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. /images/graemlins/wink.gif

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... /images/graemlins/grin.gif

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

[ML4L]

Hey emanon,

I might be mistaken, but I don't think that you can apply the simple $2 concept to the $3 or $n cases. This stems from the fact that, under the $2 case, if a player has won once, he will either win or be tied after the next flip. But, beyond $2, you get into situations (e.g. after having won the first flip) where you don't encounter both alternatives at the same time.

To my knowledge, the only way to solve the $3 or $n problem is by using methodology similar to that which elindauer used to solve the $2 problem. I'll try to post the solution sometime tomorrow (unless someone beats me to it...).

ML4L

[ML4L]

David,

You never posted an answer or critique for the "Way Tougher Two Round Game Theory Problem" that you posted a few months back. Pink Bunny and I (and a couple others) thought that we got the answer, but there was still some doubt. Anyway, if you happen to get a chance, I've been curious about how close we got. Or, others are obviously welcome to bring the thread back and post their thoughts...

Thanks.

ML4L

[DrSavage]

Let's say probability of winning a freezeout is x
you win immidiately if you flip tails twice and you are in the exact same situation if you flip one tails and one heads
therefore:
x = 0.6 ^ 2 + (0.6 * 0.4 * 2) x
or
x = 0.36 + 0.48 x
therefore x = 0.36/0.52 = 0.6923

[Ilovephysics]

hmm.. it's amazing how simple it is when you just 'think' about it for a second.. I knew there was some reason I took linear algebra, I just didn't know it until know.

[DrSavage]

I did a calculation for 3$ with algebra method while riding subway home and arrived at the very same number, namely
0.6^n/(0.6^n + 0.4^n). So it appears that the formula must be correct, but i still don't see why. I have a problem understanding why probability of having n more wins is 0.6^n, if somebody will get me through this the formula would be correct

[David Sklansky]

[ QUOTE ]
David,

You never posted an answer or critique for the "Way Tougher Two Round Game Theory Problem" that you posted a few months back. Pink Bunny and I (and a couple others) thought that we got the answer, but there was still some doubt. Anyway, if you happen to get a chance, I've been curious about how close we got. Or, others are obviously welcome to bring the thread back and post their thoughts..

ML4L

[/ QUOTE ]

I don't know the answer to that one.

[Terry]

This is a simple game to simulate. (No comments needed on the quick & dirty code.)

================================================== ===========================
RANDOMIZE TIMER: DIM NumFlips, Awon, Bwon, i AS LONG
PRINT "Working..."

BRofA = 2: BRofB = 2: Awon = 0: Bwon = 0

FOR i = 1 TO 20000000

Flip = INT(RND(1) *10)
IF Flip < 6 THEN BRofA = BRofA + 1 ELSE BRofB = BRofB + 1
IF Flip < 6 THEN BRofB = BRofB - 1 ELSE BRofA = BRofA - 1

IF BRofA = 0 THEN Bwon = Bwon + 1: BRofA = 2: BRofB = 2
IF BRofB = 0 THEN Awon = Awon + 1: BRofA = 2: BRofB = 2

NEXT i

PRINT "Player A (60% advantage) won "; Awon; " out of "; Awon; Awon + Bwon; "completed games."
PRINT 100 * (Awon / (Awon + Bwon)); "%"
================================================== ==========================

Repeated runs give 69.2% with the next digit varying from 0 to 5.

[slider77]

I would ratio the winning streaks required for each player.

P(Player 1 Streak) = .6^2 = .36
P(Player 2 Streak) = .4^2 = .16

The game must start at the same "node point" - i.e. the start of the game and a win/loss are at the same point.

So, P(Player 1 Wins) = .36/(.36+.16) = .692

[Ilovephysics]

Could someone re-post this problem (or at least provide a link to the old post)?

[emanon]

I'm quite sure the answer is right. I also wrote a quick excel program to make sure.

hmm, let me see if I can think of a better way of explaining why... its a bit of a mish-mash of stuff, maybe someone like brucez can give a clearer picture if this doesn't help.

1. There are only 2 events which matter:
i. A has 3 more wins than B
ii. B has 3 more wins than A

2. Regardless of the number of flips to reach event i or ii, the difference between Winner_victories and Loser_victories = 3. ie-->
Wins - Loses = 3.

3. In $2
(Wins -2) = Loses
So the only thing that mattered was reaching 2.
In $3
(Wins -3) = Loses
So the only thing that mattered is reaching 3.

4. Therefore, just as we did in the $2 case, we need only concern ourselves with calculating the probabilities of reaching the requisite number of wins. The steps taken to get there (ie, the path) is irrelevant. All that matters is reaching 3.
Thus we the prob (+3) and prob(-3) is the total prob space with which we are concerned.
---------------

Now, to give an example of why this doesn't neccessarily work in a tournament:
Depending on relevant stack sizes, etc, a player may change his actions to adjust his EV/SD. This means that my probability changes depending on the results of the flips(cards), and we can now no longer ignore the path taken to acheive victory.

[ML4L]

Hey emanon,

You're absolutely right. I took a few minutes this afternoon to do the algebraic method and got the same solution that you did. I did understand the concept of the path being irrelevant; for some reason, I still didn't think that your equation was correct...

Thanks for sharing your method; it will save me a lot of time and effort the next time that I'm confronted with a problem to which I can apply it.

ML4L

[ML4L]

Search:

"Way Tougher Two Round Game Theory Problem"
Last 6 months
General Theory

ML4L

[Copernicus]

[ QUOTE ]


Now, to give an example of why this doesn't neccessarily work in a tournament:
Depending on relevant stack sizes, etc, a player may change his actions to adjust his EV/SD. This means that my probability changes depending on the results of the flips(cards), and we can now no longer ignore the path taken to acheive victory.



[/ QUOTE ]

I believe your solution to the 3 problem is correct, the path to net 3 up or down doesnt matter, just that you get there.

Another way to express your reservations about tournament applicability is that the "flips" are not independent events. P(w) varies with stack size, which is, of course, dependent on prior "flips". (On the surface the path to a set of stack sizes shouldnt matter, just the stack sizes themselves. However, in reality the path can matter. A player with a stack of x TCs (no pun intended) may very well behave differently to your actions if his recent flips had him going from a large stack down to x or a smaller stack up to x.)

[slider77]

Basically, the system has no memory.

Once it clears out - i.e. the number of wins and losses are equal - it's like nothing happened in the past.

The same method is used to value stock options with a stock having a probability og going up P(u) or down, P(d). When an Up and Down occur to start, you are right back on the center line. The difference with stock options though is that there is a finite time element. So where the sequence is at the end of that time determines the value of that option for that "run" of the simulation.