Question #3a8c6

1 Answer
BeeFree
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" through "D_6# ...

#P(D_3)=(1-13/34)^2(13/34)#

#P(D_4)=(1-13/34)^3(13/34)#

#P(D_5)=(1-13/34)^4(13/34)#

#P(D_6)=(1-13/34)^5(13/34)#

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 " through " D_6#

#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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