Cute Applicable Math Question

[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).

[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

[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

[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

[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.