Gambler's Fallacy

HendrossHendross Posts: 762 Critical Contributor
edited April 2017 in Theories and Statistics
Just a friendly reminder that, you are never due/owed anything in independent trials.

The odds of flipping heads 21 times in a row is 1 in 2,097,152. The odds of the next flip being tails is 50%

Conversely, given 21 attempts, there is a 99.9999523162841796875% chance of getting a heads, the odds of the next flips being tails is still 50%

Comments

  • rossmonrossmon Age Unconfirmed Posts: 130 Tile Toppler
    I've always loved this.
  • rossmonrossmon Age Unconfirmed Posts: 130 Tile Toppler
    My other favorite is the game show paradox:
    There are 3 doors. 2 have a goat behind, and 1 has a car.
    You choose a door. Then, the host opens a different door and shows a goat.
    You have the option to change your answer. Do you keep your original door? or switch to the 3rd door?
  • DaveR4470DaveR4470 Posts: 931 Critical Contributor
    rossmon wrote:
    My other favorite is the game show paradox:
    There are 3 doors. 2 have a goat behind, and 1 has a car.
    You choose a door. Then, the host opens a different door and shows a goat.
    You have the option to change your answer. Do you keep your original door? or switch to the 3rd door?

    This is a great one, indeed. It's fairly simple to explain on a very, very basic level: there are three possible iterations of the set (car/goat/goat, goat/car/goat, and goat/goat/car); although revealing a goat eliminates one of them (and therefore you'd think that the odds your door would have a car would be 50/50), it does NOT eliminate half of the "losing" iterations. There are still two potential outcomes that have a goat under door 1. Ergo, you have a 1/3 statistical chance of winning the car. Because you know a goat is under door 3, there is only one potential outcome that has a goat under door 2. Ergo, your odds of winning if you switch are 2/3.

    This, however, is very, very, VERY difficult to prove mathematically, and the odds actually vary quite widely based on what conditions you assume. (The base scenario assumes that the host MUST pick a goat AFTER you select your door.)
  • simonsezsimonsez Posts: 4,663 Chairperson of the Boards
    Trying to convince someone of the right answer to the Monte Hall puzzle can be a great way to ruin a friendship.
  • rossmonrossmon Age Unconfirmed Posts: 130 Tile Toppler
    simonsez wrote:
    Trying to convince someone of the right answer to the Monte Hall puzzle can be a great way to ruin a friendship.
    I'm quite sure!
  • HendrossHendross Posts: 762 Critical Contributor
    I've lost about 50% of friends explaining it, those who got it find it easier with 100 doors and opening 98 of them.

    "Imagine that instead of 3 doors, there are 100. All of them have goats except one, which has the car. You choose a door, say, door number 23. At this point, Monty Hall opens all of the other doors except one and gives you the offer to switch to the other door. Would you switch? Now you may arrogantly think, “Well, maybe I actually picked the correct door on my first guess.” But what’s the probability that that happened? 1/100. There’s a 99% chance that the car isn’t behind the door that you picked. And if it’s not behind the door that you picked, it must be behind the last door that Monty left for you. In other words, Monty has helped you by leaving one door for you to switch to, that has a 99% chance of having the car behind it. So in this case, if you were to switch, you would have a 99% chance of winning the car."
  • Nate7671Nate7671 Age Unconfirmed Posts: 47
    Mind blown........

    Coin one I knew - door one is new to me....
  • mrflopwelliganmrflopwelligan Age Unconfirmed Posts: 125
    Mythbusters tackled the game show one - pretty eye opening. But they confirmed it!
  • WarbringaWarbringa Posts: 1,157 Chairperson of the Boards
    Yeah it is true but the main reason people stick to their first choice is that they view the 2 doors as a new data set, at which point yes the probability is 50/50 that your door is the winning door, and go with their "gut" instinct that their original pick is right.
  • blinktagblinktag Age Unconfirmed Posts: 157 Tile Toppler
    If you haven't seen it, the Wikipedia article on the Monty Hall
    Problem is quite thorough, and explains which assumptions are necessary for the various conclusions to hold (blind host versus devious host, etc.)

    https://en.m.wikipedia.org/wiki/Monty_Hall_problem
  • morph3usmorph3us Age Unconfirmed Posts: 859 Critical Contributor
    blinktag wrote:

    Good article. I love this bit from it:

    "Pigeons repeatedly exposed to the problem show that they rapidly learn always to switch, unlike humans."
  • Pwuz_Pwuz_ Posts: 1,168 Chairperson of the Boards
    But my Gut! It's mostly right half the time!
  • PolaresPolares Posts: 2,495 Chairperson of the Boards
    Hendross wrote:
    I've lost about 50% of friends explaining it, those who got it find it easier with 100 doors and opening 98 of them.

    "Imagine that instead of 3 doors, there are 100. All of them have goats except one, which has the car. You choose a door, say, door number 23. At this point, Monty Hall opens all of the other doors except one and gives you the offer to switch to the other door. Would you switch? Now you may arrogantly think, “Well, maybe I actually picked the correct door on my first guess.” But what’s the probability that that happened? 1/100. There’s a 99% chance that the car isn’t behind the door that you picked. And if it’s not behind the door that you picked, it must be behind the last door that Monty left for you. In other words, Monty has helped you by leaving one door for you to switch to, that has a 99% chance of having the car behind it. So in this case, if you were to switch, you would have a 99% chance of winning the car."


