A Golden Probability

Probability Level 3

You have in front of you two identical chests ( on the exterior ). You are told that one chest contains $100$ silver coins and the other contains $50$ silver and $50$ gold coins. You are then blindfolded and asked to pick a chest at random and draw a coin: Its a silver coin - its yours to keep, so you place it in your pocket. To your surprise you are then given a second chance! If you draw a gold coin you keep the entire contents of the chest; if not, you get the coins you drew. What is the ( what I'm naming ) relative likelihood " $\text{R}$ " you draw a gold coin if you switch chests vs. stay for the next draw?

Calculate:

$\text{R} = \frac{ P ( \text{Gold}| \text{Switch} ) }{P ( \text{Gold}| \text{Stay} ) }$

$\text{R} = \frac{a}{b}$ , where $a$ and $b$ are positive coprime integers.

Enter $a + b$

The answer is 149.

1 solution

Eric Roberts
Apr 18, 2021

Calculate $P ( \text{Gold}|\text{Stay} )$

$\quad$

On the first draw the probability that you chose the the chest containing $50/50$ Gold & Silver mix is:

$\displaystyle \begin{aligned} P( \text{G\&S}|\text{Silver} ) &= \frac{ P( \text{G\&S mix} ) \cdot P( \text{Silver} ) }{ P( \text{G\&S mix} ) \cdot P( \text{Silver} ) + P( \text{ All Silver } ) \cdot P ( \text{Silver} ) } \\ \quad \\ \displaystyle &= \frac{ \frac{1}{2} \cdot \frac{1}{2} }{ \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot 1 } \\ \quad \\ \displaystyle &=\frac{1}{3} \end{aligned}$

Intuitively you can arrive at this by only considering the coins ( not the chest ) and noting that mixed chest has $\frac{50}{100+50} = \frac{1}{3}$ of all Silver coins.

Thus;

$P ( \text{Gold}|\text{Stay} ) = P( \text{G\&S}|\text{Silver} ) \cdot P( \text{Gold in Mix} ) = \left( \frac{1}{3} \right)\cdot \left( \frac{50}{99} \right)$

Calculate $P ( \text{Gold}|\text{Switch} )$

$\quad$

Using the compliment the probability you initially drew a coin from the chest containing $100$ Silver coins is:

$P( \text{All Silver}|\text{Silver} ) = \frac{2}{3}$

So the probability of drawing a gold after the switch ( noting that $P( \text{Gold in Mix} ) = \frac{50}{100}$ since the first draw was from the other chest ) :

$P ( \text{Gold}|\text{Switch} ) = P( \text{All Silver}|\text{Silver} )\cdot P( \text{Gold in Mix} ) = \left( \frac{2}{3} \right) \cdot \left( \frac{50}{100} \right)$

Finally;

$\text{ R } = \frac{ P ( \text{Gold}|\text{Switch} ) }{ P ( \text{Gold}|\text{Stay} ) } = \frac{99}{50}$

and

$a + b = 99 + 50 = 149$

$\quad$

P.S. If I have this problem wrong please speak up or report.

The only wrong thing is you should have asked for the probability, not the odds. Probability is a/b, odds is n:k (n-to-k) =n/(n+k). For example, 1:1 odds=1/(1+1)=1/2

Andrei Zonga - 1 month, 3 weeks ago

The thing is... what I'm asking isn't the probability either ( it cant be- its greater than 1 ). Perhaps I have misinterpreted the definition of "Odds". If we roll a 6 sided dice the Odds of rolling a 6 are 1:5

$\text{Odds} = \frac{ P( \text{ rolling 6} )}{ P( \text{ not rolling 6} ) } = \frac{ \frac{1}{6} }{ \frac{5}{6} } = \frac{1}{5} = 1:5$

So why in this case are the Odds not $\frac{99}{50}:1$

In the meantime I've changed the wording to asking for the relative likelihood, and provide that definition. Maybe that bandages the issue?

Eric Roberts - 1 month, 3 weeks ago

A Golden Probability

The answer is 149.

1 solution

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