Cute Applicable Math Question

[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.