The Monty Hall Problem
You might be familiar with this famously regarded paradox before in your life, The Monty Hall Problem.
To provide a prior context, this game was played in a 1960’s Canadian show called ‘Let’s make a deal, hosted by a person called Monty Hall. The show contestants had the opportunity of winning a huge prize only based on a simple probability-based choice (and some luck). We will be explaining the same in detail in this blog. Treat this as a game if you haven’t read about this before.
Let’s say we have 3 inverted cups in front of you, namely A, B, and C. And one of them has a piece of diamond underneath it, and the other two don’t.
Since we are your host, we know which one of the cups has a reward you would want to grab. And we offer you to tap on the cup you think possesses the diamond. Let’s say you choose cup A. So far so simple, a one-third probability for the reward.
Now here’s a twist you’d like. Out of the remaining 2 cups B and C, we lift cup C and you find there’s nothing below it. we give you another opportunity, either switch to cup B or stay at your original choice, i.e., cup A. Take a minute to think and make a decision before you read further.
Before revealing the answer, we would like to pose some questions that might have popped up in your head. Does it really matter if you switch or stay? It’s 50-50, isn’t it? One cup gives you a diamond the other one doesn’t, that’s just a simple chance.
How about we tell you that there’s a definite strategy to this paradox that increases your probability of winning? Switch. Switch every time.
To put things in perspective, initially, when we gave you this choice, you had a one-third chance of winning with cup A and a two-third chance of winning with cup B and cup C put together. But when we flicked open cup C, suddenly with cup A, you still had a one-third chance of winning. However, with the unopened cup B, your probability of winning the diamond suddenly shot up to two-thirds. Voila! How did this happen? Let’s boil it down a bit further.
To simplify this problem, let’s consider that all of us know which cup has the diamond underneath it, say cup B. We will now encompass all different ways a person can operate with the “Switch Every Time Strategy”. Now if we ask a person to choose one of the 3 cups, he has 3 options to do so.
Choose cup A. Now we as the host would naturally pick open cup C because obviously, that’s the empty one. Now we ask the person again, “Would you like to switch or stay?” With the switch option, the player wins the diamond with cup B. Keep in mind here that the player was initially wrong with the choice of his cup.
Choose cup B. Now here it does not matter which cup we open because both A and C are empty, hence we flick open cup C (the outcome would have no difference even if you open cup A). If the player switches and chooses cup A, the player returns home empty-handed although he was initially correct.
Choose cup C. This is technically the same as the Choose Cup A strategy. We as the host turn open cup A. With the switch strategy, the player again chooses cup B and wins. Note that, even in this scenario, the player was initially wrong.
To sum up the narrative, with the “Switch Strategy”, you win 2 out of 3 times. However, scroll up again to note that if you would have relied on the “Stay Strategy” (i.e. staying with your original choice), you would have won only 1 of 3 times.
Clearly, not a 50-50 chance.
If you still haven’t understood the logic behind this, don’t worry, we will be taking another shot to help you comprehend this by modifying the game a little. Instead of having 3 cups, this time we offer you 100 cups, but only one of them has the diamond. Seems a relatively tough choice, but you are offered to guess which one of these 100 cups behold the precious item. With a poor chance of 1% of winning, supposedly you choose cup number 100.
This time we blow away 98 other cups which don’t possess the reward, leaving cup number 50. And now you could feel what’s happening. In the beginning, you had the mere probability of 0.01 to win. With the switch strategy, the probability of winning is an astounding 99% or 0.99 when all it seems to the naked eye is the mere option of choosing 1 out of 2.
What’s even more fascinating is that every time the player was wrong in his/her initial choice, the player would always bag the prize by switching. And when the wind of luck is blowing against you, you might lose with switching, i.e. choosing the correct option initially. Of course, the strategy doesn’t guarantee a win, but when observed with the large sample size, you would certainly notice that the odds shoot up instantly.
So, in case you are offered a choice like this in the future, don’t forget to switch and pray to God that you make the wrong choice initially.
Recommended » Benford's Law