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Someone is sick and extremely bored. To pass the time, he decides to roll a $6$-sided die until he gets $6$ $6$'s in a row. If it takes him $5$ seconds for every roll of the die, how long is he expected to be rolling dice until he gets his $6$ $6$'s?

I believe I have a correct answer to this problem, but I'm more curious if there is a more efficient way of solving it than I did, so I'd appreciate if you could attempt to solve it yourself and compare it with my approach.

My approach to solving this problem was to first find the expected number of die tosses and multiply that result by $5$

Let's call the expected number of tosses $X$. We can say: $X = \frac1{6^6}(6) + \frac5{6^6}(X+6) + \frac5{6^5}(X+5) + \frac5{6^4}(X+4) + \frac5{6^3}(X+3) + \frac5{6^2}(X+2) + \frac5{6}(X+1)$ Solving for X gives us $55981$ expected tosses resulting in a total $279905$ seconds.

shrizzy
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    I think the number of expected rolls is $6^1+6^2+6^3+6^4+6^5+6^6 = \frac{6^7-6}{5}=55986$ which is slightly more than your number and satisfies your equation. I suspect you may have made a slight arithmetic error when you solved it. – Henry Feb 18 '23 at 01:53
  • Related: https://math.stackexchange.com/questions/596785/expected-length-of-generating-a-pattern-throwing-dice – Henry Feb 18 '23 at 02:00
  • I’m probably wrong:The odds of tossing the same result 6 times in a row are $\frac{1}{6^6}$. The odds that the result was a particular one are $\frac 16$. The inverse of that number (279936) represent the probable number of tosses. – WindSoul Feb 18 '23 at 02:31

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The only possible approach is to first find the expected number of die tosses and multiply the result by $5$, so I assume you're looking for different ways to find the expected number of die tosses.

Here's a somewhat well-known, nice solution:

You offer a roulette-style game where players can bet any amount on on any number one through six, and then you roll a die and give a 5 to 1 payout when their number comes up (i.e. this is a fair game).

Your customers are peculiar: they all play the exact same strategy. They bet on $6$ every time, and "let it ride" until they lose. In other words, if they win the first round and get paid $5$, plus their original bet back, they bet $6$ the following round. If they win again, they have $36$ in front of them and they bet it all on the following round, and so on.

Also, each round, exactly one new customer shows up. You decide at the outset that you will shut down the game if and only if some player wins six times in a row.

This is a fair game (and since your losses are bounded, the optional stopping theorem applies) so your expected earnings must be zero.

If the game goes $T$ rounds, you get $T$ initial bets from each customer. Only the last six customers leave with anything. The last customer that arrived leaves with $6,$ the second to last with $36,$ the third to last with $6^3,$ and so on. So the expected value of $T$ must be $6+6^2+6^3+6^4+6^5+6^6,$ which is close to, but slightly off from your answer. (Check your algebra.)

spaceisdarkgreen
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    This is adaptable to the case where they all assume the pattern is going to be non-constant such as $131131$ having an expected $6+6^3+6^6$ rolls - I call it Conway's algorithm. – Henry Feb 18 '23 at 02:14
  • @Henry I was certainly aware of its adaptability but not of its provenance! – spaceisdarkgreen Feb 18 '23 at 02:15