Domino Probability 42?

Question

There are 28 total dominoes, 4 players with partners sitting across the table from each other. Each player draws 7 dominoes. If player A draws the following 7 dominoes:
(4-4, 4-6, 4-1, 6-6, 6-5, 1-1, 0-0) ( Normally A two mark hand where player A calls "fours's" as "trumps"

Note : There are 4 "four's" left.

What is the probability that an opposing player will draw the (4-5) AND at least 2 of the remaining 3 four's (4-0) (4-2) (4-3)

Answer 1

There are 21 dominoes left.

Whoever draws the 4-5, we want the probability that their other six dominoes contains at least two of the three specified dominoes.

They have six of the remaining 20.

Prob all three is therefore 6/20 * 5/19 * 4/18.
Prob exactly 2 of three is 6/20 * 5/19 * 17/18 * 16/17 * 15/16 * 14/15 * 3 = 6/20 * 5/19 * 42/18

So we have 6/20 * 5/19 * 46/18
= 1/2 * 1/19 *23/3

But it must be an opposing player, so multiply this by 2/3
= 23/171
=0.1345 approx
.

Answer 2

opposing players hold 14 dominoe and your partner has 7

each player has a 1:3 chance of having the (4-5) so 2:3 chance that an opponent has it!
and each of those players have roughly a 1:3 chance of getting each of the other three dominoes.

The guy holding 4-5 has only 6 more dominioes versus the other guys 7. so its really 6/20 not 7/21 but that only reduces chance a few percentage points.

so 2/3 x 3( 1/3) = 2/3 x 1 = 2/3 or 66.6% of the time, or two out of three hands!

Answer 3

This is a selection-without-replacement problem, so you use a hypergeometric distribution.

We have seven of the dominoes accounted for, so there are 21 unknown dominoes. The probability of a certain player drawing the (4-5) domino in their hand of seven is 7/21 = 1/3. The probability of that same player then having at least two of the remaining three 4s is:

hypgeomdist(num,draw,cop,dec)
where num = number of copies of the desired item you wish to find, draw = the number of items you're drawing, cop = the number of copies remaining in the deck, and dec = the total number of items remaining in the deck. Since the (4-5) is accounted for, there are 20 dominoes in the deck. So, we have:

probability of getting two + probability of getting all three
= hypgeomdist(2,6,3,20) + hypgeomdist(3,6,3,20)
= 0.1842 + 0.0175
= 0.2017

So, the probability of both events occurring for a SINGLE player is
1/3 * 0.2017 = 0.06723

However, you have two opponents. The probability of either ONE of them drawing the (4-5) and at least 2 of the remaining fours is twice our result:
2 * 0.6723 = 0.1345

There is a 13.45% chance that either opponent will simultaneously have the (4-5) and at least 2 of the remaining fours.

PS: The hypgeomdist() function is found in MS Excel. It works as I described it.