Division of a stake

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Zelda22

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Solve the problem of Dividing a Pot if A needs two more points and B needs four more points in order to win the pot.

How much of the $100 should go to A? You may give your answer to the nearest dollar.

Can someone explain to me how to solve this? Thanks
 
This is what I tried

for B winning

ABBBB= (1/2)^5 =1/32
BABBB= (1/2)^5 =1/32
BBABB= (1/2)^5 =1/32
BBBAB= (1/2)^5 =1/32
BBBB. = (1/2)^4=1/16

1/32 + 1/32 + 1/32 + 1/32 + 1/16= 3/16

B chances of winning = 3/16
A chances of winning = 1 -3/16= 13/16

A is winning 13/16 * $100= 81.25

Answer A get $81
 
Zelda22 said:
This is what I tried

for B winning

ABBBB= (1/2)^5 =1/32
BABBB= (1/2)^5 =1/32
BBABB= (1/2)^5 =1/32
BBBAB= (1/2)^5 =1/32
BBBB. = (1/2)^4=1/16

1/32 + 1/32 + 1/32 + 1/32 + 1/16= 3/16

B chances of winning = 3/16
A chances of winning = 1 -3/16= 13/16

A is winning 13/16 * $100= 81.25

Answer A get $81
Click to expand...
Can you please explain why you have considered B winning 4 out of 5. Why 5?
 
Zelda22 said:
Answer A get $81
Click to expand...
Did you bother to read the link I provided?
This is one of the most famous stories an the history of mathematics. It remained unsolved for about four+ centuries.
Solved in the eighteenth century in correspondence between Pascal & Fermat.
The solution is found by realizing the answer lies in the number of possible unplayed games.
If five more games were played would there certainly be a winner: [imath]A\text{ or }B~?[/imath]
[imath]\therefore\left(\sum\limits_{k = 0}^3 \dbinom{3+2-1}{k}:\sum\limits_{k = 4}^5 \dbinom{3+2-1}{k}\right)=\dfrac{26}{6}[/imath]
 
pka said:
Did you bother to read the link I provided?
This is one of the most famous stories an the history of mathematics. It remained unsolved for about four+ centuries.
Solved in the eighteenth century in correspondence between Pascal & Fermat.
The solution is found by realizing the answer lies in the number of possible unplayed games.
If five more games were played would there certainly be a winner: [imath]A\text{ or }B~?[/imath]
[imath]\therefore\left(\sum\limits_{k = 0}^3 \dbinom{3+2-1}{k}:\sum\limits_{k = 4}^5 \dbinom{3+2-1}{k}\right)=\dfrac{26}{6}[/imath]
Click to expand...
A has more probability of winning. I submited my answer and it's correct.
 
Harry_the_cat said:
Can you please explain why you have considered B winning 4 out of 5. Why 5?
Click to expand...
I considered all possible sceneries that B could win. (they keep playing unless A gets the two points that it needs to win)


If A wins the first game, B still can win the pot, if win the next 4 games.
If B wins the first game, A the second, B still has a chance if wins the next 3 games,
if B wins the first two games, A wins the third, B still can win the next two games,
if B wins the first 3 games, A wins the fourth, B can still win if wins the last game.
If B wins 4 games in a row, he gets all the points he needs and wins.

Those are the five sceneries of B winning.

After I got B's chance of winning, then I subtracted that from 1 and got A's probability of winning. Then I multiplied by 100. So, A will get $81 of the pot.
 
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I have one question, if A is winning 13/16 * $100= 81.25 then why is the Answer A get $81?
 
Zelda22 said:
This is what I tried

for B winning

ABBBB= (1/2)^5 =1/32
BABBB= (1/2)^5 =1/32
BBABB= (1/2)^5 =1/32
BBBAB= (1/2)^5 =1/32
BBBB. = (1/2)^4=1/16

1/32 + 1/32 + 1/32 + 1/32 + 1/16= 3/16

B chances of winning = 3/16
A chances of winning = 1 -3/16= 13/16

A is winning 13/16 * $100= 81.25

Answer A get $81
Click to expand...
I do not accept your work for one second. Who said that the probability of A getting a point equals the probability of B getting a point = 1/2. Why not p(A gets a point) = 2/3 and p(B gets a point) = 1/3???
 
Steven G said:
I do not accept your work for one second. Who said that the probability of A getting a point equals the probability of B getting a point = 1/2. Why not p(A gets a point) = 2/3 and p(B gets a point) = 1/3???
Click to expand...
Both have an equal probability of 1/2
 
Zelda22 said:
The question asked to round it.
Click to expand...
Seriously, if you want help you need to state the exact problem! Even your comment that the problem asked to round the answer is not complete. If you round $81.25 to the 100th place, you get $81.25, to the nearest 10th place you get $81.3, to the nearest dollar would be $81 and to the nearest $10 would be $80.
 
Steven G said:
Did I misread the problem or maybe you failed to state the probabilities?
Click to expand...
No miss-reading but a failure to note that instructions say to round to the nearest dollar.
Player A gets[imath]\frac{26}{32}[/imath] parts. Player B gets[imath]\frac{6}{32}[/imath] parts
 
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