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 October 2nd, 2013, 07:07 PM #1 Newbie   Joined: Oct 2013 Posts: 1 Thanks: 0 Gambling Probability Question A casino offers a new game called �Fair Go� as follows: A player tosses a coin. If the coin is a head, then the player is entitled to select a card from a standard pack of 52 playing cards. If the coin is a tail, he is not entitled to select a card. A round consists of tossing a coin and then selecting a card (if entitled). Cards are not to be replaced in the pack after each round. A player wins a game if he obtains exactly 4 hearts in 6 rounds, otherwise the �house� wins. The game of 6 rounds cost \$3.00 ton play and the player receives a generous \$120 if he wins. Is this a fair game as the casino claims? Any help will be appreciated. Last edited by skipjack; August 7th, 2017 at 07:55 PM.
 October 2nd, 2013, 07:41 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,415 Thanks: 1025 Re: Gambling Probability Question The way I read that, if 1st 3 flips are tails, round is over right then...yes?
 October 2nd, 2013, 10:33 PM #3 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: Gambling Probability Question Hi scottylancaster, and welcome to the forums. I think that the game is grossly unfair. Let's first calculate the odds of winning if each card is replaced, since it is (I think) much easier. The odds of getting a heart on each roll (since the cards are replaced) is: $\displaystyle \frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}$ and the odds of not getting a heart on each roll is: $\displaystyle 1-\frac{1}{8}=\frac{7}{8}$ So the odds of getting exactly 4 hearts is: $\displaystyle \frac{1}{8} \cdot \frac{1}{8} \cdot \frac{1}{8} \cdot \frac{1}{8} \cdot \frac{7}{8} \cdot \frac{7}{8} \cdot \frac{6!}{4! \cdot 2!}$ $\displaystyle =\frac{49}{8^6}\cdot \frac{720}{24 \cdot 2}=\frac{49}{262144}\cdot 15=\frac{735}{262144}\approx 0.0028$ I think that this is correct, and computer simulations give approximately this probability of winning (about 0.00278 or so). The expected value is: $\displaystyle -3(1-0.0028 )+120\cdot 0.0028=-2.6566$, which is really bad. If the cards are not replaced, since 4 hearts need to be accumulated to win, the ratio of hearts reduces faster than the other cards' ratio does. So we expect the winning percentage to drop if the game is played as stated. Indeed, the computer simulations show that without replacement, the winning percentage drops to about 0.002. The expected value is: $\displaystyle -3(1-0.002)+120\cdot 0.002=-2.754$, which is worse of course. Hopefully this will be enough (calculating the replacement statistics, showing that they are very unfair and that they will reduce without replacement). I would not want to attempt the non-replacement calculation but someone else on the forum might. Last edited by skipjack; August 7th, 2017 at 07:59 PM.
 October 3rd, 2013, 03:10 AM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Gambling Probability Question The game would actually be quite fair if you just got to pick 6 cards and needed exactly 4 hearts to win. The odds of that would be about 2.6%, which is about 40:1. The coin tossing (as shown above) puts it very much in favour of the Casino.
 June 22nd, 2017, 02:00 AM #5 Newbie   Joined: Jun 2017 From: Philippines Posts: 1 Thanks: 0 The probability of winning most casino games is between 1%-4%. At the end of the day, the house almost always wins... so I wouldn't play that game if I were you. Last edited by skipjack; August 7th, 2017 at 08:01 PM.

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