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Full problem text: The odds of the Celtics winning the NBA championship in the $2026/2027$ season are $1:2$($\frac13$ chance to clarify). The odds of the Mavericks winning the NBA championship that same year are $1:3$($\frac14$ chance to clarify). What is the probability that another team besides the Celtics or the Mavericks will win the championship.

I believe the answer is $\frac5{12}$ but my math teacher disagrees and says it is $\frac12$.

I have even written code to simulate this:

import random


def simulate_games(n_simulations):
celtics_wins = 0
mavs_wins = 0
neither_wins = 0


for _ in range(n_simulations):
    outcome = random.random()
    if outcome < 1/3:
        celtics_wins += 1
    elif outcome < 1/3 + 1/4:
        mavs_wins += 1
    else:
        neither_wins += 1

total = celtics_wins + mavs_wins + neither_wins
return {
    "Celtics win": celtics_wins / total,
    "Mavs win": mavs_wins / total,
    "Neither win": neither_wins / total
}


results = simulate_games(10000000)
print("Simulation Results:")
for outcome, prob in results.items():
print(f"{outcome}: {prob:.5f}")

Who is right, me or my math teacher?

Integreek
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Ranbeer Arora
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    How did your teacher get $6/12$? I would think of the events as disjoint, let $C$ be Celtics win, $M$ be Mavs win, and in that case I agree with your answer of $5/12$. – Pavan C. May 14 '25 at 15:09
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    @PavanC. The teacher considered either team winning (incorrectly) as independent events. Using the standard formula for both teams not winning then gives $2/3 * 3/4 =1/2$ as the probabaility. – Ingix May 14 '25 at 16:06
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    I wonder what would happen if you politely discussed the entire interchange (your doubts, your program, your question here and our answers) with your teacher. Best would be they admit error and thank you for helping them learn. If you're willing to try, report back (perhaps with an [edit] to the question or an answer of your own) and let us know what happened. – Ethan Bolker May 14 '25 at 18:10
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    I wonder whether the teacher intentionally argued for the wrong answer to encourage OP to think deeper and justify his answer. If so, great teacher! If not, then he (the teacher) is still great for offering an even more important lesson in humility and accepting when you're wrong. Not sure which I'm rooting for, but wonderful either way! – Robert May 16 '25 at 14:09

3 Answers3

27

Let $A$ denote the event of the Celtics winning the championship, and $B$ denote the event of the Mavs winning the championship. We know that $$P(A\cup B)+P(A\cap B)=P(A)+P(B)=\frac13+\frac14=\frac7{12}.$$ The teams cannot simultaneously win the championship, so $P(A\cap B)=0$. Then, the probability that some other team wins the championship is $$1-P(A\cup B)=\boxed{\frac5{12}}.$$ I believe the mistake your teacher made is assuming $A$ and $B$ are independent. In that case, we would have $P(A\cap B)=P(A)P(B)=\frac1{12}$, so that $1-P(A\cup B)=\frac12$.

DanDan面
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    One could look at the following related problem: The odds of the Celtics winning the NBA championship in the 2026/2027 season are 1:2 (1/3 chance to clarify). The odds of the Florida Panthers winning the NHL championship that same year are 1:3 (1/4 chance to clarify). What is the probability that neither the Celtics nor the Panthers win a championship? For this alternate problem the assumption of independence seems reasonable and the teachers answer would be correct. – quarague May 15 '25 at 11:33
  • Of course, even if the events of different teams winning were independent, the teacher would still be wrong, because the event you're actually asked about (another team winning) would no longer be the same as $(A\cup B)^c$. – Especially Lime May 15 '25 at 13:16
  • @EspeciallyLime: In the title, OP gives the problem as “What is the chance neither team wins?”; in the question body, they describe it as “…the probability that another team besides the Celtics or the Mavericks will win…”. So I don’t think we can know which version they were originally given. – Peter LeFanu Lumsdaine May 15 '25 at 13:22
  • @PeterLeFanuLumsdaine they explicitly describe the version in the body of the question as the "full problem text", so I think it's at least clear which version OP is claiming they were given. – Especially Lime May 15 '25 at 13:31
  • @PeterLeFanuLumsdaine: I'm not sure I see the distinction. Inevitably exactly one team wins the title, so how is "neither $C$ nor $M$ wins" and "some team other than $C$ or $M$ wins"? – Brian Tung May 16 '25 at 03:38
  • It seems A losing includes B winning, not as a certainty but as a possibility, and vice versa. We remove the overcounted ((A lose, B wins), (B lose, A wins)). Not full dependence but partial dependence. Non liquet – Hudjefa May 19 '25 at 09:15
  • @BrianTung: Indeed, the two alternative formulations are equivalent! But I was responding to Especially Lime, who was pointing out that if the events of M and C winning were independent (as the teacher mistakenly assumed) rather than mutually exclusive, then we couldn’t be assuming that exactly one team wins, so the two formulations would no longer be equivalent. – Peter LeFanu Lumsdaine May 19 '25 at 10:50
2

  1. In probability, "and" represents a multiplication if the events are independent, and "or" represents addition if the events are mutually exclusive.

  2. De Morgan's laws: not(A and b) = not A and not B, not(A or B) = not A and not B.

So if we have "and" and independence, we can immediately multiply. If we have "or" and independence, then we can use multiplication along with negation. And vice versa with mutually exclusive.

Here, we want the probability that the Celtics lose and the Mavericks lose. We can't use addition because don't have an "or", and we can't use multiplication because we don't have independence. But we can use de Morgan's laws to say:

(Celtics lose and Mavericks lose) =
(not Celtics win and not Mavericks win)=
not (Celtics win or Mavericks win)

(Celtics win or Mavericks win) gives us "or" with mutual exclusivity, so we can add $\frac13$ and $\frac14$ to get $\frac7{12}$. So if (Celtics win or Mavericks win) has a probability of $\frac7{12}$, then not (Celtics win or Mavericks win) has a probability of $\frac5{12}$.

Acccumulation
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Hmmm... Usually these questions are worded for independent events where both can happen or neither can happen and the probability of one happening (or not) has nothing to do with the other such as ... oh, let's say there is a $\frac 13$ chance that it rains on saturday and a $\frac 14$ chance you brother in law will ask for money on saturday then then chances that neither happens is $P(\text{first not happening})\times P(\text{second not happening}) = (1- \frac 13)(1-\frac 14) = \frac 23\times \frac 34 = \frac 12$.

So I'm assuming that is what your teacher is going for.

But that's dead wrong. These are mutually exclusive events. They can't both win. For mutually exclusive events (which are NEVER independent) we have that the sum of all the probabilities add to one so

$P(\text{celtics win}) + P(\text{mavericks win}) + P(\text{anything else}) =1$ so

$\frac 13 + \frac 14 + P(\text{anything else}) =1$ so

$ P(\text{anything else}) = 1-\frac 13 -\frac 14 = \frac 5{12}$.

=====

It's funny. Usually when people make probability mistakes they make them the other way. They'll say something like "If there's a 50% chance of rain on Saturday and a 75% chance of rain on Sunday, then that means there is a 125% chance of rain this weekend."

Shaun
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fleablood
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