My Math Forum Coin Flipping: Permutation vs. Combination

 February 19th, 2010, 07:07 PM #1 Newbie   Joined: Mar 2008 Posts: 15 Thanks: 0 Coin Flipping: Permutation vs. Combination I appologize in advance. It was a struggle for me to write this question. I've been thinking about Gambler's Fallacy. As I understand things, every time you flip a coin or roll a dice, the odds reset. The coin or dice have no memory of the past. So if you flip a coin twice the odds aren't 1/4 to get two heads in a row, they are 50/50. The coin remains independent. If you flip a fair coin twice, on the first toss you have "2" possible Permutations? H T On the second toss "4" possible permutations HH HT TH TT The first toss has "2" possible results. A 50% chance of Heads. The second toss has "4" possible results. Still with a 50% chance of Heads. By the second toss, each Permutation has the same chances. HH 1/4 HT 1/4 TH 1/4 TT 1/4 The second throw has 2/4 Heads and 2/4 Tails. So the chances are 50% to get heads. If you increase the number of throws, the odds are the same for each Permutation. Out of 5 flips TTTTT is just as likely as THTHT. Although I'm not exactly sure these conclusions are correct, but I've been thinking about Combination versus Permutation. When you roll one six sided dice the result is independent. When you roll two sided dice, each result is independent. A dependency can be created by adding both together to get a range of 2-12.
February 19th, 2010, 08:42 PM   #2
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Re: Coin Flipping: Permutation vs. Combination

Quote:
 Originally Posted by Brimstone I appologize in advance. It was a struggle for me to write this question. I've been thinking about Gambler's Fallacy. As I understand things, every time you flip a coin or roll a dice, the odds reset. The coin or dice have no memory of the past. So if you flip a coin twice the odds aren't 1/4 to get two heads in a row, they are 50/50. The coin remains independent.
If you flip one coin and get H, the chance of getting HH with a second flip is 1/2. If you haven't flipped either yet, the chance is 1/4.

The gambler's fallacy is thinking that, if you've already flipped (say) HHH, you're more likely to get T than H because it's 'due'. Of course this is false: as you say, the coin has no memory.

 February 19th, 2010, 10:11 PM #3 Newbie   Joined: Mar 2008 Posts: 15 Thanks: 0 Re: Coin Flipping: Permutation vs. Combination Ok thanks for the answer. I think what throws me off is "the probability of any sequence is the same" or "every outcome observed will always have been equally as likely as the other outcomes that were not observed for that particular trial, given a fair coin. Therefore, just as Bayes' theorem shows, the result of each trial comes down to the base probability of the fair coin: 1/2." If the question is asked, "what are the odds the coin comes up heads on the second toss". The answer is 2/4. This would be looking for the idependence. When it is asked what are the odds of two TT by the second toss, they would be creating Dependence on the first throw. A coin toss is an Independent variable. The first and second. How do you "test" or demonstrate Independence?
February 20th, 2010, 02:20 PM   #4
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Re: Coin Flipping: Permutation vs. Combination

Quote:
 Originally Posted by Brimstone How do you "test" or demonstrate Independence?
I don't understand the question. Can you give an example of the sort of situation you're unsure about, and what would be true if it was independent vs. dependent?

 February 21st, 2010, 08:23 AM #5 Newbie   Joined: Mar 2008 Posts: 15 Thanks: 0 Re: Coin Flipping: Permutation vs. Combination Let me start over. Instead of labeling a coin Heads and Tails, lets call it 1 and 2; a two sided dice. If I flip it one time there are two possible outcomes: 1 and 2. If I flip it twice there are four possible outcomes: 11 12 21 22 If I combine the numbers I get a range from 2-4. I fully grasp that the odds of getting a total of "2" are 1/4 I fully grasp that the odds of getting a total of "4" are 2/4 I fully grasp that the odds of getting a total of "3" are 1/4 A bell curve is formed in the number distribution. Just like when 2d6 are rolled. The confusion stems from independence. It is declared a result of "1" is "pass" and a result of "2" is fail. Flip the coin once and you have a 50% to pass. Flip the coin three times and you have a 50% chance to pass on each attempt because the coin is Independent. You don't have a 1/8 chance of passing on the third try. There is "no memory" of past results. Another way to put it is playing a game like D&D. A player has a 50% skill to hit with a sword. First round the player has a 50% skill chance to hit. Lets say he rolls a hit. Second round the player still has a 50% skill chance. Since the dice have no memory, the chance is still the same as the first round. How is this expressed? Each time the coin (or dice) are used the odds are reset. Fundamental Counting Principle? Another example roulette wheel. Player 1 bets on black one time. Player 2 bets on black ten times. It doesn't matter because each spin of the wheel resets the odds. There is independence. Every permutation is just as likely.
February 21st, 2010, 10:40 AM   #6
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Re: Coin Flipping: Permutation vs. Combination

Quote:
 Originally Posted by Brimstone Instead of labeling a coin Heads and Tails, lets call it 1 and 2; a two sided dice.
A small quibble, because I can't help myself: that's "die", not "dice". "Die" is singular, "dice" plural.

