# A Scrabble game has identical

A Scrabble game has identical copies of each letter tilenumbering as follows: A-9, B-2, C-2, D-4, E-12, F-2, G-3, H-2, I-9,J-1, K-1, L-4, M-2, N-6, O-8, P-2, Q-1, R-6, S-4, T-6, U-4, V-2,W-2, X-1, Y-2, Z-1 and Blanks-2. Seven tiles are randomly sampledwithout replacement from the above 100 tiles.If a letter appearsexactly once in this seven-tile sample, we call it a singleton inthis problem.

a. Consider a letter that has n copies in total (e.g., for theletter E, n=12). What is the probability that this letter is asingleton?

b. What is the expected value of the number of singletons?(Count “blank” as a letter.)

Answer:

Suppose there are n copies of a letter ‘x’,

Then the probability of tile being the specific letter ‘x’ =n/Total sample size = n/100.

Since every tile has the probability of being letter ‘x’ =n/100. **This is an example of bernoulli trial.**

Thus the probability of finding 1 letter in the sample size 7follows a binomial distribtuion,

The probability is given by :

The probability of letter being a singleton is

=

b.)

The probability value for each letter is as given below:

Letter |
Success probability(p) |
(1-p) |
Probability ofsingleton |

A | 0.09 | 0.91 | 0.357758 |

b | 0.02 | 0.98 | 0.124018 |

c | 0.02 | 0.98 | 0.124018 |

d | 0.04 | 0.96 | 0.219172 |

e | 0.12 | 0.88 | 0.390099 |

f | 0.02 | 0.98 | 0.124018 |

g | 0.03 | 0.97 | 0.174924 |

h | 0.02 | 0.98 | 0.124018 |

i | 0.09 | 0.91 | 0.357758 |

j | 0.01 | 0.99 | 0.065904 |

k | 0.01 | 0.99 | 0.065904 |

l | 0.04 | 0.96 | 0.219172 |

m | 0.02 | 0.98 | 0.124018 |

n | 0.06 | 0.94 | 0.289745 |

o | 0.08 | 0.92 | 0.339559 |

p | 0.02 | 0.98 | 0.124018 |

q | 0.01 | 0.99 | 0.065904 |

r | 0.06 | 0.94 | 0.289745 |

s | 0.04 | 0.96 | 0.219172 |

t | 0.06 | 0.94 | 0.289745 |

u | 0.04 | 0.96 | 0.219172 |

v | 0.02 | 0.98 | 0.124018 |

w | 0.02 | 0.98 | 0.124018 |

x | 0.01 | 0.99 | 0.065904 |

y | 0.02 | 0.98 | 0.124018 |

z | 0.01 | 0.99 | 0.065904 |

Blanks | 0.02 | 0.98 | 0.124018 |

Expected value of number of singletons

= sum of probabilities of all the letters being singleton

= **4.93572**