Cute Applicable Math Question

[David Sklansky]

This is related to my Tournament Question Part 2.

Suppose you are playing a two dollar coin flip freezeout at a dollar a flip. The coin is weighted 60-40 in your favor. Show me a SIMPLE way to calculate the EXACT probability you will win the freezeout. This question is not open to smarty pants like BB King, Tom Wiedeman, Pink Bunny, Majorkong, Bruce Z and the like.

[elindauer]

Well, I don't know if you consider this simple, but here goes...


First, here's some notation:

[31] means player one has 3 dollars, player 2 has 1 dollar.

I'll use this notation to represent the 2x2 matrix of probabilities for possible outcomes for each player (1st place and 2nd place). In English, [31] represents a 2x2 matrix:

{a b} <-- player 1 outcomes
{c d} <-- player 2 outcomes

where a is the probability of player 1 finishing first, b is the chance player 1 finishes second, and the second row represents the same quantities for the second player.


Some reflection should reveal that

a+b = c+d = a+c = b+d = 1,

and that these matrices obey simple rules for addition, multiplication by constants, etc.


OK, now to the math. We're trying to calculate [22], really, we've just been asked to find position (a) of [22]. Here's how:


.6 of the time, player 1 wins the first toss. .4, he loses. We can write this out like this:

[22] = .6*[31] + .4*[13]

or, to make it a little more generic:

[22] = p*[31] + (1-p)*[13] (a)

where p is the probability of a win in any trial. So... what are [31] and [13]? Well, they break down very simply as well, just like [22] did:

[31] = p * [40] + (1-p) * [22]
[13] = p * [22] + (1-p) * [04]


Note that we could substitute into (a) to get an equation in terms of only [22], [40], and [04]. But what are [40] and [04]? Well, in these states the game is over, we have a winner. ie:

[40] is
{1 0} <-- player one always wins and never loses from here
{0 1} <-- player two always loses...

and [04] is
{0 1}
{1 0}


So, we substitute into (a), do some elementary algebra, and produce:

[22] = 1/(1-2p(1-p)) *
{ p^2 (1-p)^2}
{(1-p)^2 p^2 }

Confusing? A bit, but if we plugin p=.6, it all simplifies to something like this:

[22] = 1/.52 *
{.36 .16}
{.16 .36}

That is, the probability of player 1 winning the game is .36/.52, or just under 70%.


-Eric

[BruceZ]

Am I barred from the tournament question 2 too? I was going to look at that one.

[ML4L]

After two flips, you will have won 36% (0.6^2) of the time, lost 16% of the time (0.4^2), and tied the other 48%, after which you begin again. Thus, it should be clear that the probability of winning the freezeout is just the ratio of wins to losses: 36%/16%, or 2.25 to 1 (around 69%).

A fun question to try is to find the probability of winning a $3 freezeout (the only method that I know of for solving that one doesn't qualify as "simple").

ML4L

[elindauer]

I prefer your solution for it's simplicity, but mine for it's, ummm, robustness. [img]/images/graemlins/wink.gif[/img]

You can use the same method I describe for solving the 2$ freezout to solve your proposed 3$ version. I would describe this method as simple, but tedious.

That is:

[33] = p*[42] + (1-p)*[24]
[42] = p*[51] + (1-p)*[33]
[51] = p*[60] + (1-p)*[42]
[24] = p*[33] + (1-p)*[15]
[15] = p*[25] + (1-p)*[06]

Five equations, five unknowns ([60] and [06] are known quantities). Solve with standard linear algebra.

[Timer]

(10 x .6) - (10 x .4) = +2

[emanon]

Ok. Here goes...
For a quick solution it is important to recognize 2 facts:
1. There must be a winner.
2. The winner must win twice in a row.

P(I win twice in a row) = .6^2 = .36
P(I lose twice in a row) = .4^2 = .16
--> P(there is a winner for a given pair of coin flips) = .36 + .16 = .52

Since there must be a winner, we know that we can ignore the area of the probability space where there is no winner.
Using bayes theorem:
--> P(I win) = .36/.52 = .692308

[emanon]

Pls read my $2 answer first.

For $3, the following is true:
1. There must be a winner.
2. The winner must win 3 more times than the loser.
3. The path taken to reach 3 more victories is irrelevant. ie, whether you win in 3 flips, or 63 flips, you still need 3 more victories to win than your opponent.

Therefore:
P(A has 3 more victories) = .6^3 = .216
P(B has 3 more victories) = .4^3 = .064
P(A)/(PA+B) = .216/(.216+.064) = .771429

This extends to the case for $n.

Incidentally, this makes this question slightly less relevant to David's original Tournament 2 question. In the coin flip case, there is no path dependence.

However, in a tournament, there is path dependence as your ability to pay blinds, call bluffs, etc is all affected by your stack size, thus affecting the probabilities depending on the path taken. (this is already pointed out in that question by other posters).

[David Steele]

Here is one way:

Lets call the chance you have to win Pw

In the first 2 flips there are 4 possible results

0 0 lose match
1 0 even with Pw chance again
0 1 even with Pw chance again
1 1 win match


Adding up the the total chance and set = to Pw:

0 + (.6 x .4)Pw + (.4 x .6)Pw + .6x.6 = Pw

Solving for Pw = .36/(1-.24 -.24) = .36/.52
69.2%


D.

[ML4L]

Hey elindauer,

Your method is essentially the one that I was alluding to with regard to solving the $3 problem; maybe I'm weird for not categorizing linear algebra as simple... [img]/images/graemlins/grin.gif[/img]

I actually started to do the $2 problem the hard way, but then stopped to think for a minute and realized that it wasn't necessary. But I was glad to see that I wasn't the only one whose instinct was to take that approach...

ML4L