Quote:
 Originally Posted by Brimstone The confusion stems from independence. It is declared a result of "1" is "pass" and a result of "2" is fail. Flip the coin once and you have a 50% to pass. Flip the coin three times and you have a 50% chance to pass on each attempt because the coin is Independent. You don't have a 1/8 chance of passing on the third try. There is "no memory" of past results.
OK, so where is the confusion?

Quote:
 Originally Posted by Brimstone Another way to put it is playing a game like D&D. A player has a 50% skill to hit with a sword. First round the player has a 50% skill chance to hit. Lets say he rolls a hit. Second round the player still has a 50% skill chance. Since the dice have no memory, the chance is still the same as the first round. How is this expressed? Each time the coin (or dice) are used the odds are reset.
Still not understanding your issue. So two chances to hit, overall possibilities: 25% two hits, 50% one hit, 25% zero hits. Each chance is independent.

Quote:
 Originally Posted by Brimstone Another example roulette wheel. Player 1 bets on black one time. Player 2 bets on black ten times. It doesn't matter because each spin of the wheel resets the odds. There is independence. Every permutation is just as likely.
Each bet is independent. Different permutations have different odds, though, since the chance of winning is not 50/50 in roulette. (I had to look up the odds online -- 18/38 for American-style wheels and 18/37 for European-style wheels -- but it's obvious that they're not even odds because the casino needs to make money.) The single most likely permutation is lose-lose-lose-lose-lose-lose-lose-lose-lose-lose (though this is not very likely!).

 February 21st, 2010, 12:36 PM #7 Newbie   Joined: Mar 2008 Posts: 15 Thanks: 0 Re: Coin Flipping: Permutation vs. Combination Ok. The RPG system has healing rolls of d100%. Depending on the severity your "Healing Rate" will be higher or lower. Like a 80% chance on a minor wound. If the d100% dice land on a "0" or "5" it is considered a critical of some sort. Example: 60% skill and you roll a "35", so that is a Critcal Success 60% skill and you roll a "41", so that is a Marginal Success 60% skill and you roll a "83", so that is a Marginal Failure 60% skill and you roll a "70", so that is a Critical Failure One method for the "healing process" is to roll daily. If you roll Critical Success the wound heals "2" points. If you roll a Marginal Success the wound heals "1" point. If you roll a Marginal Failure the wound doesn't heal any points. If you roll a Critical Failure the wound becomes Infected. The second method for the "healing process" is to roll for "5" day chunks. If you roll Critical Success the wound heals "10" points. If you roll a Marginal Success the wound heals "5" point. If you roll a Marginal Failure the wound doesn't heal any points. If you roll a Critical Failure the wound becomes Infected. If a character has a "20" point wound, does the "5 day chunk" method make it less likely to become infected? (rolling a critical failure) My observation is that both methods have the same chance of infection since each "trial" or "healing test" is independent of the previous roll.
February 21st, 2010, 01:27 PM   #8
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Re: Coin Flipping: Permutation vs. Combination

Quote:
 Originally Posted by Brimstone If a character has a "20" point wound, does the "5 day chunk" method make it less likely to become infected? (rolling a critical failure)
Much less likely. You'll make about 1/5 as many rolls.

February 26th, 2010, 11:40 AM   #9
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Re: Coin Flipping: Permutation vs. Combination

I came across this.

Quote:
 The differences causing the bickering are obviously -- and this has been pointed out several times -- those between ordered sets (permutations) and un-ordered sets (combinations). The ordered set HTTHH is different from the ordered set HTHHT; but if they're considered to be un-ordered sets, they're identical. Thus, the ordered set HTHHT is equally probable as HHHHH. The un-ordered set HTHHT (or HTTHH or any of the other combinations) is more probable than HHHHH, precisely because of those various combinations. Now, if only ordered sets are under consideration, there is still a tendency for people to be more "surprised" by the occurrence of HHHHH than by, say, HTHTH -- but this is merely psychological, just as the Gambler's Fallacy (which might motivate one to choose T as the next in the latter sequence) is psychological. Each is equally rare, when ordered sets are considered.
http://www.sciencechatforum.com/view...13891&start=30

The way I'm understanding things...

If you throw 2d6 (one red the other green) and read them individually that would be an ordered set.

If you throw 2d6 (one read the other green) and read them together that would be an un-ordered set.

Isn't rolling 1d100 20 times and testing a skill an ordered set?

February 26th, 2010, 07:00 PM   #10
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Re: Coin Flipping: Permutation vs. Combination

Quote:
 Originally Posted by Brimstone If you throw 2d6 (one red the other green) and read them individually that would be an ordered set. If you throw 2d6 (one read the other green) and read them together that would be an un-ordered set.
I don't know what you're saying. If you count order it's ordered/permutation; if not, unordered/combination. Sums are not the same as unordered:

1,1,3 is the same combination as 1,3,1, but they're different as permutations
1,1,3 has the same sum as 1,2,2, but they're different as combinations

Quote:
 Originally Posted by Brimstone Isn't rolling 1d100 20 times and testing a skill an ordered set?
I don't know exactly what you're saying, but I think the answer is "no".

If you're rolling with one of the two methods you described, stopping when
* you recover all hit points, or
* you roll a critical failure

there is a particular distribution in terms of number of days required or eventual outcome.

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