Probabilites, I’m awful at those.

by Gilbert Keith

I’m unable to prove the example given in this seemingly simple problem:

A bag contains one red disc and one blue disc. In a game of chance a player takes a disc at random and its colour is noted. After each turn the disc is returned to the bag, an extra red disc is added, and another disc is taken at random.

The player pays £1 to play and wins if they have taken more blue discs than red discs at the end of the game.

If the game is played for four turns, the probability of a player winning is exactly 11/120, and so the maximum prize fund the banker should allocate for winning in this game would be £10 before they would expect to incur a loss. Note that any payout will be a whole number of pounds and also includes the original £1 paid to play the game, so in the example given the player actually wins £9.

Find the maximum prize fund that should be allocated to a single game in which fifteen turns are played.

I’m hoping that writing this out thoroughly will help:

The probability of drawing any blue on the nth turn should be 1/(n+1)

Case 1: max # turns = 3; ==> need to draw either 2 or 3 blues in 3 turns;

Sub case 1.1: I draw 3 blue:

The probability of this should be (1/2)*(1/3)*(1/4) = (1/24)

sub case 1.2:  I draw 2 blues:

sub case 1.2.1: I draw {B,R,B}

The probability of this should be (1/2)*(2/3)*(1/4) = (2/24)

sub case 1.2.2: I draw {B,B,R}

The probability of this should be (1/2)*(1/3)*(3/4) = (3/24)

sub case 1.2.3: I draw {R,B,B}

The probability of this should be (1/2)*(1/3)*(1/4) = (1/24)

So the total probability of drawing at least 2 blues should be 7/24 (if you are drawing a max of 3).

Moving on to the example cited:

The probability of drawing any blue on the nth turn should be 1/(n+1)

Case 1: max # turns = 4; ==> need to draw either 3 or 4 blues in 4 turns;

Sub case 1.1: I draw 4 blue:

The probability of this should be (1/2)*(1/3)*(1/4)*(1/5) = (1/120)

sub case 1.2:  I draw 3 blues:

sub case 1.2.1: I draw {B,B,B,R}

The probability of this should be (1/2)*(1/3)*(1/4)*(4/5) = (4/120)

sub case 1.2.2: I draw {B,B,R,B}

The probability of this should be (1/2)*(1/3)*(3/4)*(1/5) = (3/120)

sub case 1.2.3: I draw {B,R,B,B}

The probability of this should be (1/2)*(2/3)*(1/4)*(1/5) = (2/120)

sub case 1.2.4: I draw {R,B,B,B}

The probability of this should be (1/2)*(1/3)*(1/4)*(1/5) = (1/120)

So the total probability of drawing at least 3 blues should be 11/120. SUCCESS.

Okay, I think I see the pattern. If the # of blues is 1 less than the total number of turns, then your probability for picking up that number is (1/((N+1)!))*f (x) where x is the difference between the max # of blues and the # you’re on.
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