# Thumbs Up or Down on my logic

#### hAltonJones

##### New Member
Very rusty on my statistics and will appreciate validation or correction on my thinking.

"Fair" playing field - 88 players (18 in red shirts, 70 in green shirts ) - 22 winners. Probability no red shirt wins is (I hope):

66/88 * 65/87 * 64/86 ... 49/71 = 0.00283, i.e. (66!/48!) / ( 88!/70!) Odds of 353 to 1.

Am I on the right track?

#### rogojel

##### TS Contributor
hi,
there are some assumptions behind your formula, that might be right or wrong. It seems like you have 22 trials and at the end of each trial the winner leaves the game, everybody else stays. Then the probability that the first game will be won by a black shirt would be 70/88 (not 66/88 BTW). The next trial will be run with 87 participants from which there are 18 red shirtd so the probability that a black shirt wins will be 69/87, for the third game it will be 68/86 and so on. The probability of all of this happening will be the product of the individual probabilities 70/88*69/87*...*(70-21)/(88-21)

regards

#### hAltonJones

##### New Member

There are 88 players. There are 22 prizes. 18 of the 88 have on red shirts.

What is the probability that after 22 prizes are randomly selected that none of the prizes went to someone wearing a red shirt?

Sorry for my confusion.

#### BGM

##### TS Contributor
$$\frac {\displaystyle \binom {88 - 18} {22} \binom {18} {0}} {\displaystyle \binom {88} {22}}$$

#### hAltonJones

##### New Member
Thank you BGM. However, some of the symbols (!, /, etc.) didn't seem to be included in your post. Would you please clarify and/or post the actual numeric result?