Question

Graham is in a casino and decides that he will bet on only even numbers in a card game. An ace is assumed to be the number 1. The game is played with a regular deck of 52 cards. The house rules are that if the card picked is an even number, Graham’s money doubles. If it is an odd number, Graham loses his money. Also, if the card picked is a face card (jack, queen, or king), the dealer replaces the card and picks again. Is this bet a fair one? Why?

Answer 1

Winning hands are {2,4,6,8,10}
Losing hands are {1,3,5,7,9}
These are all the possibilities for winning or losing.

A win can come from 4*number of unique even numbers, since there are 4 suits. So, the number of winning cards is 20.

Similarly, there are 4*5 ways to lose.

The total number of cards that matter is 20+20=40, not 52, because the face cards don't affect the outcome.

It's like you are playing with a deck of 40 cards.

It is obvious now that there is a 1/2 chance of winning or losing.

The expected amount a player will win, where X is 1 if a win 0 if a loss and M is the amount of money he must pay to play, is \(E[X]\) Since he get $M in a win and loses -$M in a loss:
$$
E[X]=P[X=1]\times2M+P[X=0]\times(-M)\\
=\cfrac{1}{2}\times M-\cfrac{1}{2}M\\
=\cfrac{M}{2}-\cfrac{M}{2}\\
=0
$$
Therefore, the game is fair because neither the player nor the house is expected come out ahead.

Make sense?

Answer 2

yes! thanks

Answer 3

In my 1st equation, that should have been (see item in RED for correction):
$$
E[X]=P[X=1]\times\color{red}{M}+P[X=0]\times(-M)\\
=\cfrac{1}{2}\times M-\cfrac{1}{2}M\\
=\cfrac{M}{2}-\cfrac{M}{2}\\
=0
$$
The rest of the results of the same.

Good luck!