Question

1

View NON-CALC.
Sam plays a game with a fair dice.
if he rolls a 1 or 2 he rolls again
if he rolls a 3 he wins
If he rolls a 4,5 or 6 he loses.
When he has to roll again,
if he rolls a factor of 6 he wins
If he rolls a number which is not a factor of 6 he loses,
a) Complete the tree diagram,
First roll
Second roll
Not a factor of 6
1 or 2
Factor of 6
DI
3
4,5 or 6
b) Is Sam more likely to win or lose?
You must calculate the probability he wins.

Answer 1

The probability of winning directly is, as you calculated, 8/36, and the probability of losing directly is (1+2+1)/36=4/36.

For the remaining cases, you need to sum over all remaining rolls. Let p be the probability of rolling your initial roll, and q=6/36=1/6 the probability of rolling a 7. Then the probability of rolling your initial roll before rolling a 7 is p/(p+q), and the probability of rolling a 7 before rolling your initial roll is q/(p+q). Thus, taking into account the probability of initially rolling that roll, each roll that doesn't win or lose directly yields a contribution p2/(p+q) to your winning probability.

For p=5/36, that's

(536)25+636=2511⋅36,

and likewise 16/(10⋅36) and 9/(9⋅36) for p=4/36 and p=3/36, respectively. Each of those cases occurs twice (once above 7 and once below), so your overall winning probability is

836+236(2511+1610+99)=244495=12−7990≈12−0.007.

Explanation:

Suppose you throw a 4 and let p(4) your winning probability. At your next roll you have a probability 3/36 of winning (you throw a 4), a probability 6/36 of losing (you throw a 7) and a probability 27/36 of repeating the whole process anew (you throw any other number). Then:

p(4)=336+2736p(4),so thatp(4)=13.

Repeat this reasoning for the other outcomes and then compute the total probability of winning as:

ptot=836+336p(4)+436p(5)+…