**Monty Hall** game was a popular game show in the 1970s.

Originally, the game win/loss was thought to be based on luck. So, it caused controversy when a Mathematician tried to solve the game using Math. Using basic probability she was able to help the player in making the right choice increasing the winning odds in their favor. As you will see, the probability didn’t really fit intuitively.

This article exists in an attempt to answer the below question.

How may times one has to play the game to ensure that the player wins at-least 2/3 of the times?

### How is the game played?

The game begins with 3 doors. Let’s say doors A, B, C. One of these 3 doors has a prize behind it. Others have something or nothing. For example, other doors might have a goat behind them or it could be empty.

- Monty (the game host) would ask the player to choose one of the three doors which may have a prize behind it.
- The player chooses one of the doors. Let’s say, the player chose D
*oor A*. - Now, Monty would open one of the doors B or C to reveal that it did not have a prize. Let’s say he opened
*Door B*. - Next, after revealing one of the wrong doors, Monty would ask the player if they would like to change their decision. So, the player is given a choice again. Do they want to stick with their original choice (
*door A*) or would they like to switch to*door C*? Suppose, the play decided to switch to*Door C*. - Lastly, Monty will open one of the closed doors. Whether the opened door had a prize or not, at this point, it is clear which door will have a prize. For the sake of this game, let’s say Monty opens Door C which also happens to be the winning door.
- In our game simulation, the player
**won**by switching the door :).

### Intuition vs Probability

Let’s see what the probability looks like at each step.

- Initially, we had started 3 closed doors. Hence the probability of any one of them having a prize behind it is 1/3. As a result, after choosing one of the doors, the player had a 1/3 chance of winning the game.
- After Monty opened one of the doors to reveal that it did not have the prize, we are left with 2 closed doors. By intuition, it seems our probability is now 1/2 – 1/2. We can either win or lose with equal probability. However, this is where Math says that our intuition is wrong :). Once a door is eliminated, the originally chosen door still has a probability of 1/3 to be a winning door. Consequently, the probability of the last remaining door to be the winning one is effectively 2/3 :-).

This is such non-intuitive thinking that even after knowing the answer, it doesn’t “**feel**” right. Going back to our Mathematician friend, she posted an article in a newspaper claiming that the player should always switch when given the choice since that door has a winning probability of 2/3. Naturally, it became a controversy even among Mathematicians :-).

To learn more about the game, here is the Wikipedia link https://en.wikipedia.org/wiki/Monty_Hall_problem

#### It still doesn’t make sense

How may times one has to play the game to ensure that the player wins at-least 2/3 of the times?

This is a common question that arises whenever we talk about probability. Let’s consider another example for clarity.

Given a coin, the chance of getting heads or tails is 50-50. Does it mean, that if I toss the coin 4 times, I can be certain that I will get 2 heads and 2 tails?

The answer is of course **NO. **Sometimes, you might get 4 heads or 4 tails in a row. You cannot be certain. Don’t take my word, try it.

So, what is the magic number of iterations (games) one should play to ensure 2/3 chance of winning in this game of choices?

My initial intuition was it’s got to be in 10,000 ranges. Since we now know our intuition can be widely off, I decide to create a simulation of the game and run it.

### Game Simulation Rules

A single execution of the simulation works like this.

- You start with 3 numbers. 1, 2 & 3.
- Using randomness, you choose one of them to be the winning number.
- Now, the player chooses one number randomly
- Next, Monty chooses one number excluding the winning number or the player chosen number
- Now player switches the number
- Check if the new number was our winning number. If yes, the player won. If no, the player lost

After running this simulation a dozen times, I found the following.

- We start to converge toward 2/3 = 66.6% probability by the time we reach ~100 games.
- If we continue to run the simulation more than 200 times, we are almost always within 65 & 67 percent probability.
- If we let it run a million times, you would see a straight line indicating 66.66% wins.

### Conclusion

At best, you need to play over 100* times, to ensure you won 66% of the times.

**Based on our simulation*

### Simulation Demo

Here is the link to the simulation that I published. Feel free to play with it. Click on the chart to see the probability at any given iteration time. Full Demo Here

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