# Winning rate in a hybrid game

So we are playing a game, first we toss 2 coins, if we get:

• 2 consecutive heads (HHH, HHT, THH) when we win,

• 1 head then we lose

• 2 non-consecutive head (HTH) then we advance to the next game

Next game: rolling 2 dices, we win if we get the difference between 2 dices is exactly 1.

I'm trying to calculate the winning rate for this game.

My approch is to use conditional probability: I make A as an event when it's HTH, B as an event when the difference of 2 dices is 1. I have the probability when winning the second games after losing the first game is :

Pr[B|A] = Pr[B union A] / Pr[A]

how do I add it up to the probability when I win by having 2 consecutive heads? Do I add them or multiply them together?

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You add. The probability that you win without rolling any dice is $\frac{3}{8}$. But there is also a chance to win by getting $2$ non-consecutive heads (probability $\frac{1}{8}$) and then rolling a difference of $1$ (probability say $a$). Thus the probability that you win the composite game is $\frac{3}{8}+\frac{1}{8}\cdot a$.

Now it remains to calculate $a$.

Remark: It was not specified what happens if we get TTT. The above solution assumed that we also lose in that case.

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