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Ann and Ben are playing a game consisting of multiple rounds.
The first person to reach 10 points wins the game.
Each round has only one winner. Assume the rounds are independent,
and the probability of Ann winning each round is 0.6.
a) Calculate the probability that Ann wins the game if the winner of each round earns 2 points.
b) As in (a), but if the winner also won the previous round, they earn an additional 1 point.
c) Calculate the expected number of rounds needed for Ann to win in scenario (a).
d) Calculate the expected number of rounds needed for Ann to win in scenario (b).
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I will provide here the solutions for parts (a) and (c).
Part (a)
First winning configuration for Ann is AAAAA, symbolizing that Ann wins in all 5 rounds, in a row.
The probability for this scenario is, obviously, .
Next winning configurations for Ann are five 6-round BAAAAA, ABAAAA, AABAAA, AAABAA, AAAABA.
They symbolize all 6-letter combinations of letters A and B, where there is only one B
and the last letter is A.
The probability for these five scenarios is .
Next winning configurations for Ann are all = 3*5 = 15 7-letter combinations
of letters "A" and "B", where there are two "B" and 5 "A" in any order, but the last letter is "A".
The probability for these five scenarios is .
Coefficient = 15 is the number of placing 2 letters B in 6 possible positions for them.
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| At this point, the idea of the further move should be clear. |
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Next winning configurations for Ann are all 8-letter words of letters A and B with 5 "A" and 3 "B",
where "A" is in the last position.
They will add the probability for these scenarios = 35*0.6^5*0.4^3.
Here = 35 is the number of placing 3 letters B in 7 possible positions for them.
Next winning configurations for Ann are all 9-letter words of letters A and B with 5 "A" and 4 "B",
where "A" is in the last position.
They will add the probability for these scenarios = 70*0.6^5*0.4^4.
Here = 70 is the number of placing 4 letters B in 8 possible positions for them.
Now the last calculation is to compute the sum of 5 addends
P = 0.6^5 + 5*0.6^5*0.4 + 15*0.6^5*0.4^2 + 35*0.6^5*0.4^3 + 70*0.6^5*0.4^4 =
= 0.6^5*(1 + 5*0.4 + 15*0.4^2 + 35*0.4^3 + 70*0.4^4) = 0.73343232.
At this point, part (a) is solved completely.
ANSWER. The probability in (a) is 0.73343232.
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You may round it P = 0.7334.
Part (a) is complete.
Part (c)
Based on the solution for part (a) above, we can make this table
number of rounds "n" for Ann 5 6 7 8 9
to win 10 points first
Probability for Ann to win 0.6^5 5*0.6^5*0.4 15*0.6^5*0.4^2 35*0.6^5*0.4^3 70*0.6^5*0.4^4
the game in n rounds
Now the expected number of rounds for Ann to win the game is the sum of in-pair products
E = 5*(0.6^5) + 6*(5*0.6^5*0.4) + 7*(15*0.6^5*0.4^2) + 8*(35*0.6^5*0.4^3) + 9*(70*0.6^5*0.4^4) = 5.27586048.
You may round it and get the ANSWER: the expected number for Ann to win the game in scenario (a) is 5.28.
Part (c) is complete.
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If you read to the end, it means that you enriched your mind with another beautiful and instructive
problem from the probability theory.
In your post, I like very much that the problem is written/printed/presented properly and accurately.