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Bill, George, and Ross, in order, roll a die. The first one to roll an even number wins and the game is ended. What is the probability that Bill will win the game?

As far as I know, for the first roll Bill has a chance of $\frac{1}{2}$, then followed by George with a less than $\frac{1}{2}$ chance and Ross has a lesser than the second roll from George. Why is the answer $\frac{4}{7}$?

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  • $\begingroup$ Can three people all have a 50% chance of winning a game? $\endgroup$ – John Hughes Jul 2 '16 at 12:34
  • $\begingroup$ @JohnHughes Well there are 3 evens in one roll of a die. $\endgroup$ – Ralf Rafael Frix Jul 2 '16 at 12:35
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    $\begingroup$ Bill's chance of winning on the first roll, is indeed $1/2$. Your estimates of George's and Ross' chances of winning on the first roll are incorrect (see my comment above). But what if all three roll odd numbers? Bill could win on his second roll, or his third, or (very unlikely) 14th. You've not taken these chances into account. $\endgroup$ – John Hughes Jul 2 '16 at 12:38
  • $\begingroup$ @JohnHughes Sorry I can't follow your thoughts on this, but I now get a general idea that George and Ross chances on the next two rolls, after Bill, are lesser. $\endgroup$ – Ralf Rafael Frix Jul 2 '16 at 12:43
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Assume, the probability , that Bill wins , is $p$.

If Bill does not win with the first throw, then George has probability $p$ to win.

If both Bill and George fail, then the probability that Ross wins, is again $p$.

So, Bill has probability $p$, George has probability $\frac{p}{2}$ and George probability $\frac{p}{4}$ to win. So, we have $p+\frac{p}{2}+\frac{p}{4}=\frac{7p}{4}=1$. Therefore $p=\frac{4}{7}$.

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You need to account for all outcomes favorable for Bill to win: eitger ob the first toss, your 0.5 OR on 4th, i.e. all 3 previous tosses are odd, OR on 7th and so on. Can you take it from here?

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Call the winning probabilities $b,g,r$ respectively. Then clearly $b + g + r = 1$. Also, call the probabilities that they each win in the first round $b_1, g_1, r_1$. Then $b_1 = \frac{1}{2}, g_1 = \frac{1}{4}$ and $r_1 = \frac{1}{8}$. So we have no decision in the first round with chance $\frac{1}{8}$.

Now, note that $b = b_1 + \frac{1}{8}b = \frac{1}{2} + \frac{1}{8}b$, because $b$ wins with chance $b_1$ in the first round and if gets a second round (which happens with chance $\frac{1}{8}$) he has chance $b$ again, as if he started anew. This gives us $b = \frac{8}{7} \cdot \frac{1}{2} = \frac{4}{7}$.

Similarly $g = g_1 + \frac{7}{8}p = \frac{1}{4} + \frac{7}{8}p$, so $g = \frac{8}{7}\cdot \frac{1}{4} = \frac{2}{7}$.

So $r = \frac{1}{7}$, by summing to 1, or doing a similar calculation.

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