# Probability : Sampling Without Replacement - Card Hands

6. A hand of 4 cards contains one card of each suit ( I,e., one Heart, one Diamond, One Club and one spade). Simon uses this hand to play a game for which the rules are as follows: (1) When a Heart is drawn, the game stops immediately. (2) If the Diamond is drawn, the Diamond is set aside and not replaced (so that when the game continues there are only 3 cards in the hand). (3) If a black card is drawn, that black card is set aside and another Heart is put into the hand in its place (so the game continues with the same number of cards as before). Simon draws cards from the hand, one at a time, following these rules, until he draws a Heart (which may, of course, happen on the first draw, or on any subsequent draw). If it is known that Simon drew at least 3 cards while playing the game, what is the probability that the Diamond was still in the hand when the game stopped? Hint: Draw a probability tree.

## Solution This solution is **FREE** courtesy of BrainMass!

2.6. A hand of 4 cards contains one card of each suit ( I,e., one Heart, one Diamond,

One Club and one spade). Simon uses this hand to play a game for which the rules are as follows: (1) When a Heart is drawn, the game stops immediately. (2) If the Diamond is drawn, the Diamond is set aside and not replaced (so that when the game continues there are only 3 cards in the hand). (3) If a black card is drawn, that black card is set aside and another Heart is put into the hand in its place (so the game continues with the same number of cards as before). Simon draws cards from the hand, one at a time, following these rules, until he draws a Heart (which may, of course, happen on the first draw, or on any subsequent draw). If it is known that Simon drew at least 3 cards while playing the game, what is the probability that the Diamond was still in the hand when the game stopped? Hint: Draw a probability tree

On the first draw, Simon can draw a heart, diamond, spade, or club. We know he didn't draw a heart, because the game would have stopped there, so he drew either a diamond or a black card.

1) Diamond: There is a 1/3 chance he drew a diamond, in which case there was no diamond in the hand when the game stopped.

2)Black: There was a 2/3 chance he drew a black card (we know he didn't draw a heart, and 2 of the remaining 3 choices are black) Now we draw again; the hand now contains one black card, one diamond, and two hearts. We know he doesn't get a heart, so he gets either the diamond or the black card.

2a) Diamond: This will happen half the time, so there is a 1/3 chance (he got a black card first 2/3 of the time, then a diamond ½ of the time) that he got a black card and then a diamond, and there is no diamond remaining in the hand.

2b) Black card: This will happen half the time also; note that, by the same logic as above, he got a black card and then another black card 1/3 of the time. The hand now contains three hearts and a diamond.

2ba) There is a 1 in 4 chance that he now draws the diamond (cumulative 1/12 chance that he got black/black/diamond) and the hand contains no diamonds.

2bb) There is a 3 in 4 chance that he now draws a heard, so there is a diamond in hand when the game ends. (¼ chance he got black/black/heart)

Looking back (and this is easier to see in a probability tree) the combined possibilities are 1/3 + 1/3 + 1/12 + ¼ = 1 (100% of the possibilities) and only the ¼ at the end allows a diamond to remain in the hand when the game stops. Therefore, given the information above, there was still a diamond in hand at the end of the game ¼ of the time.