What are the chances of losing a game of hangman?

Question

Halfway through a game of hangman (the word guessing game) I was playing, I wanted to calculate the chances of losing the game. I have $15$ letters remaining, and $3$ spaces to fill-up (no letter repeats). I have only $2$ wrong guesses left before losing the game. What is the probability of me losing the game?

What I tried:

Since I have $3$ spaces but only $2$ wrong guesses, I can either make:

1. $2$ wrong guesses. $$Probability=\frac{12}{15}\times\frac{11}{14}$$
2. $1$ correct guess followed by $2$ wrong guesses. $$Probability=\frac{3}{15}\times\frac{12}{14}\times\frac{11}{13}$$
3. $2$ correct guesses followed by $2$ wrong guesses. $$Probability=\frac{3}{15}\times\frac{2}{14}\times\frac{12}{13}\times\frac{11}{12}$$
4. Wrong, correct, wrong $$Probability=\frac{12}{15}\times\frac{3}{14}\times\frac{11}{13}$$
5. Wrong, correct, correct, wrong $$Probability=\frac{12}{15}\times\frac{3}{14}\times\frac{2}{13}\times\frac{11}{12}$$

Adding all these individual probabilities must (if I am not wrong) give me the total probability of the event (i.e., losing) This gave me the following answer: $Probability=\frac{440}{455}\,\text{or} ~96.7\text{%}$

Is this the correct approach to solve the problem? Am I missing something here?

Answer 1

I am assuming calling any of the three letters out is considered correct and not that they have to be called out in a specific order.

The case that you left out is:

• correct, wrong, correct, wrong

Once you add it, you get the probability of $ \displaystyle \frac{451}{455}$

But using complimentary method may be easier. For me to win the game, I must call out three correct letters in at most four attempts.

Probability of me winning is,

$ \displaystyle P(W) = \dfrac{3 \choose 3} {15 \choose 3} + \dfrac {{12 \choose 1} \cdot {3 \choose 2}} {{15 \choose 3}} \cdot \frac{1}{12} = \frac{4}{455}$

The first term is the probability of winning in next three attempts. The second term is the probability of winning in four attempts - the first three attempts must have $2$ correct letters and $1$ wrong letter. The last one must be the correct letter.

Subtracting from $1$, you get the same result as your method after correction.