Wins and Losses

A person starting with \$256 makes eight bets and wins exactly four times. The wins and losses occur in random order. If each wager is for half the money she has at the time of the bet, how much money will she win or lose?
Source: NCTM Mathematics Teacher, October 2006

Solution
Suppose the eight bets result in the following arbitrary sequence of wins and losses
1\quad 256/2=128\quad\:\:\textrm{win}\quad 256+128=384
2\quad 384/2=192\quad\:\:\textrm{loss}\quad 384-192=192
3\quad 192/2=96\quad\:\:\:\:\textrm{win}\quad 192+96=288
4\quad 288/2=144\quad\:\:\textrm{loss}\quad 288-144=144
5\quad 144/2=72\quad\:\:\:\:\textrm{win}\quad 144+72=216
6\quad 216/2=108\quad\:\:\textrm{loss}\quad 216-108=108
7\quad 108/2=54\quad\:\:\:\:\textrm{loss}\quad 108-54=54
8\quad 54/2=27\quad\:\:\:\:\:\:\textrm{win}\quad 54+27=81
The gambler loses 256-81=\$175.
What if we loaded all the four wins consecutively in the beginning so that she wins big and loses little money?
1\quad 256/2=128\quad\:\:\textrm{win}\quad 256+128=384
2\quad 384/2=192\quad\:\:\textrm{win}\quad 384+192=576
3\quad 576/2=288\quad\:\:\textrm{win}\quad 576+288=864
4\quad 864/2=432\quad\:\:\textrm{win}\quad 864+432=1296
5\quad 1296/2=648\quad\textrm{loss}\quad 1296-648=648
6\quad 648/2=324\quad\:\:\textrm{loss}\quad 648-324=324
7\quad 324/2=162\quad\:\:\textrm{loss}\quad 324-162=162
8\quad 162/2=81\quad\:\:\:\:\textrm{loss}\quad 162-81=81
Alas! The wins are big and so are the losses. She still loses \$175, but why lose the same amount in the two different scenarios?
Let x = money available before a bet. After a win she will have more money, x+x/2=x(3/2), that is 3/2 as much as before; after a loss she will have less money x-x/2=x(1/2), that is 1/2 as much. We track the ups and downs of money using this multiplicative rule as follows
1\quad x/2\qquad\quad\:\:\,\textrm{win}\:\:\:\:\: x(3/2)
2\quad x(3/4)\quad\:\:\:\:\:\,\textrm{loss}\quad\, x(3/2)(1/2)
3\quad x(3/8)\quad\:\:\:\:\:\,\textrm{win}\quad\, x(3/2)^2(1/2)
4\quad x(9/16)\quad\:\:\:\:\textrm{loss}\quad x(3/2)^2(1/2)^2
5\quad x(9/32)\quad\:\:\:\:\textrm{win}\quad x(3/2)^3(1/2)^2
6\quad x(27/64)\quad\:\:\textrm{loss}\quad x(3/2)^3(1/2)^3
7\quad x(27/128)\quad\textrm{loss}\quad x(3/2)^3(1/2)^4
8\quad x(27/256)\quad\textrm{win}\quad x(3/2)^4(1/2)^4
She starts out with x=256 and ends up with
x(3/2)^4(1/2)^4=x(81/256)
=256(81/256)
=\$81
Though the wins and losses occur in random order, she always loses 256-81=\$175.

Answer: Will lose \$175

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , . Bookmark the permalink.

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