The Monty Hall problem and Deal or No Deal can both feel like “keep your choice or switch,” but their odds are built differently. In Monty Hall, the host’s informed action changes what you should infer from the remaining doors. In Deal or No Deal, opening cases usually reveals values without a host steering you toward or away from a particular alternative.
The short answer: switching wins in the standard Monty Hall setup because the host knows where the prize is and deliberately avoids it. That reasoning does not automatically transfer to a suitcase game.
The crucial difference: who controls the information?
Here is the standard Monty Hall setup:
- There are three doors: one hides a prize and two hide non-prizes.
- You choose one door.
- The host knows where the prize is.
- The host opens a different door that contains a non-prize.
- The host offers you a switch to the one unopened door.
Your first choice has a 1 in 3 chance of being right. The other two doors collectively have a 2 in 3 chance of hiding the prize. When the host removes one losing door from that pair, the full 2 in 3 chance concentrates on the one remaining unchosen door.
That is why switching has a 2 in 3 chance of winning.
For a fuller walkthrough of that reasoning, see Why the Monty Hall Problem Isn’t 50/50.
In Deal or No Deal, by contrast, the cases are generally opened according to the contestant’s choices. A low-value case and a high-value case are equally eligible to be selected if you do not know their contents. The revealed values affect your estimate of what remains, but they do not create a Monty Hall-style signal from an informed host.
Why “two choices remain” does not mean 50/50
The tempting argument goes like this:
> One door is gone. Two doors remain. Therefore each door has a 50% chance.
That skips the method used to remove the third door.
Imagine you pick Door A in Monty Hall.
| Where the prize is | Your initial pick | Door the host can open | Does switching win? |
|---|---|---|---|
| Door A | Door A | Either Door B or Door C | No |
| Door B | Door A | Door C | Yes |
| Door C | Door A | Door B | Yes |
You initially chose correctly in only one of the three equally likely cases. Switching wins in the other two.
Now change one feature: instead of an informed host, suppose a person randomly opens one of the two doors you did not choose and happens to reveal a non-prize. That event supplies information, but the probabilities are different.
If your first choice was correct, a random opener is guaranteed to reveal a non-prize. If your first choice was wrong, a random opener has only a 1 in 2 chance of avoiding the prize. Conditional on seeing a non-prize, your original choice is then 1 in 2 likely to be correct. In that altered game, the two unopened doors really are 50/50.
The lesson is not “always switch.” It is: ask how the information was generated.
What Deal or No Deal reveals—and what it does not
In a typical suitcase game, you select a case to keep closed while opening other cases. Each opened case tells you exactly which value is no longer available. That is useful information because it changes the set of possible outcomes in the unopened cases.
But it does not usually tell you that your own case became more or less likely to contain a particular value compared with another specific unopened case.
Suppose four cases contain $1, $10, $100, and $1,000. You choose one case, then independently choose another case to open. It contains $1.
Three cases remain: yours and two others. The remaining values are $10, $100, and $1,000. Before any further information, each unopened case has the same 1 in 3 chance of holding each remaining value.
If you later swap your case for one particular other unopened case, neither case has an inherent odds advantage. They are exchangeable: each has the same probability distribution over the remaining values.
That differs from Monty Hall because no one with knowledge selectively protected the prize while opening a losing option.
Conditional probability, step by step
Conditional probability means updating your estimate after observing evidence. The key question is whether that evidence was more likely under one hidden state than another.
Monty Hall
Let “stay wins” mean your first pick held the prize.
- Before the host acts, the probability that stay wins is 1 in 3.
- If stay wins, the host can open either losing door.
- If stay loses, the host has only one permissible door to open: the other losing door.
The host’s reveal is constrained by the prize location. His action is not random noise. It preserves the original 2 in 3 probability that your first pick was wrong, and switching capitalizes on that probability.
An ordinary suitcase reveal
Suppose you choose a case and then choose another case to open without knowing its value.
- The opening rule does not depend on where the top prize is.
- Seeing a low value tells you that low value is gone.
- It does not identify one remaining unopened case as the special beneficiary of a 2 in 3 probability mass.
You should update the possible values remaining, not pretend that one particular alternative case inherited the odds of all discarded cases.
Quick check: which game favors switching?
Try this before reading the answer.
You pick one of three boxes. One holds a prize. A moderator knows the contents, opens an unchosen empty box, and offers a switch. Should you switch?
A. No, because two boxes remain. B. Yes, because your original box was right only 1 in 3 of the time. C. It makes no difference because the moderator opened an empty box. D. Yes, but only if the moderator opened the box at random.
Answer: B. Your initial box has a 1 in 3 chance of holding the prize. The moderator’s informed reveal leaves one unopened alternative carrying the 2 in 3 chance that your first choice was wrong.
Why the other answers fail:
- A treats the remaining boxes as symmetric even though the moderator’s choice was constrained.
- C notices the empty-box reveal but ignores why that reveal was guaranteed.
- D reverses the logic. A random reveal changes the calculation; the standard advantage comes from an informed moderator who avoids the prize.
Now change the facts: you choose a case from three, then randomly open one of the other two and it happens to be empty. In that version, conditional on the empty reveal, staying and switching are equally likely to win. The moderator’s knowledge is the dividing line.
How this matters when deciding on a bank offer
Deal or No Deal is not just a question of which case has the top prize. The practical choice is usually between a known offer and an uncertain set of remaining values.
A sound way to think about it:
- List the values still in play.
- Treat each unopened case as equally likely to contain each remaining value, unless the rules provide additional information.
- Compute the average remaining value as a benchmark.
- Compare the offer with that average, while recognizing that risk tolerance matters.
For example, if the remaining values are $1, $100, and $1,000, the average is:
\[ (1 + 100 + 1000) / 3 = 367 \]
An offer below $367 is below the simple average remaining value. That alone does not force a decision: someone may reasonably prefer a guaranteed amount to a risky outcome. But it does show why “my original case feels lucky” is not an odds-based argument.
Do not import Monty Hall’s switching rule into this decision. In a standard suitcase setup, swapping your chosen case for one named remaining case does not improve your expected value.
FAQ
Is Monty Hall the same as Deal or No Deal?
No. Monty Hall relies on an informed host who deliberately opens a losing door. In Deal or No Deal, the contestant’s selections usually reveal values without that informed filtering, so no particular remaining case gains a switching advantage.
Why does switching work in Monty Hall but not in a suitcase game?
In Monty Hall, your initial choice is wrong 2 in 3 times, and the host’s rule exposes a losing alternative without exposing the prize. In a suitcase game, opening a case removes a value but does not normally distinguish your case from another unopened case.
If low amounts are opened in Deal or No Deal, should I switch cases?
Not merely because low amounts were opened. Low amounts being removed raises the average value among the remaining cases, but every unopened case shares that improved set of possibilities equally.
Is Monty Hall still 2 in 3 if the host picks a door at random?
No. If the host randomly opens an unchosen door and happens to reveal a non-prize, the two unopened doors are 50/50. The standard 2 in 3 result requires the host to know the prize location and always avoid revealing it.
What question should I ask before applying Monty Hall logic?
Ask: “Could the person revealing information have shown me something different if the hidden outcome were different?” If the answer is yes, their action may carry a conditional-probability signal. If not, treat the reveal as information about eliminated options, not as a reason one remaining option is privileged.