Question #3a8c6

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Question #3a8c6

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Answer 1 · 2016-01-09T18:40:15

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

Explanation:

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

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

$P (D_{1}) = 2 (\frac{13}{52}) (\frac{39}{51}) = \frac{13}{34} \approx 0.382352941$ $P(D_1)=2(13/52)(39/51)=13/34 ~~0.382352941$

[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.

$P (D_{2}) = (1 - \frac{13}{34}) (\frac{13}{34}) = 0.23615917$ $P(D_2)=(1-13/34)(13/34)=0.23615917$

Continuing for $D_{3} through D_{6}$ $D_3" through "D_6$ ...

$P (D_{3}) = {(1 - \frac{13}{34})}^{2} (\frac{13}{34})$ $P(D_3)=(1-13/34)^2(13/34)$

$P (D_{4}) = {(1 - \frac{13}{34})}^{3} (\frac{13}{34})$ $P(D_4)=(1-13/34)^3(13/34)$

$P (D_{5}) = {(1 - \frac{13}{34})}^{4} (\frac{13}{34})$ $P(D_5)=(1-13/34)^4(13/34)$

$P (D_{6}) = {(1 - \frac{13}{34})}^{5} (\frac{13}{34})$ $P(D_6)=(1-13/34)^5(13/34)$

Finally, since $D_{7}$ $D_7$ MUST be a win for one of the players, this concludes the game and the probability of getting to $D_{7}$ $D_7$ is the complement of the sum of the probabilities for $D_{1} through D_{6}$ $D_1 " through " D_6$

$P (D_{7}) = 1 - 6 \sum 1 P (D_{n})$ $P(D_7)= 1 - sum_1^6P(D_n)$

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

Hope that helped!

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