UNO PROBABILITIES

UNO CONTAINS 104 CARDS FOUR OF WHICH ARE VERY SPECIAL. SEVEN CARDS ARE DEALT TO EACH PLAYER. IN A THREE MAN GAME WHAT ARE THE ODDS THAT ONE PLAYER WILL BE DEALT THREE OF THE SPECIAL CARDS? I TRIED TO FIGURE IT OUT AND CAME UP WITH ABOUT 1 CHANCE IN 3000....HOW WRONG WAS I???

I get about 0.002215 (approximately 1 chance in 451) as the probability of any of the three players getting exactly three of the four special cards on the deal. That's 3 * combin(4,3) * combin(100,4) / combin(104,7).

BTW, I thought an Uno deck had 108 cards.

According the manufacturer, there are 108 cards in a standard UNO deck. There are (combin 108,7) ways to make a seven card hand out of 108 cards (27,883,218,168).

The odds of any one of three players getting any three of the four special cards amongst the seven dealt is (I think, someone with a surer grip on probabilities will need to double check this set up): The ways to make a three card hand out of four possible cards (combin(4,3)) * The ways to make a three card hand out of seven cards in a hand (combin(7,3)), all divided by the number of ways to make a seven card hand out of 108 total cards (combin(108,7)). This is multiplied by three players to get 1 in 66,388,615. Note that this also includes the hands that also have the fourth special card.

Thank you very much for replying. Your approach is probaly more correct than mine. I tried a simple approach as follows:On the first two cars received the chances of receiving a special one are 8/104 Assuming one card is gone the chances on the next two cards are 6/104 and on the last three 6/104. Probability of all three events is the product of the individual probabilities or approximately 300/1000000 or 3/10000 or 1/3000. Quite a difference from 1/66388615. Does anyone know the fallacy in my simplistic approach???

Thank you for replying..Someone else mentioned there were 108 my grandson told me 104. My calculation of 1/3000 was for a particular player to get the three special cards so my odds would be 1/1000 for one anyone of the three to the three special cards.. This is a lot closer to 1/451 then another estimate of 66 million to one. Did you see his post and my response to how I calculated...I suspect your approach which I don't understand is probably better than mine but what about his??

Quote: btily123

... I tried a simple approach as follows:On the first two cars received the chances of receiving a special one are 8/104 Assuming one card is gone the chances on the next two cards are 6/104 and on the last three 6/104. Probability of all three events is the product of the individual probabilities or approximately 300/1000000 or 3/10000 or 1/3000...

Your answer IS a good approximation for the case of getting one special card in the first two cards, one special card in the second two cards, and one special card in the last three cards. Let's call this particular distribution "111", where the each number is the number of special cards in the first two cards, second two cards, and last three cards, respectively. Although, 111 is the most probable way of getting three special cards, there are seven other ways: 003, 012, 021, 102, 120, 201, and 210. Here are the probabilities for each (assuming only 104 cards in a deck):

003 0.00002
012 0.00013
021 0.00006
102 0.00013
111 0.00025
120 0.00004
201 0.00006
210 0.00004

The total is about 0.00074. Multiply this by three players to get about 0.002215 .