Rare Snapoker Hands
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
The card game Snapoker is a very complex game for four players, requiring the use of two decks of cards with all odd spades, even hearts, prime diamonds (including aces) and non-prime clubs (including aces) removed. Note that jacks are 11, queens are 12, and kings are 13. Two jokers are added to the deck and are each worth 15. The cards are dealt among the four players. A hand starts when the dealer shouts, "Let's Snapoker!" Following a frenetic minute of randomly throwing all cards onto the table, at the dealer's call of, "Heeeeeeere, piggy, piggy, piggy!" players are required to grab as many cards as they can. Once all cards have been grabbed, the total value of the cards in each player's hand is added up. Points (4, 3, 2 and 1) are awarded to players on each hand, with the highest card total receiving maximum points. In the event of a tie, the player who can drink a glass of milk through their nose the quickest will receive the greater points. A running score is kept throughout the game. The game is over when someone reaches a total of 518 points or when a fight breaks out, which is a common occurrence.
The other day I was playing Snapoker with three of the quickest-handed people I have ever known. In any hand I was lucky if I was able to grab three cards, while they grabbed the rest. On one hand I was able to grab four cards and, to my surprise, they were all the same number and same colour. What were the four cards I grabbed and what is the chance of obtaining such a four-card hand in Snapoker?
My hand could have been either four black 2's or four red 9's. All other numbers in a Snapoker deck have two black and two red cards, except for the two aces, which are both red, and the two jokers.
Now for the probability. In the case of black 2's, the odds of the first card being a black 2 would be 4/52. Given that, the odds of drawing another 2 would be 3/51 (then 2/50, then 1/49). Thus the probability of obtaining four black 2's is (4*3*2*1)/(52*51*50*49) which equates to a chance of 1 in 270,725. This probability then needs to be multiplied by 2 as four red 9's are also a possibility. Therefore, the probability of grabbing a four-card hand with the same number and colour is 2 in 270,725.
Feb 05, 2003
|Someday I an really going to try to play this game, having answered all these teasers about it. Nice teasers snaps.:-D|
Mar 03, 2003
|The odds of 52x51x50x49 would apply if the cards had to appear in a particular order. They don't. So in the case of black 2's, the odds of the first card being a black 2 would be 4/52. Given that, the odds of drawing another 2 would be 3/51. (Then 2/50, then 1/49.) This turns the odds into (4*3*2*1)/(52*51*50*49). Multiply this result by 2 again as red 9s are also a possibility.|
Mar 03, 2003
|Quax, you are talking about the order of drawing the cards (first card, second card...). There is no order, so your explanation is incorrect. You have four cards, which, being Snapoker, were all grabbed at once. All up there are 52*51*50*49 possible combinations of four-card hands. Only two of those hands have the same numbered and coloured cards. Hence the original answer is correct.|
May 02, 2003
|Snaps, Quax is correct. What you have done is said that there are 52x51x50x49 possible hands of 4 cards. This includes all permutations of each hand. But then you say there are only 2 possible hands that meet your criteria. At this point you are ignoring the permutations within the hand. So you are comparing apples and oranges and the math is incorrect.|
To simplify things, let's pretend there are 3 cards (K,Q,J) and you draw 2. There are 3x2 permutations (KQ, KJ, QK, QJ, JK, JQ) but only 3 different hands (KQ, KJ, QJ). You want to calculate the probability of having a KQ hand. You either say that there are 6 permutations of hands possible and 2 permutations of the desired draw (KQ and QK): probability of 2/6 or 1/3. OR you say there are 3 possible 2-card hands and only 1 hand meets the requirements: so probability 1/3. But you CANNOT say there are 6 permutations of hands and only one hand meets the requirements: probability of 1/6.
May 03, 2003
|Oh yeah, oops. Sorry, Quax was right. Thanks for clearing that up Gizzer. Funny that no editors picked up this error when I originally submitted it. I wonder if it is possible to change the answer? Have to ask Jake about that.|
Oct 04, 2010
|This problem is very unclear.|
First, 1 is neither prime nor composite, so it should never be removed.
Second, you need to clearly state you are adding two jokers into the combined deck. Otherwise, one could interpret your problem as adding two jokers into each deck, with a total of four jokers in the snapoker deck.
This gives us some number, N, for the total number of cards in the snapoker deck. The answer is just 2 / (N choose 4). N is 52 if you count A as prime and non-prime (hence removing 27 cards from a 52 card deck, and then adding in 2 jokers at the end to the combined deck: (52-27) * 2 + 2 = 52). But the mathematically correct N should be (52 - 25) * 2 + 2 = 56 or (54 - 25) * 2 = 58 depending on whether 2 jokers are added into the combined decks, or into each deck.
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