Two players throw (one player after the other) a pair of dice.

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M.Mashal

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Two players throw (one player after the other) a pair of dice. The first one to get a 7 wins the game. What is the probability of the player that made the first throw win the game? Repeat the problem with 3 players.
 
You are going to have to add up an awful lot of probabilities so please start now (and show us your work!!)
 
M.Mashal said:
Two players throw (one player after the other) a pair of dice. The first one to get a 7 wins the game. What is the probability of the player that made the first throw win the game? Repeat the problem with 3 players.
Click to expand...
Two days, so a little bit more of a hint. For the first question (2 players) we have (assuming fair dice)
Win on 1 throw = w = (1/4)0 1/2
Win on 3 throws = ffw = (ff)w = (1/4)1 (1/2)
Win on 5 throws = ffffw = (ff)(ff)w = (1/4)2 (1/2)
...
Win on 2n+1 throws = ... = (1/4)n (1/2)
Total probability P
P = (1/2) [ 1 + (1/4)1 + (1/4)2 + (1/4)3 + (1/4)4 + ... ]
= ???

Now do the same kind of thing for 3 players.
EDIT: SEE CORRECTION BELOW. You could treat this as an example for tossing a coin rather than rolling dice.
 
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M.Mashal said:
Two players throw (one player after the other) a pair of dice. The first one to get a 7 wins the game. What is the probability of the player that made the first throw win the game? Repeat the problem with 3 players.
Click to expand...
\(\displaystyle \begin{array}{*{20}{c}}{(1,1)}&{(1,2)}&{(1,3)}&{(1,4)}&{(1,5)}&{(1,6)}\\{(2,1)}&{(2,2)}&{(2,3)}&{(2,4)}&{(2,5)}&{(2,6)}\\{(3,1)}&{(3,2)}&{(3,3)}&{(3,4)}&{(3,5)}&{(3,6)}\\{(4,1)}&{(4,2)}&{(4,3)}&{(4,4)}&{(4,5)}&{(4,6)}\\{(5,1)}&{(5,2)}&{(5,3)}&{(5,4)}&{(5,5)}&{(5,6)}\\{(6,1)}&{(6,2)}&{(6,3)}&{(6,4)}&{(6,5)}&{(6,6)}\end{array}\)

Look at the above, there are six pairs giving sum seven. So \(\displaystyle \mathcal{P}(S=7)=\frac{6}{36}=\frac{1}{6}\)
In order for the first player to win, it must be on an odd numbered throw.
Winning on the fifth throw means, no seven the first four throws and then a seven on the fifth.

Thus \(\displaystyle \frac{1}{6}\sum\limits_{k = 0}^\infty {{{\left( {\frac{5}{6}} \right)}^{2k}}} \) is the probability of the first player wins.

Can you explain why that sum?
 
pka said:
\(\displaystyle \begin{array}{*{20}{c}}{(1,1)}&{(1,2)}&{(1,3)}&{(1,4)}&{(1,5)}&{(1,6)}\\{(2,1)}&{(2,2)}&{(2,3)}&{(2,4)}&{(2,5)}&{(2,6)}\\{(3,1)}&{(3,2)}&{(3,3)}&{(3,4)}&{(3,5)}&{(3,6)}\\{(4,1)}&{(4,2)}&{(4,3)}&{(4,4)}&{(4,5)}&{(4,6)}\\{(5,1)}&{(5,2)}&{(5,3)}&{(5,4)}&{(5,5)}&{(5,6)}\\{(6,1)}&{(6,2)}&{(6,3)}&{(6,4)}&{(6,5)}&{(6,6)}\end{array}\)

Look at the above, there are six pairs giving sum seven. So \(\displaystyle \mathcal{P}(S=7)=\frac{6}{36}=\frac{1}{6}\)
In order for the first player to win, it must be on an odd numbered throw.
Winning on the fifth throw means, no seven the first four throws and then a seven on the fifth.

Thus \(\displaystyle \frac{1}{6}\sum\limits_{k = 0}^\infty {{{\left( {\frac{5}{6}} \right)}^{2k}}} \) is the probability of the first player wins.

Can you explain why that sum?
Click to expand...
You are totally correct that the actual probability has to be taken into account. For some reason, I was thinking of a fair coin where the probability is 1/2 both ways. So we have, for the two player problem,
2n+1 trial=(fail fail)nwin = (1/6) (25/36)n
n=0 is Player one wins on his first toss
n=1 is Player one wins on his second toss
n=2 is Player one wins on his third toss
...
NOTE: This gives the same answer as pka has
 
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