# math (Probability)

#### annasilva3

Stef and marieke play a gambling game. Marieke throws a coin and wins if she throws 'head'. Stef throws a dice and wins if he rolls '6'. They take turns playing. Marieke starts, then Stef and so on until one of them wins.
1.Calculate the probability that Marieke wins
2. does anything change the results if Stef started

#### romsek

Math Team
I would start this by drawing a probability tree and seeing what it reveals. Do you know how to do that?

#### annasilva3

I don't know. But aren't there two situations, how can i do it

#### annasilva3

I would start this by drawing a probability tree and seeing what it reveals. Do you know how to do that?
I dont know

#### romsek

Math Team
I really don't understand how you never learned how to draw a probability tree.

MP - Marieke plays
MW - Marieke wins
SP - Stef plays
SW - Stef wins

Now figure out the probability of each path through the (infinite) tree that leads to MW and sum them all.

For #2 do the same thing but have start go to SP instead of MP

idontknow

#### annasilva3

I really don't understand how you never learned how to draw a probability tree.

View attachment 11098

MP - Marieke plays
MW - Marieke wins
SP - Stef plays
SW - Stef wins

Now figure out the probability of each path through the (infinite) tree that leads to MW and sum them all.

For #2 do the same thing but have start go to SP instead of MP
Is it right that if Stef begins, the chance that Marieke can win is the same . (= 6/7)

#### romsek

Math Team
Is it right that if Stef begins, the chance that Marieke can win is the same . (= 6/7)
Not quite. You did get the correct probability that Marieke wins though.

#### annasilva3

Not quite. You did get the correct probability that Marieke wins though.
Isn't it just 1/2 * (1/ (1-(5/6)*(1/2)))

#### romsek

Math Team
If we start in state SP

$$\displaystyle p[MW] = \dfrac 5 6 \dfrac 1 2 + \dfrac{5}{6} \left(\dfrac{5}{12}\right)\dfrac 1 2 + \dfrac{5}{6} \left(\dfrac{5}{12}\right)^2 \dfrac 1 2 + \dots + \dfrac{5}{12} \left(\dfrac{5}{12}\right)^k + \dots= \\ \dfrac{5}{12}\sum \limits_{k=0}^\infty \left(\dfrac{5}{12}\right)^k = \dfrac{5}{12} \cdot \dfrac{12}{7} = \dfrac 5 7$$

#### annasilva3

If we start in state SP

$$\displaystyle p[MW] = \dfrac 5 6 \dfrac 1 2 + \dfrac{5}{6} \left(\dfrac{5}{12}\right)\dfrac 1 2 + \dfrac{5}{6} \left(\dfrac{5}{12}\right)^2 \dfrac 1 2 + \dots + \dfrac{5}{12} \left(\dfrac{5}{12}\right)^k + \dots= \\ \dfrac{5}{12}\sum \limits_{k=0}^\infty \left(\dfrac{5}{12}\right)^k = \dfrac{5}{12} \cdot \dfrac{12}{7} = \dfrac 5 7$$
Thank u