I'm having trouble solving this difficult probability question with my high school understanding of the math involved. I'm wondering if anyone else can solve it and, more importantly, guide me through the math of the question?
Two people are rolling 20-sided dice to decide if an action will be taken or not.
The first roll is 50/50. If the die rolls more than 10, all future rolls get a +3 bonus.
The second roll is also 50/50. If the die roll, plus any bonus it might get from the first roll, is more than 10, the second person is considered "accepting" to the action.
The third roll is more complex. Anything less than 5 is a failure to convince the second player to perform the action. Anything more than 5 is a success to convince the second player ONLY IF the second player is "accepting". And any roll over 17 is an outright success, regardless of the second player's status.
The third roll can be repeated any number of times. But each successive roll gains a -2 penalty that increases with each roll. Thus, the second try would have a -2 penalty, the third -4, the fourth -6, and so on.
Given these parameters, what is the chance the action will be taken? And how can that chance be mathematically deduced?
I assume that by a roll "over 17" you mean a roll of at least 17, that by "less than 5" you mean 4 or less, and that by "more than 5$ you mean "5 or more." (The two statements you made about the third roll, regarding a roll of 5, will otherwise leave the result of rolling exactly 5 undefined.)
In order to convince the second player to do the action, one of the following must happen (roll numbers are in curly braces, as in $\{>10\}$): $$ \begin{array}{cll} \{\leq10\},\{\leq10\},\{\geq17\} & P = 0.5 \cdot 0.5 \cdot 0.2 &= 0.05 \\ \{\leq10\},\{\leq10\},\{<\},\{\geq19\} & P = 0.5 \cdot 0.5 \cdot 0.8\cdot 0.1 &= 0.02 \\ \{\leq10\},\{>10\},\{\geq 5\} & P = 0.5 \cdot 0.5 \cdot 0.8 &= 0.20 \\ \{\leq10\},\{>10\},\{< 5\}, \{\geq 7\}& P = 0.5 \cdot 0.5 \cdot 0.2 \cdot 0.7&= 0.035 \\ \{\leq10\},\{>10\},\{< 5\},\{< 7\}, \{\geq 9\}& P = 0.5 \cdot 0.5 \cdot 0.2 \cdot 0.3 \cdot 0.6&= 0.009 \\ \{\leq10\},\{>10\},\{< 5\},\{< 7\}, \{< 9\}, \{\geq 11\}& P = 0.25 \cdot 0.06 \cdot 0.4 \cdot 0.5&= 0.003 \\ \{\leq10\},\{>10\},\{< 5\},\{< 7\}, \{< 9\}, \{< 11\}, \{\geq 13\}& P = 0.25 \cdot 0.06 \cdot 0.4 \cdot 0.5\cdot 0.4 &= 0.0012 \\ \{\leq10\},\{>10\},\{< 5\},\cdots, \{< 11\}, \{< 13\}, \{\geq 15\}& P = 0.25 \cdot 0.06 \cdot 0.02 \cdot 0.06 \cdot 0.3 &= 0.00054 \\ \{\leq10\},\{>10\},\{< 5\},\cdots, \{< 13\}, \{< 15\}, \{\geq 17\}& P = 0.015 \cdot 0.0012 \cdot 0.06 \cdot 0.07 \cdot 0.2 &= 0.000252 \\ \{\leq10\},\{>10\},\{< 5\},\cdots, \{< 15\}, \{< 17\}, \{\geq 19\}& P = 0.015 \cdot 0.0012 \cdot 0.0042 \cdot 0.8 \cdot 0.1 &= 0.0001008 \\ \\ \{>10\},\{\leq7\},\{\geq14\} & P = 0.5 \cdot 0.35 \cdot 0.35 &= 0.06125 \\ \{>10\},\{\leq7\},\{<14\},\{\geq16\} & P = 0.5 \cdot 0.35 \cdot 0.65\cdot 0.25 &= 0.0284375 \\ \{>10\},\{>7\},\{<14\},\{<16\},\{\geq18\} & P =0.5 \cdot 0.35 \cdot 0.65\cdot 0.75\cdot 0.15 &= 0.0127969 \\ \{>10\},\{>7\},\{<14\},\{<16\},\{<18\},\{\geq20\} & P = 0..5 \cdot 0.35 \cdot 0.65\cdot 0.75\cdot 0.85\cdot 0.05 &= 0.0036258 \\ \{>10\},\{>7\},\{\geq 2\} & P = 0.5 \cdot 0.35 \cdot 0.95 &= 0.16625 \\ \{>10\},\{>7\},\{< 2\}, \{\geq 4\}& P = 0.5 \cdot 0.35 \cdot 0.05\cdot 0.85 &= 0.0074375 \\ \{>10\},\{>7\},\{< 2\},\{< 4\}, \{\geq 6\}& P = 0.5 \cdot 0.35 \cdot 0.05\cdot 0.15\cdot 0.75 &= 0.0009844 \\ \{>10\},\{>7\},\{< 2\},\{< 4\}, \{< 6\}, \{\geq 8\}& P = 0.5 \cdot 0.35 \cdot 0.05\cdot 0.15\cdot 0.25\cdot 0.65&= 0.0002133 \\ \{>10\},\{>7\},\{< 2\},\{< 4\}, \{< 6\}, \{< 8\}, \{\geq 10\}& P = 0.175 \cdot 0.0075\cdot 0.25\cdot 0.35\cdot 0.55 &= 0.0000632 \\ \{>10\},\{>7\},\{< 2\},\cdots, \{< 8\}, \{< 10\}, \{\geq 12\}& P = 0.175 \cdot 0.0075\cdot 0.25\cdot 0.35\cdot 0.45\cdot 0.45 &= 0.0000233 \\ \{>10\},\{>7\},\{< 2\},\cdots, \{< 10\}, \{< 12\}, \{\geq 14\}& P = 0.175 \cdot 0.0075\cdot 0.0875\cdot 0.45\cdot 0.55\cdot 0.35 &= 0.0000099 \\ \{>10\},\{>7\},\{< 2\},\cdots, \{< 12\}, \{< 14\}, \{\geq 16\}& P = 0.175 \cdot 0.0075\cdot 0.0875\cdot 0.2475\cdot 0.65 \cdot 0.25 &= 0.0000046 \\ \{>10\},\{>7\},\{< 2\},\cdots, \{< 14\}, \{< 16\}, \{\geq 18\}& P = 0.175 \cdot 0.0075\cdot 0.0875\cdot 0.2475\cdot 0.65 \cdot 0.75 \cdot 0.15 &= 0.0000021 \\ \{>10\},\{>7\},\{< 2\},\cdots, \{< 16\}, \{< 18\}, \{\geq 20\}& P = 0.175 \cdot 0.0075\cdot 0.0875\cdot 0.2475\cdot 0.4875 \cdot 0.85 \cdot 0.05 &= 0.0000008 \end{array} $$ The total probability is $0.600192$. In the interest of saving time, you might consider just rolling at succeeding if the number is $9$ or greater.