maths question about dominoes

Easy one.  28 dominoes in a set, of which 7 are doubles (double blank up to double six).

The rows of seven bit doesn't matter, all that matters is there are four corners, so the problem is the same as working out the probability of drawing four dominoes at random of which none is a double.

The probability of the first domino drawn being non-double is 21/28.  If that domino is non-double, the probability of the next being non-double is 20/27 (one fewer dominoes to pick from and one fewer non-double to find) etc.

So the probbaility of all 4 being non-double is (21/28) * (20/27) * (19/26) * (18/25) = 0.292

So approx 71% of the times you lay your tiles out you'll get a double on one corner, and 29% of the time you won't.

Well the chances of corner 1 NOT being a double is 21/28.
The chances of corner 1 not being a double and corner 2 not being a double is 21/28 * 20/27
The chances of corner 1 not being a double and corner 2 not being a double and corner 3 not being a double is 21/28 * 20/27 * 19/26
The chances of corner 1 not being a double and corner 2 not being a double and corner 3 not being a double and corner 4 not being a double is 21/28 * 20/27 * 19/26 * 18/25 = .2923

So the chances that at least one of the corners IS a double is 1 - .2923 = .7076

Oops just a bit too late!

Not at all - good to have independent confirmation.

thank you guys, a mate of mine says if we play for 100 shuffles he will beat me. i have the doubles he as non doubles. so i take him on with a pound a shuffle, should i be confident.

Using the binomdist function in excel:

BINOMDIST(50,100,0.7076,TRUE)=.00001

This says that if you take 100 trials where each trial has a .7076 chance of success, the chance of only having 50 or less successes= 1 in 100000

So that is the chance that you won't be in profit after 100 trials.

i thought this fred was gonna be about pizza