Off-topic but might be fun!

I was playing Scrabble with my Dad earlier today. To decide who would play first we each drew a tile - the person with the tile closest to A would start the game.

I drew an R, then Dad did the same. We put both tiles back then I drew an O, then Dad did the same. We again put the tiles back. I then drew an M, and Dad did the same!

There were 98 letters in the bag. There were 6 letter R's, 8 O's and 2 M's when we started. I've calculated the probability of each of us drawing those letters as follows, taking into account the fact that there were only 97 tiles each time Dad drew, and of course one fewer of that particular letter each time (i.e. 6 in 98 chance of me drawing an R, 5 in 97 for Dad).

Me
R - 6.1%
O - 8.2%
M 2%

Dad
R- 5.2%
O - 7.2%
M - 1%

We're wondering what the probability was of us drawing the same letter as each other on three consecutive occasions. Can anyone help? Thanks in advance! rbarker82@hotmail.com

Just take the product of his probabilities, yours are irrelevant as you just drew 3 random tiles. He matched them.

Using your numbers without checking anything, I get about 1 in 27K chance that he would match those specific 3 draws of yours (but not just any draws you might make due to the varying letter distribution).

To answer that we would need to know how many of all the letters there were in bag, not just the R, O, and M, because you could have drawn any letters. Given that you draw R, O and M, the probability that he also draws R, O, and M is the product of his probabilities or 1 in 27,076.

BTW, I've heard that when played at the highest levels, Scrabble is all about probability rather than who knows the most words. Everyone knows all of the words that can be made, and the game is about calculating the probabilities that your opponent has the tiles to make those words so that you don't play the wrong tile and allow him to make them.

Okay, I see you put the tiles back in the bag after each tie ( but not everyone does this since the official rules don't explicitly say what to do in case of a tie ).

Assuming you're using the English version, out of the 100 tiles, the distribution of tiles with at least two occurrences are given below ( with the blank represented by "[]" ):

E: 12
A,I: 9
O: 8
N,R,T: 6
D,L,S,U: 4
G: 3
[],B,C,F,H,M,P,V,W,Y: 2

Then the probability of a tie is simply [(12)(11)+2(9)(8)+(8)(7)+3(6)(5)+4(4)(3)+(3)(2)+10( 2)(1)]/[(100)(99)] = 496/9900 ~ 0.05010101010 so the probability of three ( or more ) ties is simply this number cubed or about 0.00012575910724. The odds are about 7950.710392 to 1 against, so it's not as rare as one might think.

That would be an interesting analysis. I've played with those types -- the ones who read the Scrabble dictionary. Never challenge them. That's an interesting aside.

RE: Bruce's comment. I've seen scrabble on TV somewhere and the hosts always talk about what word the person should play (given what he or she knows; the pieces in play on available to him/her) and what is still possible (either in the bag or on the opponent's rack). I think there is a computer that solves it; the hosts aren't that smart I don't think. I remember being amazed that these scrabble "pros" played the exact word they should play about 9 times out of 10. But that could be a memory error by me.