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#1
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Cute Applicable Math Question
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. |
#2
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Re: Cute Applicable Math Question
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 |
#3
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Re: Cute Applicable Math Question
Am I barred from the tournament question 2 too? I was going to look at that one.
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#4
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The Answer
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 $3 freezeout (the only method that I know of for solving that one doesn't qualify as "simple"). ML4L |
#5
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Re: The Answer
I prefer your solution for it's simplicity, but mine for it's, ummm, robustness. [img]/images/graemlins/wink.gif[/img]
You can use the same method I describe for solving the 2$ 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. |
#6
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Re: The Answer
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 |
#7
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Simple $3/general freezeout soln
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). |
#8
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Are You Sure?
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 $2 problem. I'll try to post the solution sometime tomorrow (unless someone beats me to it...). ML4L |
#9
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Re: Are You Sure?
I did a calculation for 3$ 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 |
#10
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Re: Are You Sure?
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. |
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