sum_{i=1}^{10} Z_i \right] = \sum_{i=1}^{10} E \left[ Z_i \right] = 10 E\left[ Z_1 \right] $$

(We can do these two equalities because the expected value is linear and all \$Z_i\$ are identically distributed)

This means it is enough to compute the expected value for a single coin, which leads to the following program, where the formula can be simplified to the above:

r=0.0
for c in input():r=r*(10-c)/10+(1-r)*c/10
print r*10

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Answer 2

05AB1E, 10 bytes

Uses a formula I got by plugging sympy into my Python solution.

5(/æ¦PO5(*

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$$ \sum_{A \subseteq X, A \ne \emptyset} (-5)^{1-\left| A \right|} \prod_{a\in A} a = -5 \sum_{A \subseteq X, A \ne \emptyset} \prod_{a\in A} {a \over -5} $$

where \$X\$ is the multiset \$\left\{ x_1, x_2, \cdots, x_n\right\}\$.

5(/         # divide each value by -5
   æ¦       # get all non-empty subsets
     P      # take the product of each subset
      O     # sum all values
       5(*  # multiply result by -5

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A direct port of the Python answer is the same length:

Îv1y5/-*y+

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Answer 3

JavaScript (V8), 29 bytes

x=>x.reduce((a,b)=>a+b-a*b/5)

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Based on ovs's answer.

Answer 4

Vyxal, 10 bytes

λ₌+*5/-;dR

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port of 05AB1E answer’s formula

and also, wholesomely rekted by lyxal with 5 bytes saved

Answer 5

Jelly, 7 bytes

÷5CPC×5

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$$ 5\times(1-\prod_{i=1}^n 1-{x_i \over 5}) $$

Read this from right to left to get the Jelly code. The formula is derived from the same sympy output as the 05AB1E answer. This has a nice symmetry to it as C×5 is the inverse of ÷5C.

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Jelly, 8 bytes

Port of my Python answer

+_×÷5ɗɗ/

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Answer 6

MathGolf, 9 bytes

0ë{‼+*5/-

Port of @Yousername's JavaScript answer, which is based on @ovs' answers.

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Explanation:

0         # Start with 0
 ë        # Push the input-integers as float-array
  {       # Foreach over these floats:
   ‼      #  Apply the following two commands separated to the stack:
    +     #   Sum the top two values
    *     #   Multiply the top two values
     5/   #  Divide the product by 5
       -  #  And then subtract it from the sum
          # (after the loop, the entire stack is output implicitly as result)

Answer 7

Charcoal, 16 bytes

≔⁰ηＦθ≧⁻×ι⊖∕η⁵ηＩη

Try it online! Link is to verbose version of code. Explanation: Port of @ovs' Python solution, except with a slightly rearranged formula r-=c*(r/5-1) which saves a byte when written in Charcoal.

≔⁰η

Start with zero heads.

Ｆθ

Loop through the numbers of flips.

≧⁻×ι⊖∕η⁵η

Calculate the expected number of heads.

Ｉη

Output the expected number of heads.