Question

Question #3a8c6

Answer 1

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!