# Problem 116 Solution

Let ai denote the probability if the player has \$i he will reach \$30 before losing everything. Let p the probability of winning any given bet = 9/19.

a0 = 0

a1 = 9/19*a2
a2 = 9/19*a3 + 10/19*a1
a3 = 9/19*a4 + 10/19*a2
.
.
.
a27 = 9/19*a28 + 10/19*a26
a28 = 9/19*a29 + 10/19*a27
a29 = 9/19*a30 + 10/19*a28
a30 = 1

Divide the left side into two parts:

9/19*a1 + 10/19*a1 = 9/19*a2
9/19*a2 + 10/19*a2 = 9/19*a3 + 10/19*a1
9/19*a3 + 10/19*a3 = 9/19*a4 + 10/19*a2
.
.
.
9/19*a27 + 10/19*a27 = 9/19*a28 + 10/19*a26
9/19*a28 + 10/19*a28 = 9/19*a29 + 10/19*a27
9/19*a29 + 10/19*a29 = 9/19*a30 + 10/19*a28

Rearange with 10/19 terms on the left side and 9/19 terms on the right:

10/19*(a1) = 9/19*(a2 - a1)
10/19*(a2 - a1) = 9/19*(a3 - a2)
10/19*(a3 - a2) = 9/19*(a4 - a3)
.
.
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10/19*(a27 - a26) = 9/19*(a28 - a27)
10/19*(a28 - a27) = 9/19*(a29 - a28)

Next multiply both sides by 19/9:

10/9*(a1) = (a2 - a1)
10/9*(a2 - a1) = (a3 - a2)
10/9*(a3 - a2) = (a4 - a3)
.
.
.
10/9*(a27 - a26) = (a28 - a27)
10/9*(a28 - a27) = (a29 - a28)

Next telescope sums:

(a2 - a1) = 10/9*(a1)
(a3 - a2) = (10/9)2*(a1)
(a4 - a3) = (10/9)3*(a1)
.
.
.
(a29 - a28) = (10/9)28*(a1)
(a30 - a29) = (10/9)29*(a1)

Next add the above equations:

(a30 - a1) = a1 * ((10/9) + (10/9)2 + (10/9)3 + ... + (10/9)29)

1 = a1 * (1 + (10/9) + (10/9)2 + (10/9)3 + ... + (10/9)29)

a1 = 1 / (1 + (10/9) + (10/9)2 + (10/9)3 + ... + (10/9)29)

a1 = (10/9 - 1) / ((10/9)30 - 1)

Now that we know a1 we can find a20:

(a2 - a1) = 10/9*(a1)
(a3 - a2) = (10/9)2*(a1)
(a4 - a3) = (10/9)3*(a1)
.
.
.
(a19 - a18) = (10/9)18*(a1)
(a20 - a19) = (10/9)19*(a1)

Add the above equations together:

(a20 - a1) = a1 * ((10/9) + (10/9)2 + (10/9)3 + ... + (10/9)19)
a20 = a1 * ((10/9)20 - 1)) / (10/9 - 1))
a20 = [ (10/9 - 1) / ((10/9)30 - 1) ] * [ ((10/9)20 - 1) / (10/9 - 1) ]
a20 = ((10/9)20 - 1) / ((10/9)30 - 1) =~ .3198

Michael Shackleford, A.S.A.