probability dreidel game problem Reuven and Shimon play a dreidel game. In each round, each of them
rolls the dreidel.(dreidel can fall on G,N,H,P only in equal probability) If one of them gets a G and the other not, he wins; if
both get a G or none gets it, the game continues. They play the game
11 times. Let Ri be the number of rounds in the i-th game,there are 11 games overall.
(For example, if in the rst round of the sixth game Reuven got a P
and Shimon an H, in the second both got a G, in the third { Reuven
a P and Shimon an N, and in the fourth ( Reuven a G and Shimon an
H, then R6 = 4.) Let R be the total number of rounds in all games, X
the total number of wins of Reuven and A the event whereby at some
point during the match Shimon has had more wins than Reuven.
1.E((1/R1)) =?
2.P(A \ X = 7) =?
thanks.
 A: In each round of a game, either of the three scenarios can take place - $1$. Shimon wins, $2$. Reuven wins, $3$. Result is inconclusive and the game continues.
Shimon can win a game if and only if he gets a G and Reuven gets either N or H or P. The chance of any player of getting any one of G,N,H and P is $\frac{1}{4}$ and moreover, getting any one is independent of getting another since dreidel can fall on G,N,H,P only in equal probability. Thus, $\mathbb{P}$[Shimon wins a game] = $\binom{3}{1}$ $(\frac{1}{4})^2 = \frac{3}{16}$.
With similar logic, $\mathbb{P}$[Reuven wins a game] = $\frac{3}{16}$
$\mathbb{P}$[A round ends inconclusively] = $1 - 2.\frac{3}{16} = \frac{5}{8}$
Think about a sequence of rounds which would either end conclusively (a success) or would end inconclusively (a failure). Probability of a "success" is the same as the chance of either Shimon or Reuven wins, that is, $\mathbb{P}$[Shimon wins a game]$+$$\mathbb{P}$[Reuven wins a game] = $2.\frac{3}{16}=\frac{3}{8}$. Then, the probability of "failure" = $1-\frac{3}{8}=\frac{5}{8}$
Clearly each game can be thought of a sequence of those inconclusive rounds (failures) followed by a conclusive round (success); Thus the number of rounds in any game has a Geometric distribution with "success" probability $\frac{3}{8}$
$R_1$: number of rounds in the $1^{st}$ game. This is a Geometric random variable and can take values $1,2,3,...$
$\mathbb{P}[R_1=j]=(1-p)^{j-1}p$, where $p$=Probability of "success"=$\frac{3}{8}$
Then,
$$\mathbb{E}[\frac{1}{R_1}]=\sum_{j=1}^\infty\frac{1}{j}\mathbb{P}[R_1=j]=\sum_{j=1}^\infty\frac{1}{j}(1-p)^{j-1}p = -\frac{p\mathbb{ln}p}{1-p} = 0.59 (approx)$$
$X$: total number of wins of Reuven; this is again a random quantity which follows the rule of a Binomial distribution with number of trials = $11$ and probability of success = $\frac{3}{16}$
Let me define $Y$: total number of wins of Shimon. This will have the identical distribution as that of $X$.
Given $X=7$, $X+Y$ can take $7,8,9,10,11$ since there are $11$ games in all. That is, given $X=7$, $Y$ can take $0,1,2,3,4$
$A$: [at some point during the match Shimon has had more wins than Reuven] = [$Y$>$X$]
$\mathbb{P}[A,X=7]=\mathbb{P}[Y>7,Y\le4]=0$
That is, $\mathbb{P}[A|X=7]=0$
