Words With Friends Odds

[arjay]

I regularly play Words With Friends (Online Scrabble via Facebook) with 4 people.
In the opening move of a recent game, unusually, I received 2 blank tiles in the seven that are issued. They can be used as any of 26 letters and are equal to wild cards.
There are 90 tiles available for play 2 of which are blank
I told one of the people with whom I play about this oddity.
He suggested that I asked the Answerbank members what were the odds of this happening.
Your help will be greatly appreciated

[fiction-factory]

If there are 70 tiles to pick from and two are blank then the chances that the first two you pick are both blank is 1/70 x 1/69 = 1/4830

The chances of getting 2 within the first two though are much greater since they can be the 1st and 3rd, 1st and 4th,... and so on to 6th and 7th. I'll have a go after breakfast if I can find my calculator

[fiction-factory]

Typo.

Re-read that as:

If there are 70 tiles to pick from and two are blank then the chances that the first two you pick are both blank is 1/70 x 1/69 = 1/4830

The chances of getting 2 within the first SEVEN though are much greater since they can ALSO be the 1st and 3rd, 1st and 4th,... and so on to 6th and 7th. I'll have a go after breakfast if I can find my calculator

[fiction-factory]

I think there are 21 ways of getting exactly 2 blanks from 7 tiles.
So now need to check that and factor that in. Will get calculator later

[fiction-factory]

I see there are 90 tiles so odds of first two are 1/90 x 1/89

Okay I may leave this thread now after two typos as when I get the answer a particular poster will probably again derail the thread by saying it was a lucky guess/ fourth time lucky and I don't have a clue what I'm on about

[arjay]

Thank you factor-fiction

[Cloverjo]

I'll have a go.
From 90 tiles, there are 90c7 different possible sets of 7 tiles.
Of these, there are (1/90) x (1/89) x 88c5 sets that contain two blanks. Divide this number into the first and you get the odds. Unfortunately, my calculator isn't powerful enough to crunch the large numbers.

[Cloverjo]

I've just done the arithmetic and the probablity comes out as 0.00000065.
That's if my reasoning is correct. I'm hoping fiction factory will come back, or Buenchico or jim360 will come along to put me right.

[Jonathan-Joe]

Suppose I can distinguish between the two blank tiles (say one has a small diagonal scratch, the other a horizontal / vertical one).
After arjay has taken his tiles, the first blank is either among his seven or the 83 with other players / in the pool. Probability of arjay holding is 7/90. Given that arjay holds the first, the second blank is either with arjay's remaining six or with other players / pool. Probability of arjay holding is 6/89. Probability that arjay holds both blanks is 7/90 x 6/89 = .00524.
So any one player could expect to be issued both blanks about 5 times in a thousand games.
(I give the probability as a fraction - I don't find odds so meaningful).

[Jonathan-Joe]

I mean "decimal fraction".

[Cloverjo]

I'm bumping this up, as I've seen that jim360 is around. I'd be interested to know how he'd go about solving this.