Three to Gamble
A game of chance and some skill ensues between 3 professional gamblers. In the first game, gambler A loses to gambler B and C an amount of money equaling the amount each possessed before the game. In game 2, gambler B loses, to gamblers A and C, an amount of money equaling the amount each had before game 2. In the 3rd game, gamblers A and B gain from gambler C as much money as they had before game 3. After game 3, each man checks his money and each has $24.
How much money did each gambler have before the first game?
(for this one, I have no solution ... so, I am counting on you guys to cuss, discuss and sort it out... and for those purists in the crowd, "cuss" is informal for "curse" ... heh heh)
How much money did each gambler have before the first game?
(for this one, I have no solution ... so, I am counting on you guys to cuss, discuss and sort it out... and for those purists in the crowd, "cuss" is informal for "curse" ... heh heh)
posted by Zaux at
10:05 PM
8 Comments:
new to this, but I believe;
A=39
B=21
c=12
Regards, Curtis
Hi Curtis ... most of the time, the puzzle poster has the solution. However, for this partcular one, I do not. After I have breakfast, I will work through it and see if we agree.
Yes Curtis ... I agree with your solution
3 gamblers A,B & C
A has $a
B has $b
C has $c
game 1:
A loses to B -> b
A loses to C -> c
A has (a-b-c)
B has 2b
C has 2c
Game 2:
B loses to A -> (a-b-c)
B loses to C -> 2c
A has (a-b-c)+(a-b-c)= 2(a-b-c)
B has 2b-(a-b-c)-2c= 3b-a-c
C has 2c+2c= 4c
Game 3:
C loses to A -> 2(a-b-c)
C loses to B -> (3b-a-c)
A has 2(a-b-c)+2(a-b-c)= 4(a-b-c)
B has (3b-a-c)+(3b-a-c)= 2(3b-a-c)
C has 4c-2a+2b+2c-3b+a+c= 7c-a-b
after game 3, all have $24
solve 3 equation with 3 unknowns:
4(a-b-c)=24
2(3b-a-c)=24
7c-a-b=24
solving we get:
a=39 A had $39
b=21 B had $21
c=12 c had $12
Three to gamble
1st round
A-B-C, 2B, 2C
2nd round
2*(A-B-C), 2B-(A-B-C)-2C, 4C
2A-2B-2C, -A+3B-C, 4C
3rd round
4A-4B-4C, -2A+6B-2C, 4C-(2A-2B-2C)-( -A+3B-C)
4A-4B-4C, -2A+6B-2C, -A-B+7C
*1) 4A-4B-4C=24
*2) -2A+6B-2C=24
*3) -A-B+7C=24
From *3
*4) A=-B+7C-24
sub into *2
-2*(-B+7C-24)+6B-2C=24
8B-16C+48=24
8B=16C-24
B=2C-3
sub into *4
A=-(2C-3)+7C-24=5C-21
sub into *1
4*(5C-21)-4*(2C-3)-4C=24
20C-84-8C+12-4C=24
8C=84+24-12=96
C=12
B=2C-3=24-3=21
A=5C-21=60-21=39
Answer:
A=39
B=21
C=12
Cam
Lazy method:
Work backwards
End
A=24
B=24
C=24
Previous Round (3rd)
A=24*0.5=12
B=24*0.5=12
*C=24+0.5*24+0.5*24=48
Previous Round (2nd)
A=12*0.5=6
*B=12+12*0.5+48*0.5=42
C=48*0.5=24
Previous Round (1st)
A=12*0.5=6
*B=12+12*0.5+48*0.5=42
C=48*0.5=24
Beginning
*A=6+42*0.5+24*0.5=39
B=42*0.5=21
C=24*0.5=12
Answer:
A=39
B=21
C=12
Cam
Hi Cam ... nice backwards solution
Bit sloppy with the copy paste...
Lazy method corrected.
Lazy method:
Work backwards
End
A=24
B=24
C=24
Previous Round (2nd)
A=24*0.5=12
B=24*0.5=12
*C=24+0.5*24+0.5*24=48
Previous Round (1st)
A=12*0.5=6
*B=12+12*0.5+48*0.5=42
C=48*0.5=24
Previous Round (Beginning)
*A=6+42*0.5+24*0.5=39
B=42*0.5=21
C=24*0.5=12
Answer:
A=39
B=21
C=12
Cam
OK Cam ... that was forwards and backwards ... do you have a sideways solution? ... heh heh
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