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John Doe was a rich man. He had $k + 1$ piggy banks, $k$ coins in each. In $i$-th piggy bank, $i-1$ coins are genuine and $k + 1 - i$ coins are counterfeit.

John equiprobably chooses the piggy bank and does this sequence of actions $n$ times:

  1. He shakes a piggy bank until a coin falls out (any coin can fall out with the same probability);
  2. Writes down information about whether the coin was a genuine or counterfeit;
  3. Throws the coin back into the piggy bank.

John is legitimately surprised, as all $n$ times the coin was counterfeit. What is the probability $P_{k}(n)$ that the next coin to fall out of the chosen piggy bank is also counterfeit?

I. What is the explicit formula for $P_{k}(n)$? Find the probability for $n = 2$ and $k = 5$, find $P_{5}(2)$.

II. Find $\lim_{k \rightarrow \infty} P_{k}(n)$.

Attempt

$P(\text{fake coin} \space | \space n \space \text{fake coins}) = \frac{P(\text{fake coin and} \space n \space \text{fake coins})}{P(n \space \text{fake coins})} = \frac{P(n + 1 \space \text{fake})}{P(n \space \text{fake})}$. Applying the total probability rule to each term in the ratio yields this result. $$P_k(n) = \frac{\sum_{i = 0}^{k} (\frac{i}{k})^{n+1}}{\sum_{i = 0}^{k} (\frac{i}{k})^n}$$ How to proceed with the limit? $$\lim_{k \rightarrow \infty} \frac{\frac{1}{k}\sum_{i = 0}^{k} i^{n+1}}{\sum_{j = 0}^{k} j^{n}}$$

Inter Veridium
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    This tests your ability to confront the gambler’s fallacy. You (or in this case John) can think something is “due” to happen, but that’s not how probability works. Also, the answer is embedded in the question! “we consider that the chance of falling out is also the same for every coin” – gen-ℤ ready to perish May 20 '20 at 00:01
  • @gen-zreadytoperish If we have a piggy bank with $0≤i≤k$ counterfeit coins, then the probability of $n$ counterfeit coins falling out in succession equals $(\frac{i}{k})^n$. Then we apply the Bayes' formula and find the posterior probability. – Inter Veridium May 20 '20 at 00:17
  • @gen-zreadytoperish If we presumed that $k$ approaches infinity, it would mean that the probability of counterfeit coin falling out is $1$? – Inter Veridium May 20 '20 at 00:27
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    Oh, now that I’d have to give some thought to and presently I’m about to have dinner with my dad. I think I oversimplified things. – gen-ℤ ready to perish May 20 '20 at 00:29
  • Everything is equally likely, so nothing will affected by the $n$ previous experiments. The probability of falling out an counterfeit coin at $n+1$th turn is the same as before. So, $$P_k(n)=\sum_{i=1}^{k+1} \frac{1}{k+1}\frac{k+1-i}{k} =1-\frac{k+2}{2k}$$. – Alapan Das May 20 '20 at 02:19
  • @AlapanDas Sorry, this answer is wrong. The numerical results do not support your claim. – Inter Veridium May 20 '20 at 10:09
  • @AlapanDas Besides, the conditional probability of picking the $i$-th piggy bank does not equal to $\frac{1}{k+1}$. Intuitively, the more counterfeit coins fall out, the more likely is that John has chosen the piggy bank with more counterfeit coins. – Inter Veridium May 20 '20 at 10:18
  • But you claimed that John equiprobably choses a piggy bank. There is no condition. – Alapan Das May 20 '20 at 10:36
  • @AlapanDas He chooses a piggy bank equiprobably, but the condition that $n$ previous coins were counterfeit gives us an insight into which piggy bank it actually was. For example, take $n > 0$. There's no way it was a piggy bank with $0$ counterfeit coins. – Inter Veridium May 20 '20 at 10:50
  • That isn't a condition. He got counterfeit coins all the times just by chance. That doesn't affect the next turn as you throw back the coin in its respective piggy bank. – Alapan Das May 20 '20 at 10:59
  • If a fair coin is flipped 21 times, the probability of 21 heads is 1 in 2,097,152. The probability of flipping a head after having already flipped 20 heads in a row is $\frac{1}{2}$...... Wikipedia – Alapan Das May 20 '20 at 11:02

1 Answers1

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For every nonnegative integer $n$ the summation$\sum_{i=1}^{k}i^{n}$ can be written as a polynomial of degree $n+1$ in $k$ where the coefficient of $k^{n+1}$ equals $\frac{1}{n+1}$.

See Faulhaber's formula.

Applying that to your (correct) result for $P_k(n)$ we find: $$\lim_{k\to\infty}P_k(n)=\frac{n+1}{n+2}$$

Remarkably the RHS is exactly the probability that would have rolled out for $P_k(n)$ if rule 3 (the coins are thrown back) would not be followed.

For that see this question and its answers.

drhab
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