View Full Version : Gambler's fallacy

joseph0000

07-11-2010, 12:45 PM

Hi everybody. I have a question about the gambler's fallacy application. I have this board game called "Talisman", where each player has an attribute called "fate", a number of "rerolls" that a player can use when the last dice result doesn't like, i.e., when a player throws a dice, and he/she doesn't like the number obtained, it can spend a "fate" to throw the dice again. He/she must take the new result.

Talking with my friends, I conclude that the "fate" is an illusion because the gambler's fallacy. Suppose that the player requires a "6" with its dice in a throw. He has a probability of 1/6 of obtain that number. When the player throws the dice and obtain another number then he spends a "fate" to throw the dice again, with the same probability: 1/6.

I say then that with or without fate, you have the same probability, and throwing again the dice is an illusion. Everybody argues to me that you have another "chance" of throwing, so this gives you more "chances" of obtain the number that you want to. Some understand my explanation and agree about my calculations, however, they maintain the "chance" argument, but cannot explain mathematically, the explanation is always intuitive.

Can anyone tell me if I'm wrong and why? Thanks!

that depends on your objective (payoff function).

E.g. If you only interested in whether the outcome is "6" or "not 6", (binary)

i.e. the number from 1 to 5 are indifferent to you and they are all

considered to be failure, throwing once again will give you a better chance

to obtain a success throw since your strategy will be "throw the dice until you

get a "6" eventually. So whenever you did not get a "6", throw again.

More mathematically, the probability of getting the first 6 with throwing

1 time = \frac {1} {6}

2 times = \frac {1} {6} + \frac {5} {6} \times \frac {1} {6}

= \frac {11} {36} > \frac {1} {6}

You got an extra chance to improve your outcome in the second throw.

You can say the payoff function for this game is that

f(X) = \left \{ \begin{matrix} 1 , X = 6 \\

0, X = 1, 2, 3, 4, 5 \end{matrix} \right.

and E[f(X)] is increasing when number of

available throws increases.

However, if your payoff function is not like this, say you just want to have a

bigger number, as large as possible. Then if you get a "5" in the first throw,

you maybe reluctant to throw again as you may expect you get a even

smaller number in the next throw. That's why this may give you a feeling that

even you get one more throw, it may not improve your outcome.

But you should also consider when you get a "1" in the first throw.

Then you will surely throw again if "1" is the worst result.

If it is not the worst, you need to consider the expected change to see whether

throw again or not. It depends on your own payoff function in the game.

If you already obtained the best result, you will surely terminate the throw.

joseph0000

07-11-2010, 02:16 PM

I don't understand the explanation for the two times step. What is the concept that you use for the two throws to say that has more probability? (I don't know enough about probability!)

SadieKhan

07-11-2010, 03:08 PM

I can understand where you are coming from joseph,

You are right, since the probability of getting a 6 is the same in both trials , this is a fallacy, it might be mere luck to get a 6 in second throw, as you cant take the side of the dice.

If you got a 4 and you didn't like the outcome you gamble your "fate" to get another try, but the probability is the same in both trials, Lets say you had 6 cards numbered 1 to 6 and you could take the card out if it is other than 6 and you use your fate and getting 6 numbered card`s probability will be 1/5 then and you are likely to get a better chance but in the case of dice probability remains the same in both trials.

Just my two cents.

Sadie

Dason

07-11-2010, 03:57 PM

On the second toss there is still only a 1/6 probability of getting a six.

However, this isn't really what's important. What is important is analyzing beforehand whether or not it helps to use "fate" and toss a second time.

So we want to know if the strategy to possibly use fate can increase the probability of getting what you want.

Pr(Getting a six given that you will use fate if you need to) = Pr(Get a six on first roll) + Pr(Had to use fate and you get a six on the second roll) = 1/6 + (5/6)(1/6) = 6/36 + 5/36 = 11/36

So this is greater than 1/6 so the strategy of using fate if you need to will increase your probability of getting what you want when we consider this as a strategy you will use beforehand. So using fate can help.

Now you're right in that once you know you failed the first throw then you still only have a 1/6 probability of getting it on the second but that doesn't mean fate isn't helpful.

