Maths problem

It's a 1 in 6 chance isn't it?

would rather be on your side CLYDEBANK tbh seeing as every draw you're chance is 1.5 compared your son's 3.0.

very slim imo, worked out from 5-0 to 8-3 and the chances of those 4 results add up to about 1.4%. looking at the direction of the figures chance of winning can't be much higher than 4%

3%

3% is correct (more precisely 3.03recurring, or 1/33), but I don't have a mathematical proof as to why.

The problem is a 1-dimensional random walk, with p = 2/3, and with sticky boundaries.

I hope he spends the money wisely CB.

From the central position, the chance of 5-0 is 32/243 and the chance of 0-5 is 1/243, which indicates the 32:1 ratio of winning that I found by exhaustive calculation.

What I am struggling to do is to explain why this ratio is the actual result, as after 5 trials the distribution of outcomes is biased away from this central position.

cheers bushy and others.

bushy
you sound like mr logic out of the viz

Write total bank as B, set constant stake at 1, bet pays even money so at each step bank moves up 1 with prob p, down 1 with prob (1-p)=q

Say you stop after your bank reaches some amount W

Now let R_B = Pr(Funds reach W before they reach 0, given current bank B)

Then we have to solve

R_B = q * R_(B+1) + p * R_(B-1) (***)

With boundary conditions R_0 = 0 and R_W = 1

Try solution R_B = A * k ^ B (A, k constant)

Substitute into (***) to get quadratic in k, two solutions. Substitute B=0, W to find constants

Gives

R_B = ( 1 - ( q / p ) ^ B ) / ( 1 - ( q / p ) ^ W )

(Note p = q is a special case, working comes out a little different, R_B = B / W)

In your case p = 2/3, B = 5, W = 10

I ~concur with Mr Bushy

0.030303030303030303

using a spreadsheet solution

I just played with two jacks and two queens from a deck of cards.

5 points each at start

same - different
5-5
4-6
3-7
2-8
3-7
4-6
3-7
2-8
1-9
2-8
1-9
0-10

Apologies to Eldrick if this is just a long-winded version of his method, but you could do this:

Get a series of 9 equations relating the probabilities of winning from the various scores. Eg:

P(win from 9-1) = 1/3 + (2/3 * P(win from 8-2)
P(win from 8-2) = 1/3 * P(win from 9-1) + (2/3 * P(win from 7-3)

etc.

and then just solve them simultaneously. (There are 9 equations and 9 unknowns).

I'm not familiar with any high falutin' maths stuff, so apologies for the kitchen sink approach!

Contrarian, I have the equations as:
(where P(91) = prob of winning from 9-1 ahead)

P(91)=2/3 + 1/3 P(82)
P(82)=2/3 P(91)+1/3 P(73)
...
P(28)=2/3 P(37) + 1/3 P(19)
P(19)=2/3 P(28)

Substituting back up:
P(28)=2/3 P(37) + 2/9 P(28)
P(28)=6 P(37)/7

And now there's a series where the next calc=2pd/2pd +1 where pd=previous denominator.
So P(37)=14(P46)/15 ...

Until P(91)=1022/1023

So P(55) ( the calc for winning from 5-5) is
1022/1023*
510/511*
254/255*
126/127*
62/63

which=32/33

same solution basically

when W = 2B you can rewrite the denominator as difference of two squares, then simplify to get:

R_B = 1 / ( 1 + ( q / p ) ^ B )

q/p = 1/2 and B = 5 gives:

R_5 = 1 / (1 + (1/2)^5) = 32/33 = 0.03030303...

the chances of two reds being pulled out is ....16.66.....5/1
thge cnances of two blues being pulled out....16.66......5/1
the chances of two odds balls being poulled out is.......66.66....1/2
the chances of any matched coloured balls being pulled out is 33.33 2/1

1. did your maths teacher never tell you to spend an insanely excessive amount of time reading the question before you picked up your pen?

2. i'm going to lay you 10/1 about two blue balls coming out, because i think you're special. how much would you like?