Before continuing, the reader should acquaint himself with Philip’s explanation of the basic form of the Monty Hall problem, as shown in this YouTube video:
I just had a very interesting email exchange with a reader named Johan who had some excellent questions about Monty Python-esque probability puzzles. The questions he brought up were so thought-provoking that I decided to dedicate an entire blog post to answering them. Figuring out what’s wrong with Johan’s hypotheticals is an excellent exercise in improving your probabilistic vision. I hope you enjoy it!
What if you have 4 cups on a table and then you ask your friends to take one coin and put it randomly under one cup while you are looking away. Then you pick one cup secretly not telling anybody (25% of picking right). Then you say to your friends “Ah my misstake! I meant only that there should be 2 cups!” and you remove two cups at random not selecting the one you have secretly selected
If it happens that the two cups you removed had the coin you simply go away and redo the experiment with some other group of friends. On the other hand if they were empty you continue (you have now 2 cups on the table and the coin is under one of them).
The one you secretly selected in the beginning will then still have 25% chance of containing the coin. If you then select the other cup and shows it to your friends you will have 75% to find the coin.
The thing that bothers me is that some other person could have secretly selected the other cup (the cup I did not secretly select or none of those I removed). Then he will say he has 75% chance of finding the coin under the cup I secretly selected in the beginning. But according to me, the cup he secretly selected, will have 75%. How can both cups have different probabilities depending on the observer?
What is wrong by removing the cups randomly and have some luck instead of having Monty to remove them? Or is it something else that is crazy?
First, probability can indeed vary by observer depending on what information they possess. Suppose you pick the lottery number 575,361. Since it’s out of 1 million possible numbers, you correctly assume that you have a 1/1,000,000 chance of winning the lottery. Now, suppose I work for the lottery and I’m privy to inside information that the winning number starts with the digit 5. Well, with that extra information I know that you really have a 1/100,000 chance of winning. Two different observers, two different probabilities, both correct from their own perspective. Of course, the “absolute truth probability of the universe” remains unchanged, whatever that is. Remember that probabilities are always just guesses. For example, there could be another lottery employee out there who’s privy to the first TWO digits of the winning number and his guess would be better than either of ours. So, when dealing with probabilities, we want to get the most accurate answer for the information we have, and that’s the best we can do.
Second, in your example the two players are playing subtly DIFFERENT games. Let’s take a moment to remember why the probability doesn’t change in the Monty Hall problem when the doors are opened (or the cups are removed). When you pick 1 cup out of 4, you have a 25% chance of picking the winning cup. When 2 cups are removed, and there are only 2 left, why are the probabilities now not 50-50? It’s because you never had a chance of being eliminated. Since you knew that your cup would never be removed, nothing has changed now that other cups have been removed; your cup is still only 25%. However, suppose I am the “other secret player.” I have no such assurance that my cup will not be removed; in fact, in 50% of cases I’m eliminated from the game before it even begins (because you remove my cup). Therefore, if I survive the elimination, my cup’s probability does indeed increase to 50%. To sum it up, the secret player has a 50% chance of surviving the elimination, and then a 50% chance of having the correct cup of the remaining 2, for a 25% chance total of winning the game.
So, in reality, when you play this game you will come to the conclusion that your cup has a 25% chance of winning and the other a 75% chance, and the secret player, if he even survived this long, will believe that both cups have a 50% chance of having the coin.
Okay, but that still seems like a discrepancy. How to resolve that? The key is to give the secret player more information. Let’s do the same thought experiment but assume that the secret player KNOWS what you are doing (i.e. this Monty Hall experiment). Well, as the secret player, I now know that you purposefully did not eliminate your own cup. What does this tell me? It tells me that one of the remaining cups has a 25% chance of winning and the other a 75% chance of winning, just as you know, but I still don’t know which cup is yours, so I don’t know which is which! In other words, I know that my cup has a 50% chance of being the cup with a 25% chance of winning, and a 50% chance of being the cup with a 75% chance of winning, which all calculates out perfectly to: 0.5 x 0.25 + 0.5 x 0.75 = 0.5 = 50%!!
So, even knowing that you are playing the Monty Hall game I still can’t do better than 50% unless you tell me which cup you chose and purposely didn’t eliminate. Of course, if you tell me you are playing the Monty Hall game, and you tell me which cup you chose, then we aren’t actually two different people anymore since we are both privy to the same information and come to exactly the same probability calculations.
I have a deck of cards and you will try to pick the ace of spades. First you choose one card but don’t look at it (you will then have the probability 1/52). I, the dealer, know all the cards and eliminate 50 of the remaining cards that is not the ace of spades. So now you have two cards and are allowed to change your guess. Of course you should change because you will have the propability of 51/52 if you switch.
Now consider this, we do it again:
You select one card from the deck as before (1/52). Then you select a second card and at the same time look at the remaining 50 cards. Because you are lucky none of these cards is the ace of spades! That means that you have done exactly the same work as I did before when I selected the second card for you! You are now stuck with 2 cards (one of them is the ace of spades) and you should switch because of the same reasons as before.
The big problem here is of course that you just as well could have picked 2 cards from the deck at once and then have checked the remaining cards to find out that the ace happened to be in one of your selected 2 cards. You will obviously then have a 50-50% chance and it would not matter if you change. How can that be?
One of the tricky things when dealing with probability is that it’s easy to talk yourself into something that isn’t true. You say, “That means that you have done exactly the same work as I did before when I selected the second card for you! You are now stuck with 2 cards (one of them is the ace of spades) and you should switch because of the same reasons as before,” but it turns out that that is NOT the case. In both of the examples you give in your email there is a 50% chance of each card being the ace of spades.
There are fundamentally two ways the game can be played: A) you can take two cards and then look at the deck to see if the ace remains, or B) you can pick the first card, look at the deck, and then pick the second card. Let’s go over each of these.
A) Picking two cards first is easy. If the ace remains in the deck, you’ve lost. If you don’t see the ace, each card has a 50% chance of being the ace.
B) Picking one card, looking at the deck, and then picking another is tricky. 1/52 times you won’t see the ace in the deck, in which case you’ve already got the ace and it doesn’t matter what the second card is. 51/52 times you will see the ace and you will be COMPELLED to pick it as the second card. Remember, those are the rules: you must discard 50 cards that AREN’T the ace of spades.
The first scenario you describe in your email simply isn’t possible (well, it’s ambiguous). You say, “Then you select a second card and at the same time look at the remaining 50 cards,” but what does that even mean? Either you pick it randomly, in which case you’re in scenario A above where both cards are randomly picked first, or you look at the deck first, in which case you are unable to pick the second card randomly as mentioned in scenario B.
Do you see the problem? It is PHYSICALLY impossible to both look at the deck and see the ace of spades and AT THE SAME TIME randomly pick the second card. Those aren’t the rules of the game. The only way you can be sure to randomly select both cards is to use scenario A, which is not how the game is played and leads to a 50-50 chance for each card.