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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:

enter image description here

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

Community
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A_for_ Abacus
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  • So, I read your final solution and I am confused. When you define B₂W₂ be the outcome next draw — do you mean "next Black's draw", or the "White's draw?" IOW, is it the 2nd or the 3rd turn (as we know White draws 2nd and Black draws 3rd). Also, what even means $P(B₁B₂)$…? I read it as "probability of drawing black twice in a row", which makes no sense, because if B₁ happened, Black won the game, B₂ would be impossible, IOW $P(B₁B₂) = 0$. – Hi-Angel Aug 14 '22 at 13:35

2 Answers2

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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*}

Donald Splutterwit
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  • I guess the main takeaway from this is that it is impossible to break probability recursion. IOW you can't calculate the exact probability as P(B) would infinitely recurse into itself, and you have to stop it at some point. You can't extract an exact number out of any kind of recursive probability. – Hi-Angel Aug 14 '22 at 14:44
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    Actually, after a chat discussion it turned out, the problem is solvable in finite numbers. Basically, you just ignore the recursion (apparently you can do that) by replacing $P(B)$ with a variable $x$. Then we have $x=p+pqx$. Then we substitute: $p = ½$, $pq = ¼$. Then $x=p+pqx ⇔ x = ½ + ¼x ⇔ 4x = \frac{4}{2} + \frac{4}{4}x ⇔ 4x = 2 + 1x ⇔ 3x = 2 ⇔ x = \frac{2}{3}$ – Hi-Angel Aug 16 '22 at 19:21
  • What is the exact mathematical reason for the recursive equation? Intuitively it seems fine but intuition is not a rigorous mathematical argument and in this case it seems very hand-waving. In fact, you are assigning a probability to a set although a.) you don't know how the set looks like and b.) you don't know how the probability measure is defined. So I would expect to set up a proper probability space to solve this question. – Philipp May 17 '23 at 15:07
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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.

Arthur
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  • I guess it won't work for probability that are dependent on previous draws such as $P($Black Wins on the $9$th draw$)$. – A_for_ Abacus Mar 02 '18 at 00:10
  • @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). – Arthur Mar 02 '18 at 06:28