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?

Question

Xander Guzman

Mathematics

31 May, 19:59

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?

+4

Answers (1)

Know the Answer?

Lincoln · Answer 1

If we exclude face cards, there are 5 even numbers (2,4,6,8,10) and 5 odd numbers (1,3,5,7,9). Face cards are excluded because in this scenario they are not counted as odd nor as even number and if Graham picks face card he doesn't gain or lose his money.

Therefore there are 50% chance that he will get money (5 out of 10 cards will double his money because out of 10 numbers 5 are even) and 50% chance that he will lose money. If 50% chance to win is considered fair bet than this one is fair.