If a die is tossed 10 times and six of the tosses are 1s, what is the expected number of 2s in the 10 tosses?

Question

If a dice is tossed $10$ times and six of the tosses are $1$s,

What is the expected number of $2$s in the $10$ tosses?

I don't know if I'm overthinking this or not. Do I need to include the $6$ tosses that rolled a $1$ or am I just using the $4$ rolls that are left with the probability that rolling a $2$ is $\frac{1}{5}$, since no more $1$s can be rolled?

Answer 1

Assuming it's a fair die, you'd only worry about the remaining 4 rolls.

This would give you 4(1/6), with probability of rolling a 2 as 2/3.

Answer 2

It depends on how you interpret the question.

If you want to know the number of expected $2$'s in $10$ tosses (i.e. the number expected before doing any trials) then the answers is $\frac{1}{6} \times 10 = 10/6 = 1+ \frac{4}{6}$ths or $1.\overline6$ reccuring. This is the expected number of $2$'s in any set of $10$ tosses using a fair die. (You could argue that even if you rolled $10$ $1$'s, the expected number of $2$'s would still be $1.\overline6$)

If you are asking the number of expected $2$'s in a set of $10$ tosses where there have been $6$ rolls so far and all are $1$'s (and that you could potentially roll more ones), then it is $4 \times \frac{1}{6} = \frac{4}{6} = 0.\overline6$ recurring (again assuming that you are using a fair die)

If all $10$ tosses have been carried out by someone else (ie all results are concluded) and exactly $6$ were $1$'s (no potential for any more) then you are correct in assuming $\frac{1}{5}$th chance per trial for a $2$. In this case the expected number would be $4 \times \frac{1}{5} = \frac{4}{5} = 0.8$

However - this is assuming the die has not been tampered with.

Answer 3

If a fair die is used, we can indeed use the knowledge that exactly six $1$s were thrown. Each remaining value has a probability of $\frac{1}{5}$ of being rolled, so if $X$ is the number of times a $2$ is rolled, we find:

$$E[X] = 4 \cdot \frac{1}{5} = 0.8$$

Alternatively, using the definition of $E[X]$ along with the binomial distribution:

$$E[X] = \sum_{i=0}^{4} i P(X = i)$$

$$= 1 \cdot {4 \choose 1} \cdot \frac{4^3}{5^4} + 2 \cdot {4 \choose 2} \cdot \frac{4^2}{5^4} + 3 \cdot {4 \choose 3} \cdot \frac{4^1}{5^4} + 4 \cdot {4 \choose 4} \cdot \frac{4^0}{5^4} = \frac{4^4 + 12 \cdot 4^2 + 12 \cdot 4 + 4}{5^4} = \frac{500}{625} = 0.8$$