Red and Blacks

Logic Level 2

A betting game is played by Albert and Bernard.

Albert will mix 2 red cards and 1 black card and open them, one by one, to Bernard.

Bernard will bet a sum of money at the start of the game so when each card is revealed, the amount of money changes.

If it is a red card, the amount of money will increase by 50%
If it is a black card, the amount of money will decrease by 50%

After all the 3 cards has been revealed, the game will end.

However, after the first game, Albert said this game was unfair, because Bernard will gain 50% more money at the end. (100% - 50%)

So the game is played again after the 1st time, but with 1 red card and 2 black cards. Bernard betted the same amount of money as in the first game.

Bernard said this game is now fair, because he would gain 50% at first, then lose 50% later.

At the end of both games, did Bernard gain or lose any money?!

Yes, he lost 50% of his betted money No, he had the same amount of money as compared to before the 2 games Yes, he gained 50% of his betted money Yes, he lost 75% of his betted money It depends on the draw of cards Yes, he gained 75% of his betted money Yes, he gained 25% of his betted money Yes, he lost 25% of his betted money

1 solution

Winston Choo
Sep 8, 2018

For every black card, the betted amount would be $\frac{1}{2}$ $x$

For every red card, the better amount would be $\frac{3}{2}$ $x$

For the first game, the amount of betted money in the end is $\frac{3}{2}$ ^2 * $\frac{1}{2}$ * $x$ = $\frac{9}{8}$ $x$

For the second game, the amount of better money in the end is $\frac{1}{2}$ ^2 * $\frac{3}{2}$ * $x$ = $\frac{3}{8}$ $x$

So Bernard betted a total of 2 $x$ and received only $\frac{9}{8}$ $x$ + $\frac{3}{8}$ $x$ = $\frac{3}{2}$ $x$ back.

So he lost 25% of his betted money.

Red and Blacks

1 solution

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