Would you rather rely on your original intuition or believe in the supremacy of probability?
What does “trust your gut” mean? This works when you are given a scenario without any credible information/hint. You have to trust your intuition and make a choice. This is a blind scenario, meaning you have no other choice but to go with one choice.
But what if you are offered a choice within your choice? Would you take the odds or stick with your guts? Let us decode this dilemma of options and what is the best strategy to make your bets.
The scenario is called Monty Hall Problem– a situation in which you are offered three choices, one gets you a big reward, and the other two are duds.
You make an initial choice based on your intuition.
Then the host reveals one of the other two options you did not choose. That option was the wrong one.
Then the host gives you another chance– to either stick with your original choice or switch to the other remaining option.
Here is the “best solution” and not the “sure-shot solution” to the Monty Hall Problem– go for the switch option.
Let us dive deeper into the problem and why it asks you to doubt your first gut feeling.
What Is the Monty Hall Problem in Simple Terms?
Monty Hall hosted the popular television show, Let’s Make A Deal on CBS. The game show’s format was straightforward– make a choice and earn huge rewards. Let’s get it straight; it is a sort of gamble. But, over the years, people tried to figure out whether there is a pattern to this gamble.
Let’s see the Monty Hall problem with an example–
The year is 1965. Monty Hall, the enigmatic host, invites you to the stage of “Let’s Make A Deal.” Now you are standing in front of three doors. One door has a luxury car behind it (worth millions). The other two doors have a donkey each (still worth it, but not as much).
Mr. Hall asks you to make a choice; obviously, you are betting on the car. There is a 1 in 3 chance that you would choose the car. You choose option A.
| DOOR A: Your choice ⅓ chance | DOOR B: Unknown ⅓ chance | DOOR C: Unknown ⅓ chance |
The host says, “good choice!” and proceeds to reveal one of the two options (i.e., Either Door A or Door C).
Suppose he revealed DOOR C. And what do you know; there was a donkey behind it.
| DOOR A: Your choice | DOOR B: Unknown | DOOR C: Donkey |
Now, while you sigh relief that you didn’t choose DOOR C, the host throws another option. He asks you a simple question, “Would you like to switch your choice to DOOR B?” And here comes the actual problem.
What Is the Solution to Monty Hall’s Problem Explained?
You are in a dilemma, but still, it narrowed down to 50-50 choice, hasn’t it?
Well, probability works in mysterious ways. Because now you have a clue.
The clue is that– The Host is the key.
He is aware of which door consists of the winning prize. And to make for an entertaining TV show, he revealed an option that he was sure was a dud. Now that he has offered you another chance, doesn’t it feel like he knows there is some weightage to DOOR B; since he purposely did not reveal that door?
So here is your best bet– go for the switch option because the chances are not evidently 50%. The probability that DOOR B would have the car is actually ⅔ or 66.66%; whole sticking to your option still has a chance of only 33.33% of winning.
This is also known as “CONDITIONAL PROBABLITIY.”
| DOOR A: Your choice ⅓ chance | DOOR B: Unknown ⅔ chance | DOOR C: Donkey |
Why Is the Chance Not 50-50 in the Monty Hall Problem?
How is the likelihood, not 50-50, but the scale is tipping to one side?
Instead of 3 doors, let’s assume there are 30 doors. Again, Mr. Monty wants to make a fun show, so he reveals all 28 dud doors one by one until only one door remains.
So now, there are two doors– one is your original choice, and the other was filtered out of 29 selections. Which number would you trust more, 1 or 29?
This exact same logic applies to three choices as well. Hence when two options are remaining, the odds are not 50-50, but the possibility of the other option is slightly higher. Why? – because the hot purposely did not reveal that door.
Here is a simple probability chart– These are three possible events of choosing each door as your first choice. In the three options, you either decide to switch or stick to your original option depending on which door the host reveals first.
| Your original choice | DOOR A | DOOR B | DOOR C |
| Probability 1 | STICK | SWITCH | SWITCH |
| Probability 2 | SWITCH | STICK | SWITCH |
| Probability 3 | SWITCH | SWITCH | STICK |
If we tally the number of times the SWITCH option yields a winning car to the number of times sticking to your first intuition–
The odds are switching wins 6 out of 9 times- ⅔ chance.
While sticking to the original choice is only 3 out of 9 times- ⅓ chance.
So what do you think is the best-case scenario?
Such a decision is also called “COUNTER INTUITIVE.”
What would be your decision? Do you agree with this theory? Or do you believe in “Trust Your Gut?”
Let us know in the comments section below.
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