Some probabilities in card game

1. Sep 19, 2009

bigwuying

$$\sum$$The question is:
You are involove in a game where two card are dealt consectively. Supposethat the dealer pays you $3of the second card dealt is a club, regardless of the first card and that you pay him$1 if the second card in not a club and the first card is not an ace. (othewise, no money changes hands.) Use $$\sum$$$$f_{i}$$$$p_{i}$$ to compute the mean value of the money you win per game if you play it many time.

The ans is 5.8cent

I have spent a lot of time on that, but i still don't understand.

2. Sep 19, 2009

mathman

List all different cases and probabilities

Second card a club - p1=1/4

Second card not a club and not an ace 36/52, first card not an ace 47/51
P2=9x47/(13x51)

Second card not a club and an ace 3/52, first card not an ace 48/51
p3=3x16/(52x17)

payoff= 3p1 - (p2+p3)

I got 5.76 cents.

3. Sep 19, 2009

bigwuying

The following is wrong reasoning,but i still can caculate the answers

second card a club- p1=1/4

First card not a club and not an ace 36/52, second card not a club 38/51
p2=(36X38)/(51X52)

First card is a club and not an ace 12/52 , second card not a club 39/51
p3=12x52/(39/51)

mean = 3p1-(p2+p3) =5.77

My course TA told me it was wrong reasoning, frustrate me a lot.

4. Sep 20, 2009

mathman

Off hand I would say your TA is wrong.

5. Sep 20, 2009

Phrak

I get the same answer, .0576(9) taking the first card first, and the second card dependent. Perhaps the TA didn't understand the ordering you used?

6. Sep 21, 2009