*I STILL DON'T GET THE ANSWERS PROVIDED. PLEASE HELP! In a game, the participant rolls a pair of dice. If the result is a $7$, he wins. If the outcome is a number $n$ different from $7$, he continues rolling the dice over and over again. If he gets $7$, he loses. If he gets another $n$, he wins.
- Find the probability that the participant wins this game?
- Find the probability that his final roll will be a $7$?
- Let $A$ denote the number of times the participant wins and $B$ denote the number of times he got a $7$. suppose he has played $100$ games, what are the means and standard deviations of $A$ and $B$?
- Calculate $P(A\geq 77)$
- Calculate $P(B\geq77)$
both of the last questions assume that he plays this game $100$ times.
I personally think we need to use TWO-STAGE Experiment method to solve this problem i think it's necessary for me to clarify this a bit. for example, if he gets 7 on the first try, he wins. now if he gets 6 on the first try, he continues rolling the pair of dice until he gets 6 again (he wins) or he gets 7 (he loses). this is called crapless craps game. the probability of getting 7 in a roll is 6/36, of getting 6 or 8 is 5/36, of getting 5 or 9 is 4/36, of getting 4 or 10 is 3/36, of getting 3 or 11 is 2/36, of getting 2 or 12 is 1/36. i don't know where should I go next. i think we need some conditional probabilities here since there are many stages in getting the probability of winning this game
Draw up a tree of the possible outcomes in the game.
Sum up all the ways you can win: $p_{win}=\frac{1}{6}+\frac{2}{36}*\{\frac{1*1}{7}+\frac{2*2}{8}+\frac{3*3}{9}+\frac{4*4}{10}+\frac{5*5}{11}\}=\frac{6557}{13860}\approx0.47$
Sum up the ways your final roll is a seven: $p_{seven}=\frac{1}{6}+\frac{2}{36}*\{\frac{1*6}{7}+\frac{2*6}{8}+\frac{3*6}{9}+\frac{4*6}{10}+\frac{5*6}{11}\}=\frac{9613}{13860}\approx0.69$
When you consider 100 independent games we should consider a binomial distribution to evaluate the mean and standard deviation. We have $E=np, SD=\sqrt{n(1-p)p}$, so after getting some numerical results:
$E_{win}=47.3, SD_{win}=5.0$ and $E_{seven}=69.4, SD_{seven}=4.6$
As for the probabilities of winning/ending with seven at least 77 times: for this you need to either evaluate the cumulative distribution function exactly or do some approximation. With Mathematica I calculate the likelihood as $P(x \ge 77)=1-CDF[dist,76]$ to be: $P(A\ge 77)=1.25*10^{-9}$ and $P(B\ge 77)=0.058$.