News2 mins ago
What are the odds?
14 Answers
I chose this category to post in a hope that I would get answers from mathematical , probability solving type people.
I play the game Words With Friends (unofficial Scrabble) on my phone and always have the maximum number of games going. The other day I realised the seven tiles the game had generated for me were the letters in my surname.
Can anyone work out roughly what the chances are of that?
I play the game Words With Friends (unofficial Scrabble) on my phone and always have the maximum number of games going. The other day I realised the seven tiles the game had generated for me were the letters in my surname.
Can anyone work out roughly what the chances are of that?
Answers
celtic's answer is correct if you are asking: my last name is ABCDEFG; what are the odds of drawing those letters at random from a bag containing the letters ABC...XYZ?
But, doesn't it also depend on the probability of drawing a certain letter? There is on;t one Z in a Scrabble set but many E's. So if you last name is ZAXJAQK, the probability would be far lower than if it was DEARTIS.
But, doesn't it also depend on the probability of drawing a certain letter? There is on;t one Z in a Scrabble set but many E's. So if you last name is ZAXJAQK, the probability would be far lower than if it was DEARTIS.
Ok, so there are 98 tiles in a scrabble set, of which you have:
9 I
12 E
9 A
6 R
4 L
1 Z
3 G
Prob of drawing an I is 9/98; prob of drawing an E is 12/97 (only 97 tiles left after you draw first), prob(A) = 9/96 ... prob (G) = 3/92. The Z kills you since there is only one in the set.
Changing the required order of draws would affect the total probability only slightly (any of these numbers divided by 98, 97, 96 etc are pretty much the same).
So the probability = (9/98) x (12/97) x ... x (3/92) = 1.003 x 10^-9. A very slim chance!!
9 I
12 E
9 A
6 R
4 L
1 Z
3 G
Prob of drawing an I is 9/98; prob of drawing an E is 12/97 (only 97 tiles left after you draw first), prob(A) = 9/96 ... prob (G) = 3/92. The Z kills you since there is only one in the set.
Changing the required order of draws would affect the total probability only slightly (any of these numbers divided by 98, 97, 96 etc are pretty much the same).
So the probability = (9/98) x (12/97) x ... x (3/92) = 1.003 x 10^-9. A very slim chance!!