In summary:

If you decide you'll never use fate then you will only ever have a 1/6 probability of getting a 6.

If you decide you'll use fate when necessary then you increase the probability of getting a 6 in that turn to 11/36 which is greater than 1/6.

maybe i should say you are not "forced" to take the second throw.

E.g. I simplified the game as a coin tossing game. Whenever you get a head,

you will win the game.

If you are given 1 toss only, you have one half chance to win.

Say you are forced to toss 2 times now and only the result of the second toss

will be taken into account, the result of the first toss is irrelevant.

(independent tosses) So you still have on half chance to win.

But if you are not forced to toss the second time, i.e. you have the freedom to choose take the second toss or not. If you already get a head in the first

toss, you win and choose not to take the second toss; else you have a an

extra chance: you can forget about the first toss and try your luck on the

second toss.

joseph0000

07-11-2010, 10:45 PM

I like the coins sample, it is used in the gambler's fallacy. The coins are like a two sided dice, so the results of a toss are 1 and 2, both with a 50% chance of success.

Let's suppose that you have fate and that the desired result is a 2.

Throwing always two times the coin, the possibilities are (1, 1), (1, 2), (2, 1) and (2, 2). It is easy to see that we have a 75% chances to obtain a 2 throwing two times.

However, that's not what we are doing with the fate. With the fate, we throw the first time and obtain a 1, then the possible results for our throw are (1, 1) and (1, 2), because (2, 1) and (2, 2) are actually impossible (we already obtain a 1 in the first throw). So between (1, 1) and (1, 2) the chances are 50%, and spending the fate point doesn't improve our probability of success. If you obtain a 2 thanks spending fate, it is just luck and it was not a result of the fate help.

Is this reasoning correct? Maybe I miss something.

Dason

07-11-2010, 10:54 PM

You're sort of mixing viewpoints here. You're trying to look at things from both the viewpoint of before throwing any die and at the same time after throwing one and not getting the desired result. Read my previous post if you don't understand what I'm saying.

Another way to think of it is this: You throw the first die, you don't get what you want. Should you use fate? You argue that since there is still only a 50% chance it doesn't matter but think about if you don't use fate... you have a 0% chance of getting what you want. Clearly fate is helping you here.

TheEcologist

07-11-2010, 11:18 PM

You're sort of mixing viewpoints here. You're trying to look at things from both the viewpoint of before throwing any die and at the same time after throwing one and not getting the desired result. Read my previous post if you don't understand what I'm saying.

Another way to think of it is this: You throw the first die, you don't get what you want. Should you use fate? You argue that since there is still only a 50% chance it doesn't matter but think about if you don't use fate... you have a 0% chance of getting what you want. Clearly fate is helping you here.

Dason is right to put it in a game theorectical perspective. Weigh the risk with the reward in mind - though that is technically more strategy than stats.

This does however remind me of a previous post you might find interesting:

http://www.talkstats.com/showthread.php?t=8205

joseph0000

07-11-2010, 11:25 PM

Yes Dason, thanks. I've read your post before. However, I don't understand this (using coins, is easier for me ;):

Pr(Getting a 2 given that you will use fate if you need to) = Pr(Get a 2 on first roll) + Pr(Had to use fate and you get a 2 on the second roll) = 1/2 + (1/2)(1/2) = 1/2 + 1/4 = 3/4

I can use the same logic with a casino's roulette: I have money to bet two times, but I only bet the second time if I lose the first one.

If I bet to reds in the previous roll and fail, then using the rest of the money (fate money) to bet red in the second will improve my chances to 3/4, in the next roll. This conclusion cannot be true, the casinos will go to ruin! I can be even clever and just watch four blacks and bet to reds in the fifth time!

You right about 0% chances when I fail, and fifty using fate after fail. However, having fate or not, my chances of obtain a 2 are 50% and fate doesn't count for this.

Dragan

07-12-2010, 12:44 AM

Dason is right to put it in a game theorectical perspective. Weigh the risk with the reward in mind - though that is technically more strategy than stats.

This does however remind me of a previous post you might find interesting:

http://www.talkstats.com/showthread.php?t=8205

Very good.

I was wondering if someone would find this review.:)

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