Found the following problem related to "tricky" problems from elementary mathematics:
Mr. White and Mr. Green were playing cards. Mr. White made the suggestion: “Let’s play for money. For each game we play the bet would be 50% of all the money you have in your pocket.”
When Mr. Green hesitated, Mr. White added: "Your skills are greater that mine in this game; usually you win this game more frequently than I do."
Mr. Green agreed. He said he had 32 dollars. The initial bet for the first game was 16 dollars. They played 7 games, four of which were won by the better player, Mr. Green.
How much money did Mr. Green have after these seven games?
What was checked
After research it turns out that Math Stack Exchange platform does not recently contain any notices related to such problem. Because of that I figured out that for the purpose of expansion of the content of this platform I am going to post such question with my attempt (provided below).
Attempt of solution
Given model of game implies:
- increase of size of posession by a factor of $\frac{3}{2}$ in the case of win
- decrease of size of posession by a factor of $\frac{1}{2}$ in the case of defeat
Since there are $4$ wins and $3$ defeats and since we are only interested in seeing final outcome from given session of $7$ games, then using implications from rule of product from combinatoris, the total quantity of money from the session of $7$ completed games in the case of Mr. Green is:
$$ \left(\frac{3}{2}\right)^{4} \cdot \left(\frac{1}{2}\right)^{3} \cdot 32 = \frac{81}{16} \cdot \frac{1}{8} \cdot 32 = \frac{81}{4} = 20 \frac{1}{4} [\text{dollars}] $$
Think that this solves the problem.
Aim
If anyone would post a feedback regarding given attempt to this problem, then I think that this would target for enriching a content of Math Stack Exchange platform. Simple as that as this is neither homework nor academic coursework nor assessed problem.
Your answer is totally correct, don't listen to the comments, and here's why: multiplication is commutative! You've figured out that losing is multiplying by $\frac 1 2$ and winning is $\frac 3 2$, thus, since multiplication is commutative, order doesn't matter! Here's an example: $((((((32×\frac 1 2)×\frac 3 2)×\frac 3 2)×\frac 1 2)×\frac 3 2)×\frac 1 2)×\frac 3 2$ $=32×(\frac 3 2)^4×(\frac 1 2)^3=20+\frac 1 4$
QED