I don't undstand how the textbook come up with recursive forumulas.

For example,

Consider the following gambling game for two players, Black and White. Black puts $b$ black balls and White puts $w$ white balls in a box. Black and White take turns at drawing at random from the box, with replacement between draws until either Black wins by drawing a black ball or White wins by drawing a white ball. Supposed black draws first.

Calculate $P($Black wins$)$

Textbook Answer:

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I'm not sure how they knew their equation encompasses all ways Black could win.

I going to assume this is their reasoning:

The first draw could only be be Black or White.

$P($Black wins$)$

$= P($Black wins$|B)P(B) + P($Black wins$|W)P(W)$, obviously this encompasses all ways Black could win.

$= P($Black wins$|B)P(B) + ( P($Black wins$|WW)P(WW) + P($Black wins$|WB)P(WB) )$

$= P($Black wins$|B)P(B)$ + ( 0 + $P($Black wins$|WB)P(WB) )$

$= P($Black wins$|B)P(B)$ + $P($Black wins$|WB)P(WB)$


And how to set-up recursive probability equations in general + when to use them?

Edit 1:

I used everyone's feedback and came up with an in-depth solution. I think the logic is sound. For anyone that need it:

enter image description here


We can use a recursive method here because if the game continues past the first two draws, then the situation is exactly the same as at the beginning of the game. Thus the probability of black winning from that point cannot be different from the probability of black winning from the start.

  • $\begingroup$ I guess it won't work for probability that are dependent on previous draws such as $P($Black Wins on the $9$th draw$)$. $\endgroup$ – A_for_ Abacus Mar 2 '18 at 0:10
  • $\begingroup$ @A_for_Abacus It works a bit. For black to win on the ninth draw, there must first be four rounds of $WB$, then a $B$, so the probability is $P(WB)^4\cdot P(B)$. If you, for instance, take away replacement then you have ruined the possibility of thinking recursively here (you can still do a recursive argument, but it's much more work). $\endgroup$ – Arthur Mar 2 '18 at 6:28

Black wins or continues playing if either he picks a black ball ($p$) or he picks a white ball & then White picks a black ball ($qp$); at this stage he will be faced with exactly the same situation so \begin{eqnarray*} P(B)= p +pq P(B). \end{eqnarray*}


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