# Probability and the Monty Hall problem [duplicate]

In this video is explained that during the Monty Hall problem you have a $\frac {2}{3}$ probability of winning if you always switch and a $\frac {1}{3}$ probability of winning if you never switch.

I understand the reasoning but it just feels wrong. Because if we can assume that the host always shows a goat after the first try, which implies that there are just 2 options left, you could still choose both options by switching or not.

Doesn't that conclude that there is always a $\frac {1}{2}$ probability of winning?

• It does "feel" wrong, but feelings are sometimes a poor substitute for reasoning. This is one of those times. There are lots of explanations on the web (you found one). You're unlikely to get a genuinely new explanation as an answer here. You could start at wikipedia: en.wikipedia.org/wiki/Monty_Hall_problem – Ethan Bolker Oct 25 '15 at 13:55
• @EthanBolker I understand, but I do not get why the probability is not $\frac {1}{2}$ if you look at it this way. I hope someone could explain this. – Tim Oct 25 '15 at 13:57
• There are indeed two options left. But they are not equally likely to win. – Ethan Bolker Oct 25 '15 at 14:01
• Already answered here: Monty Hall Three-Door Puzzle – Morgan Rodgers Oct 25 '15 at 14:02
• I don't think that this question is a duplicate or that the answers to the other question address the question asked here: Does choosing at random lead to $P(\text{winning})=\frac 12$? The answers to the other question show that the two remaining doors have probabilities of $\frac 23$ and $\frac 13$ of concealing the prize, not equal probability. But, equal probability is not the issue. Choosing one of the two doors at random leads to $P(\text{winning})=\frac 12$ regardless of the $\frac 23$, $\frac 13$ result: any $p, 1-p$ will give the same $\frac 12$ for $P(\text{winning})$. – Dilip Sarwate Oct 25 '15 at 17:26

Your calculation that choosing at random (with probability $\frac 12$ each) whether to switch or stay put gives a probability $\frac 12$ of winning the prize is perfectly correct.

Let $A$ be the event that your first choice is indeed the door that conceals the prize. Then, $P(A) = \frac 13$ and $P(A^c) = \frac 23$. Note that $A^c$ is the event that one of the two doors not chosen conceals the prize. Once one unchosen door has been opened, you flip a fair coin and switch or stay put according as the coin shows Heads or Tails. If $W$ is the event that you win the prize, then

\begin{align} W &= (A, T) \cup (A^c, H)\\ P(W) &= P(A)P(T) + P(A^c)P(H)\\ &= \frac 13\times \frac 12 + \frac 23\times \frac 12\\ &= \frac 12. \end{align}

The "always switch" strategy can be put into this framework by using a double-headed coin that always turns up Heads, leading to

\begin{align} P(W) &= P(A)P(T) + P(A^c)P(H)\\ &= \frac 13\times 0 + \frac 23\times 1\\ &= \frac 23 \end{align}

and similarly for the "always stay put" strategy, the chances of winning are $\frac 13$. I leave it to you as an exercise to figure out what the probability of winning is if you choose to toss a biased coin that turns up Heads with probability $p, 0 < p < 1$ and to show that this probability is smaller than the probability of winning with the $p = 1$ "always switch" strategy.

Rework the problem with 100 doors and 99 goats. You pick a door, the host then opens 98 doors you didn't pick all with goats. So of the original 100 doors only two are left, the door you picked and a door the host went through a bit of effort not to open.

So how does this feel? Now to me, I know when I pick one door at random out of 100 I'm going to get a bad door. I'm not that lucky. So I know the host is going to open all the other goat doors and leave the prize door closed. So should I switch. Of course, I only had a 1 in 100 chance of picking the correct door in the first place so there is a 99 in 100 chance that the goat was behind another door. Whatever door it was was the door the host kept closed. So there is a 99 in 100 chance that the prize is behind the other door.

The trick is to realize this isn't the probability of the prize being in a specific other door (the door wasn't specified). It's the probability the the prize wasn't behind your door.

The way I like to think about it intuitively is to think about the chances that your first pick, which is not revealed right away, was not the prize: $\frac{2}{3}$. Then you are shown one among the remaining two that is not the prize (so $\frac{2}{3}$ of the time, there will only be a single remaining possibility for this second non-prize). For me, setting it up this way makes it intuitive to see that switching will win $\frac{2}{3}$ of the time while retaining the original choice is essentially a $\frac{1}{3}$ gamble that your initial choice happened to be the prize.