math (Probability)

Mar 2020
20
0
Netherlands
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
Sep 2015
2,958
1,673
USA
I would start this by drawing a probability tree and seeing what it reveals. Do you know how to do that?
 

romsek

Math Team
Sep 2015
2,958
1,673
USA
I really don't understand how you never learned how to draw a probability tree.

Clipboard01.jpg

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
 
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Mar 2020
20
0
Netherlands
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
Sep 2015
2,958
1,673
USA
If we start in state SP

\(\displaystyle p[MW] = \dfrac 5 6 \dfrac 1 2 + \dfrac{5}{6} \left(\dfrac{5}{12}\right)\dfrac 1 2 + \dfrac{5}{6} \left(\dfrac{5}{12}\right)^2 \dfrac 1 2 + \dots + \dfrac{5}{12} \left(\dfrac{5}{12}\right)^k + \dots= \\
\dfrac{5}{12}\sum \limits_{k=0}^\infty \left(\dfrac{5}{12}\right)^k = \dfrac{5}{12} \cdot \dfrac{12}{7} = \dfrac 5 7\)
 
Mar 2020
20
0
Netherlands
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 :)