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The game is as played:

To preface you can roll a die at most three times. You roll the die the first time, if you get a six you get \$6 and you roll the die again, if you don't get a six the game stops and you get \$0. You roll the die a second time if you get a six you get \$36 plus the \$6 you got last roll and you roll again, if you don't roll a six you then you only get the \$6 you made the first die roll. If you roll a six on the last roll then you get \$216 plus \$36 from the second roll and \$6 from the first roll. If you don't roll a six on the third roll, you just get whatever you made the previous two rolls. What is the expected return?

The way I tried solving it:

$E(X) = \frac{1}{6}(6) + \frac{5}{6}(0) + \frac{1}{36}(36) + \frac{1}{6}\frac{5}{6}(6) + \frac{1}{216}(216) + \frac{1}{36}\frac{5}{6}(36+6) = \$4.8$

Is this correct?

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    $\begingroup$ Welcome to Cross Validated. Can you please edit your post and write your attempts at solving the problem? If your question is clear and focused on your specific difficulty and you show your effort in solving the problem, it's more likely to get good and helping answers. By the way, take the opportunity to take the Tour, if you haven't done it already. See also some tips on How to Ask, on formatting help and on writing down equations using LaTeX / MathJax. $\endgroup$ – Ertxiem - reinstate Monica Oct 31 '19 at 10:12
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    $\begingroup$ What do you mean by "keep going until you roll another 6, stop if it wasn't 6", which seems contradictory to me? $\endgroup$ – gunes Oct 31 '19 at 10:42
  • $\begingroup$ If you roll [6, 6, 2] and you receive 6+36, then your attempt seems correct. If you only receive 36, then your attempt is wrong. $\endgroup$ – KaPy3141 Oct 31 '19 at 13:51
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    $\begingroup$ Because "dice" is a plural, it sounds like you are rolling an unspecified number of two or more dice each time--but how many? $\endgroup$ – whuber Oct 31 '19 at 21:32
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    $\begingroup$ did it help? I change it from dice to die, my bad $\endgroup$ – Ryan Day Oct 31 '19 at 21:35
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There are 2 ways that you could do this (well actually probably more), but your solution combines these together so you are over counting the results of the first 2 rolls.

Based on your description I assume that if I rolled 6, 6, 1 then my total would be \$42 (\$6 from first roll, \$36 from second roll, \$0 from 3rd roll).

Option 1 is to treat each roll independently and work out the expected winnings from each roll, then sum the expectations. So we expect to win $1 from the first roll and since later rolls will not take that away we leave it at \$1. We have a 1 in 36 chance of winning the second roll (since we need to win the 1st to even have the opportunity to roll the 2nd time), so the expected contribution of the 2nd roll is \$1. Similarly the expected contribution of the 3rd roll is \$1. Summing the 3 together gives \$3 total.

$E(x) = (\frac16 (6) + \frac56 (0)) + (\frac1{36} (36) + \frac{35}{36} (0)) + (\frac{1}{216} (216) + \frac{215}{216} (0)) = 3$

The other approach is to look at total winnings. We can win \$0, \$6, \$42, or \$258. To win \$6 you must roll a 6 on the first roll and anything but a 6 on the 2nd, so that probability is $\frac16 \frac56$. Working out similar results gives:

$E(x) = \frac{5}{6} (0) + \frac16 \frac56 (6) + \frac{1}{36} \frac{5}{6} (6 + 36) + \frac{1}{216} (6 + 36 + 216) = 3$

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