# Thread: probability and expected gain

1. ## probability and expected gain

i dunno how to do this question. could anyone possibly help? thx in advance.

A banker throws three fair, ordinary dice. If the result is a total of 11 points or above, or three "1", or three "2", or three "3", he wins $900. Otherwise he loses$900.
What is his expected gain?

i wonder if there are other methods other than listing out all events.

2. No. You need to know each three dice permutation that results in 11 or more.

The "three 1's, 2's or 3's" one is easy.

3. Some useful facts:

$P(all 1's)+P(all 2's)+P(all 3's) = 3(\frac{1}{6})^3$

$P(X_1+X_2+X_3 >11) = 1 - P(X_1+X_2+X_3<11)$

4. Hello, littleprince327!

A banker throws three fair, ordinary dice.
If the result is a total of 11 or above, or three 1's, or three 2's, or three 3's, he wins $900. Otherwise he loses$900.
What is his expected gain?

With three dice, there are $6^3 \,=\,216$ possible outcomes.

The totals range from 3 to 18.

$\text{Note that: }\;\underbrace{3,4,5,6,7,8,9,10}_{\text{108 ways}}\, , \, \underbrace{11,12,13,14,15,16,17,18}_{\text{108 ways}}$

Then: . $P(\text{sum }\geq 11) \:=\:\frac{108}{216}$

Also: . $P(\text{three 1's or three 2's or three 3's}) \:=\:\frac{3}{216}$

Hence: . $P(\text{win}) \:=\:\frac{108}{216} + \frac{3}{216} \:=\:\frac{111}{216} \:=\:\frac{37}{72}$

. . and: . $P(\text{lose}) \:=\:\frac{35}{72}$

Therefore: . $E \;=\;\frac{37}{72}(+900) + \frac{35}{72}(-900) \;=\; +\,25$

He can expect to win an average of \$25 per game.

5. Originally Posted by Anonymous1
Some useful facts:

$P(all 1's)+P(all 2's)+P(all 3's) = 3(\frac{1}{6})^3$

$P(X_1+X_2+X_3 >11) = 1 - P(X_1+X_2+X_3<11)$
you left out the 11...

$P(X_1+X_2+X_3 \ge 11) = 1 - P(X_1+X_2+X_3<11)$