I have done some simple flipping coin problems but I am not sure how to solve this one. This is from a review sheet for my final.

In each round of a game, three cards are dealt from the standard deck of $52$ cards. If at least two cards are of the same suit then you get $2$ dollars, otherwise you lose $3$ dollars.

(a) What is the probability of winning one round of this game?

(b) What is the probability of losing one round?

(c) What is the expected gain (or loss if its negative) per round?

  • $\begingroup$ b is trivial, since you can only win or lose. It seemed easier to me to calculate the probability of getting 3 different suited cards (therefore losing) $\endgroup$ – Dleep May 18 '15 at 0:36

It is easier to answer part (b) first. To lose, we need the cards drawn to be distinct suits. There are $4$ suits, so we choose one of $4$ suits to leave out. There are $3!$ ways for the chosen suits to come up, and each suit has $13$ cards. So there are $$ 4\times 6\times 13^3 $$ ways to lose, out of $52\times51\times50$ possible ways to draw the cards. Thus the probability of losing is $$ \frac{4\times 6\times 13^3}{52\times51\times50}=\frac{169}{425} $$ Now that we know part (b), it is easy to find part (a). Namely, the probability of winning is just $1$ minus the probability of losing. So the probability of winning is $$ \frac{256}{425} $$ For part (c), we want to find the expected gain. We multiply the probability of each event by the value won (or lost) from that event, and then add the results. So this gives $$ E[X]=\frac{256}{425}\times 2-\frac{169}{425}\times 3=\frac{1}{85} $$

  • $\begingroup$ thanks, where did you get the $4$ and the $6$ $\endgroup$ – Csci319 May 18 '15 at 0:42
  • $\begingroup$ We needed 3 suits to come up, out of 4 possible suits. The 4 is just me choosing which suit to leave out. The 6 comes from 3!. Namely, it is the number of ways to order the suits. For instance, if the suits that came up were clubs, spades, and hearts, I could have CSH, CHS, SCH, SHC, HCS, or HSC. $\endgroup$ – TomGrubb May 18 '15 at 0:44

I'll try to make it simple to understand. For (b) assuming you have n cards of different suit in your hand, the probability of getting another one is:

$x = $ Nº of cards on the deck of a different suit than any of yours.

$y = $ Remaining cards on the deck.

$n = $ Amount of cards in your hand.

$$ P(n) = \frac{x}{y} = \frac {13*(4-n)}{52 - n}$$

Since you have to draw 3 cards, it can be calculated as:

$$ P(0)P(1)P(2) = \frac {52}{52} \frac {39}{51} \frac {26}{50} = \frac {169}{425} $$


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