A prize is hidden behind one of three doors. You choose the door where you think the prize is hidden. But before the door is opened, one of the other 2 remaining doors is opened to reveal no prize. You must now decide to keep your original door choice or to switch to a different door in hopes of revealing the prize! Sounds easy? Beware! In this game of chance, things aren’t always as obvious as they seem!

This a great game to help explain how conditional probability works. The conditional probability of an event *B* is the probability of that event occurring given that an event *A* has already happened. The activity is based on a popular TV game show known as “Let’s Make a Deal” that began in 1963 in the US. It is also known to mathematicians as the “Monty Hall” problem, after Monty Hall who was the original game show host for over 30 years.

(Suggested for Grades 6 – 12. Adjustably suitable for younger grades.)

__Materials and Setup (per pair of players): __

- 3 cups, labeled
*A*,*B*, and*C* - One small object (the “prize”) that easily fits inside a cup without getting stuck.
- 1 tally sheet per game (tally sheet template: Make a Deal with Monty!)

This two-player game is played in 2 rounds of 20 trials each.

Choose one player to be the “Host.” The Host always knows where the prize is hidden.

The other player is the “Guest.” The Guest does not know where the prize is located.

Place the 3 cups upside down in a row in front of the Guest on a table.

__How to Play the Game: __

**Step 1:** The Host asks the Guest to look away and to not peek. When the Guest is not looking, the Host places the prize under one of the cups.

**Step 2: **The Guest begins each trial by pointing at or saying the letter of the cup where they think the prize is hidden (the Guest does not touch the cups).

**Step 3:** Before revealing the Guest’s choice, the Host lifts up one of the other two cups (one that does not contain the prize) and shows its empty contents to the Guest.

**Step 4:** The Host offers the Guest an opportunity to switch their initial cup choice for the other cup that still hasn’t been revealed.

**Play the game 20 times using each of the following two scenarios:**

- The Guest always stays with their original choice and does not switch cups.
- The Guest always switches their original choice when offered a chance to do so.

**Scenario 1****:** The Guest always keeps their original choice.

- Played this way, the Host reveals the contents of the Guest’s cup each time a new game is played for 20 trials (repeat Step 1 to Step 4 above). On a trial, if the Guest’s cup has the prize, the Guest records a check on the tally sheet for that trial under the corresponding column.
- After 20 trials, add up the number of times the prize. Then divide that number by the total number of trials = 20. The resulting number will be the “experimental probability” of getting a prize when the Guest always stays with the initial choice.

**Scenario 2****:** The Guest always switches their original choice.

- Play as in Step 1 but in this scenario the Guest always switches cups when offered to do so.
- After 20 trials, add up the number of times the prize was found for each trial. Then divide that number by the total number of trials = 20. The resulting number will be the “experimental probability” of getting a prize when the Guest always switches their initial choice.

What’s going on?

There are no guarantees on any given turn for getting a prize. But if this game is played many times, it turns out that switching an initial choice is the best strategy! Why? Because the conditional probability of an event is the probability that the event will occur given the knowledge that another event already occurred.

In the Monty Hall problem, once one of the empty cups is eliminated, and players see the 2 remaining cups, they usually think there is a 1 in 2 chance that they have picked correctly. But, in terms of probability, their first choice has a 1 in 3 chance of being correct. So, the calculated conditional probability for ending up with the prize in Scenario 1 is 1/3. After the first cup is eliminated, if a player switches their first choice, the chances of winning increase from to 2 to 3, meaning that the calculated conditional probability for ending up with the prize in Scenario 2 is 2/3. Let’s examine this in some more detail!

**Scenario 1****: **

The Guest Chooses Cup *A* and does not switch on each trial:

**Explaining the results in the Scenario 1 chart****: **This is an example of when the Guest stayed with Cup *A*. Crossed out italicized words in the boxes indicate cups that have been eliminated by the Host. There is one “win” and 2 “lose” results. Likewise, if charts were shown for the Guest choosing to stay with Cups *B* and *C* respectively, the same end results would appear. Therefore, there is a 1-in-3 chance of winning the prize by always sticking with your first choice.

**In Scenario 2****:**

The Guest chooses Cup *A* and then switches on each trial:

**Explaining the results in the Scenario 2 chart****: **This is an example of when the Guest at first chose Cup *A*, but then switched cups when offered a chance to do so for each trial. There are two “wins” and one “lose” result. This means, if the Guest switches each time after one of the empty cups is turned over, they have a 2-in-3 chance of winning. Again, if this chart were made for a different starting cup, the same results would occur in the end; the Guest would win 2 out of 3 times! So, switching your initial choice is the best option in playing this game (but not the most intuitive!)

Questions:

- Compare the results for the your experimental probabilities based on 20 trials of each scenario with the calculated conditional probabilities. Did the experimental probabilities come close to the calculated conditional probabilities? Explain.
- Compile the results from several more rounds of trials. Look at the results after playing a number of times. How do the experimental and calculated conditional probabilities compare?

**Other ideas for playing this game: **

- Change the number of cups. For instance, use 4 cups with one hidden prize. Let the Guest choose the first cup, but before it is revealed, the Host shows the contents of 2 of the remaining 3 cups before the final selection is made. Does this change the overall end game?
- Explore other conditional probability scenarios. For example, determine the probability of 2 dice summing to 10 given that both rolled values are odd.

**For younger students****:** Try playing with 2 cups at first. The Host doesn’t reveal where the prize is. Play in the same fashion and discuss what happens each time they either win or lose. Could they improve their chances? How? Compare this to tossing a coin. Can they improve their chances of always getting a head or a tail? How about tossing two or more coins? This opens much discussion about choices, and whether or not what happens before another trial affects the outcome.