From cfb256077bb0f1743aae4b7b0bf1152fa76e9c96 Mon Sep 17 00:00:00 2001 From: Gaurav Borse Date: Fri, 27 Mar 2026 18:47:14 +0530 Subject: [PATCH] Improve clarity and grammar in Monty Hall explanation ### Beginner Note This problem is a classic example where human intuition fails in probability. Although it appears that both remaining doors have equal chances, Bayesian reasoning reveals that switching doors significantly increases the probability of winning. --- examples/Monty_Hall.ipynb | 20 ++++++++++++++++---- 1 file changed, 16 insertions(+), 4 deletions(-) diff --git a/examples/Monty_Hall.ipynb b/examples/Monty_Hall.ipynb index 191d3365b1..87b4a4afe9 100644 --- a/examples/Monty_Hall.ipynb +++ b/examples/Monty_Hall.ipynb @@ -17,7 +17,7 @@ "\n", "Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, \"Do you want to pick door No. 2?\" Is it to your advantage to switch your choice?\n", "\n", - "By intution it seems that there shouldn't be any benefit of switching the door. But using Bayes' Theorem we can show that by switching the door the contestant has more chances of winning.\n", + "By intuition, it may seem that there is no benefit in switching doors. However, using Bayes' Theorem, we can show that switching actually increases the probability of winning.", "\n", "You can also checkout the wikipedia page: https://en.wikipedia.org/wiki/Monty_Hall_problem" ] @@ -26,9 +26,21 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "### Probabilistic Interpretetion:\n", - "So have 3 random variables Contestant $C \\in \\{1, 2, 3\\}$, Host $H \\in \\{1, 2, 3\\}$ and prize $P \\in \\{1, 2, 3 \\}$. The prize has been put randomly behind the doors therefore: $P(P=1) = P(P=2) = P(P=3) = \\frac{1}{3}$. Also, the contestant is going to choose the door randomly, therefore: $P(C=1) = P(C=2) = P(C=3) = \\frac{1}{3}$. For this problem we can build a Bayesian Network structure like:\n", - "\n" + ### Probabilistic Interpretation + +We define three random variables: + +- Contestant (C): The door initially chosen by the contestant. +- Host (H): The door opened by the host (always revealing a goat). +- Prize (P): The door hiding the prize. + +Each of these variables can take values from {0, 1, 2}, representing the three doors. + +Since the prize is placed randomly: +P(P=0) = P(P=1) = P(P=2) = 1/3 + +Similarly, the contestant initially picks a door at random: +P(C=0) = P(C=1) = P(C=2) = 1/3 ] }, {