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Suppose that we roll a fair die until a 6 comes up or we have rolled it 10 times. What is the expected number of times we roll the die?

The answer to this is about 5.03.

However, if you remove the restriction of rolling it 10 times, the answer is 6.

I'm having a difficult time understanding conceptually why it would be different for infinite amount of rolls vs a max of 10 rolls. Both of the expected values are under 10, so why does it change so much?

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    $\begingroup$ The expected value is an average. It is certainly possible to roll a die ten times and never get a six, but the maximum possible value is $10$, while in the infinite case, it is possible to have to roll the die $100$ or $1000$ or $1,000,000$ times. So the expected value is going to be not only different, but greater. $\endgroup$ – Thomas Andrews Mar 18 '15 at 1:24
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"So much" seems a bit exaggerated: it only goes up from $5.03$ to $6$. The probability that you need to throw the die more than $10$ times to get $6$ is the probability that a non-$6$ turns up on all of the first ten trials. That is $\left(\frac 5 6\right)^{10}\approx 0.1615\ldots$ or about $1/6.19\ldots$ --- nearly one in six. And if you do need more than $10$ trials, there's quite a good chance you'll need more than $11$ and more than $12$. Indeed, given that you need more than ten trials, there is a probability of $\left(\frac 5 6\right)^3 = 0.5787\ldots$ that you'll need more than $13$ trials.

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The expected number of rolls should be smaller, intuitively, since it can only be 10 or less as opposed to arbitrarily large on there infinite sequence case.

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The expected value is $$\sum_{n \ge 1} n \cdot \mathbb{P}(\text{game ends on the $n$th roll}).$$ We want to see how this sum differs in the two games (with restriction and without).

For $1 \le n \le 9$, the corresponding probability $\mathbb{P}(\text{game ends on the $n$th roll}).$$ $ is the same in both games.

However, in the first game, the remaining probability mass is weighted by $n=10$: $$\mathbb{P}(\text{game ends on the $10$th roll}) = 1 - \mathbb{P}(\text{game ends before the $10$th roll}),$$ whereas in the second game, the remaining mass is split up and weighted by $n=10$, $n=11$, and so on.

\begin{align*} 10 \cdot \mathbb{P}(\text{first game ends on $10$th roll}) &= \sum_{n \ge 10} 10 \cdot\mathbb{P}(\text{first appearance of $6$ is on $n$th roll})\\ &< \sum_{n \ge 10} n \cdot \mathbb{P}(\text{first appearance of $6$ is on $n$th roll})\\ \end{align*}

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If we let $p=\frac16$ be likilhood of a successful roll the odds of us rolling $k$ consecutive rolls followed by a $6$ is $$P(k) = (1-p)^{k}p$$ Since it took $k$ failures (each with likelihood $(1-p)$) and one success ($p$).

Let's first consider the infinite case, $$E[K]=p\sum_{k=1}^\infty k\cdot(1-p)^{k-1}=p\sum_{k=1}^\infty k(1-p)^{k-1}$$

Notice however, that when $k = 10$ our summation changes: $$E[\min(K,10)] =p \sum_{k=1}^{9} k\cdot(1-p)^{k-1} + 10\ p\sum_{k=10}^\infty (1-p)^{k-1} < E[K]$$

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