# Question 3a8c6

Jan 9, 2016

Find the probability of someone winning for deals 1 through 6 ...

#### Explanation:

Let ${D}_{n}$ for $n = 1 , 2 , 3 , 4 , 5 , 6$ be a winning deal

Also, note that Deal 7 MUST be a win if it gets that far.

P(D_1)=2(13/52)(39/51)=13/34 ~~0.382352941

[Note: multiply by 2 because either player could win]

Now, moving on to the the second deal, this assumes the first deal neither player won.

P(D_2)=(1-13/34)(13/34)=0.23615917

Continuing for ${D}_{3} \text{ through } {D}_{6}$ ...

$P \left({D}_{3}\right) = {\left(1 - \frac{13}{34}\right)}^{2} \left(\frac{13}{34}\right)$

$P \left({D}_{4}\right) = {\left(1 - \frac{13}{34}\right)}^{3} \left(\frac{13}{34}\right)$

$P \left({D}_{5}\right) = {\left(1 - \frac{13}{34}\right)}^{4} \left(\frac{13}{34}\right)$

$P \left({D}_{6}\right) = {\left(1 - \frac{13}{34}\right)}^{5} \left(\frac{13}{34}\right)$

Finally, since ${D}_{7}$ MUST be a win for one of the players, this concludes the game and the probability of getting to ${D}_{7}$ is the complement of the sum of the probabilities for ${D}_{1} \text{ through } {D}_{6}$

$P \left({D}_{7}\right) = 1 - {\sum}_{1}^{6} P \left({D}_{n}\right)$

The table below summarizes the probability distribution and the expected value for D which is equal to approximately 2.5 deals

Hope that helped! 