    But I don't think people really understands the problem more easily with 100 doors, they just chose to change more because now in their mind it goes from 1/99 originally to 50/50 after the 98 doors are opened (because then there is just two doors, so then it is 50% for them), so it is better to change because now chances look higher on the second time (50% > 1%), but not because they understand that if they change they will have 99% chance of winning, which is the difficult part to grasp.

    I always struggle with this problem, the 50/50 always come like the solution and I have to remember that Monty Hall will always open doors with goats, this is very important, always will ask you to switch (not just sometimes) and it is all part of the same problem, not two independent problems. But it is one of those super counterintuitive problems, even for Mathematicians (you just need to read the first batch of replies the author got when the paper was presented).
  • DismasDismas Posts: 3 Just Dropped In
    DaveR4470 wrote:
    rossmon wrote:
    My other favorite is the game show paradox:
    There are 3 doors. 2 have a goat behind, and 1 has a car.
    You choose a door. Then, the host opens a different door and shows a goat.
    You have the option to change your answer. Do you keep your original door? or switch to the 3rd door?

    This is a great one, indeed. It's fairly simple to explain on a very, very basic level: there are three possible iterations of the set (car/goat/goat, goat/car/goat, and goat/goat/car); although revealing a goat eliminates one of them (and therefore you'd think that the odds your door would have a car would be 50/50), it does NOT eliminate half of the "losing" iterations. There are still two potential outcomes that have a goat under door 1. Ergo, you have a 1/3 statistical chance of winning the car. Because you know a goat is under door 3, there is only one potential outcome that has a goat under door 2. Ergo, your odds of winning if you switch are 2/3.

    This, however, is very, very, VERY difficult to prove mathematically, and the odds actually vary quite widely based on what conditions you assume. (The base scenario assumes that the host MUST pick a goat AFTER you select your door.)


    Myth busters tested this.... always switching won something like 80 percent of the time
  • hangehange Posts: 6 Just Dropped In
    While I fully understand the Monte Hall theory there is something it doesn't take into account.

    After the door is opened to remove one of the losing options the choice of whether to move to a new door or not is now 50/50.

    So the initial choice has a 1/3 chance in winning. That much is obvious.

    The second choice has a 1/2 chance of winning if you change doors. Also obvious.

    However, based on the new choice of switching or staying you now have a 1/2 chance of winning no matter which option you choose.

    I'm sure I'll be flamed to death for this but I'm right if you think about it.
  • DayvDayv Posts: 4,449 Chairperson of the Boards
    I shouldn't really be encouraging a necro thread like this, but the key reason it's not 1/3 or even 1/2 is because the host is aware of where the goats are, and whether or not you selected the non-goat prize. Two-thirds of the time, you selected a goat, so the host only has one other goat to show you and you can now switch to the not-goat using the reversed odds of your original choice. However, one-third of the time you have selected not-goat, so it doesn't matter which prize the host reveals, because you're going to see a goat either way.

    It's that simple: the host's choice is fully informed and not random*, and inverts your original guessing odds only if you switch.

    On the other hand, if you really want a goat, hold on to your original choice. Goats are pretty rad.
    giphy.gif
    * Dayv's pedantic corner: ok, sure, the host's selection of a goat could be random when the host has two unselected goats to choose from, but really he's always going to pick the cuter one. The host gets to keep that goat.
  • BGonBGon Age Unconfirmed Posts: 5
    I'll echo the sentiment others have said - don't try explaining this to your friends/family.

    Well said, DayvBang, especially the emphasized words.

    For those that still don't get it, here's a quick YouTube video: Monty Hall Problem for Dummies.
  • hangehange Posts: 6 Just Dropped In
    It isn't that I still don't get it. I was just pointing out the logical fallacy.

    Again. Original choice. You'll be right 1 out of 3 times. Obvious.

    Now one wrong choice is removed.

    Second choice. Key word here is choice. Either switch door or stay.

    Choose to switch equals 50/50
    Choose not to switch is now also 50/50.

    Your odds ONLY increase from the original choice if there is NO second choice. As in, after you select the original door the host will remove one wrong answer ONLY if you agreed to switch without being given that second choice.
  • MarkersMakeMarkersMake Posts: 392 Mover and Shaker
    hange wrote:
    It isn't that I still don't get it. I was just pointing out the logical fallacy.

    Again. Original choice. You'll be right 1 out of 3 times. Obvious.

    Now one wrong choice is removed.

    Second choice. Key word here is choice. Either switch door or stay.

    Choose to switch equals 50/50
    This is your mistake.

    Regardless of what you choose to do, the car is behind the remaining door 2/3 of the time. You should probably switch if you want to win it.
  • ejm04ejm04 Age Unconfirmed Posts: 47 Just Dropped In
    edited April 2017
    Here are the scenarios of the Monty Hall case.

    You pick Door 1, then Monty reveals a goat:

    Door1Door2Door3SwitchStay
    CarGoatGoatLoseWin
    GoatCarGoatWinLose
    GoatGoatCarWinLose

    Switch 67% Stay 33%


    You pick Door 2, then Monty reveals a goat:

    Door1Door2Door3SwitchStay
    CarGoatGoatWinLose
    GoatCarGoatLoseWin
    GoatGoatCarWinLose

    Switch 67% Stay 33%


    You pick Door 3, then Monty reveals a goat:

    Door1Door2Door3SwitchStay
    CarGoatGoatWinLose
    GoatCarGoatWinLose
    GoatGoatCarLoseWin

    Switch 67% Stay 33%

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