a prisoner sentenced to death is offered a chance to live by the emperor.

Question

He is asked to remove three coins from a pouch that contains 1 gold, 4 silver, and 3 copper coins. A gold is worth 2 silver coins. A silver is worth 4 coppers. If the total value of the coins picked is at least 9 coppers, then he is executed. What are the chances of him living?

Answer 1

So here we have 1 gold = 2 silvers = 8 coppers.

The gold coin plus any other coin will make his total at least 9 coppers, so he must not draw the gold.

Similarly, two silver coins plus any other coin will make his total at least 9 coppers, so he must not draw two silvers.

Therefore, the only safe combinations are copper-copper-copper and copper-copper-silver.

 

This combination problem does NOT involve replacement; that is, there are eight coins in the pouch on the first draw, seven on the second, and six on the third.  So the probability of drawing any one coin is a fraction whose numerator is the number of particular coins left, and whose denominator is the total number of coins left.  For example, the probability of drawing the gold coin on the first draw is 1/8.

 

So his chance of drawing three coppers is (3/8)*(2/7)*(1/6)=1/56,

his chance of drawing silver-copper-copper is (4/8)*(3/7)*(2/6)=1/14,

his chance of drawing copper-silver-copper is (3/8)*(4/7)*(2/6)=1/14,

and his chance of drawing copper-copper-silver is (3/8)*(2/7)*(4/6)=1/14.

 

Adding these together gives his chance of living:  1/56 + 1/14 + 1/14 + 1/14 = 13/56--not great!

 

Notice also, that the last three combinations all have the same probability.  It does not matter whether he gets the silver coin first, second, or third.  So we can actually calculate all three of these at once.  There are three ways to draw 1 silver and 2 coppers so the probability of that combination is 3*(3/8)*(2/7)*(1/6) = 3/14.

 

Answer 2

There are only two outcomes that avoid execution: removing three copper or removing two coppers and one silver. There are 8 coins, so the total number of removals of three coins, without replacement, is 8x7x6 = 336. There are 3x2x1 = 6 ways to pick three coppers and there are 3x2x4x3 = 72 ways to pick 2 coppers and one silver in any order. Thus there are 78 successes / 336 equally possible outcomes = 13/56.