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Need help! My stupid friend thinks coin flips are 50-50!

  1. Jan 18, 2005 #1
    Need help! My friend thinks coin flips are 50-50!

    Alright, I've been trying to convince my friend that the outcomes of a coin flip x times in a row affects the x+1'th time. If you flip a coin 4 times and they are all heads, the 5th time is more likely to be a tail because if the coin is even, over time there should be as many tails as there are heads. I even tried to prove it to him using math:
    P(5 heads in a row) = .5^5 = 0.03125
    P(4 heads in a row) = .5^4 = 0.0625

    Which directly translates into: 5 heads in a row is less likely to happen than just 4 heads in a row so if you flip a coin and get 4 heads, the 5th time is more likely to be a tail.

    My friend just doesn't understand this, he's saying that even when I'm calculating the probabilities, I'm using .5 for each flip so the probability should be .5 every time, but he's obviously wrong, because the probabilities are different! Please help me convince him that my way is right...after all why does he think when they do coin flips, they do best out of 3 times???
     
    Last edited: Jan 18, 2005
  2. jcsd
  3. Jan 18, 2005 #2
    I have a feeling this is going to be an interesting thread. :biggrin:

    P.S. If you remove the insulting adjective before the word friend in your title you might get more consideration for reasonable replies. :wink:
     
    Last edited: Jan 18, 2005
  4. Jan 18, 2005 #3
    How do I do that?

    Never mind...figured it out.
     
  5. Jan 18, 2005 #4

    Hurkyl

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    Try computing P(4 heads followed by a tails). How does it compare to P(4 heads followed by a heads)?
     
  6. Jan 18, 2005 #5
    They're equal, but how does that prove that what I said is true?
     
  7. Jan 19, 2005 #6

    Hurkyl

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    It doesn't: it supports your friends position.
     
  8. Jan 19, 2005 #7

    Chronos

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    Statistically, the odds of n number of consecutive results, given a 50-50 outcome probability, is 2^(n-1). Don't bet with your stupid friend unless you don't mind getting sheared like a cashmir goat.
     
  9. Jan 19, 2005 #8

    matt grime

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    I didn't realize it was a rule that you must toss a coin "best of 3".
     
  10. Jan 19, 2005 #9
    Waaaait wait a second here. So you all are saying that my friend is right?
    But the probabilities say that if you flip a coin 6 times, you should get about 3 heads and 3 tails, so if you flip it 5 times and get 5 heads the 6th time should almost deffinately be a tail.
     
  11. Jan 19, 2005 #10

    chroot

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    Your friend is right, and you are wrong, without question. The coin flips are entirely independent. Everytime you flip a coin, you have a 50-50 chance of getting both heads or tails, no matter what happened before.

    Wrong, without question. The probability is that half the time you flip a coin, you'll get heads, and half the time you'll get tails, on average. That's all.

    Your logic is actually what suckers people into playing games like roulette. In casinos, roulette tables have a big flashy sign above them that lists the results of the last ten spins of the wheel. Passing gamblers see that a wheel has spun, say, ten blacks in a row, and they can't help but think there's a higher chance to spin red next. That false hope helps make the casino richer.

    To illustrate your thought process, let's look at the possible results of spining a roulette wheel twice:

    BB
    BR
    RB
    RR

    Note that there are two possible sequences with one black and one red spin, but only one each of all black or all red. This leads people -- including yourself -- to mistakenly think that sequences that have an equal mixture of reds and blacks are more likely than sequences of all red or all black. This is not true!

    Now let's look at the possibilities with 10 spins:

    All blacks (1 possibility)
    Some mixture of blacks and reds (1,022 possibilities)
    All reds (1 possibility)

    Do you see what's happening here? There are 1,024 possibilites, only two of which are all black or all red. That leads people -- including yourself -- into thinking that a sequence of all reds or all blacks is highly improbable. It's not! The sequence BBBBBBBBBB is no less probable than the sequence RBRRBRBRBB. Sure, one looks more "random," but, in fact, both sequences have exactly the same probability. Every possible sequence of spins has a 1/1,024 chance of happening.

    Richard Feynman one explained this misconception to a class by announcing that he had just come in from the parking lot, where he saw a car with the license plate ARW457. He expressed his amazement that, of all the millions and millions of license plates in the state of California, he had seen that one!

    Sometimes people refer to "the law of large numbers" when dealing with probabilities. Only if you flip the coin a large number of times can you be certain of getting 50% heads and 50% tails. If you flip it just once, obviously you don't -- you get either 100% heads or 100% tails. Only if you flip the coin an infinite number of times, in fact, are you guaranteed of getting 50% heads and 50% tails.

    Your friend is the smart one, Peng, and you are the one who is "stupid." Coin flips are independent. No matter what you flipped in the past, the probability of each flip is 50-50. The universe does not conspire to make coins work differently from one flip to the next. Even if you flipped a coin ten billion times and it came up heads every time, there is no greater chance of it being tails the next. Your friend is a smart one -- stick around, and you might learn something from him/her.

    - Warren
     
  12. Jan 19, 2005 #11

    chroot

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    No, it doesn't. What you actually compared was the probability of any individual 5-flip sequence to the probability of any individual 4-flip sequence. Since there are more possible 5-flip sequences than possible 4-flip sequences, it's clear that each invidual sequence's probability should be smaller.

    Why don't you instead compare these two sequences:

    P(HHHHH) = 0.5^5 = 0.03125
    P(HHHHT) = 0.5^5 = 0.03125

    - Warren
     
  13. Jan 19, 2005 #12

    Hurkyl

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    Here are some other ways of internalizing this fact:

    You've flipped 5 coins, and got 5 heads. Now, you are performing an experiment with one coin flip -- and you know that one coin flip has a 50% chance of being heads.

    getting 5 heads and 1 tail is more likely than getting 6 heads, because there are more ways to get 5 heads and a tail:

    HHHHHT
    HHHHTH
    HHHTHH
    HHTHHH
    HTHHHH
    THHHHH

    vs

    HHHHHH

    However, because you started with 5 heads, you've ruled out all of those other possibilities that made 5 heads and a tail more likely than 6 heads.


    And to help internalize your mistake (which, as chroot mentioned, is common), you rationalized that on 6 coin flips that you are most likely to get 3 heads and 3 tails... but that's absolutely impossible once you've seen 5 heads, so you should know that there was a kink in your reasoning.
     
  14. Jan 19, 2005 #13
    yes. coin tossing contains both markov and martingale properties, and are entirely independent from one another. This same principle works with payoff, volatility and drift terms. Your expected payoff will be what you currenty have up to [tex] P_n_-1 [/tex] days will just be the current amount you have.
     
    Last edited: Jan 19, 2005
  15. Jan 19, 2005 #14

    Galileo

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    Wouldn't it be nice if Peng was right?

    I would toss a coin 5 times till I got 4 heads in a row on the first 4 throws of the sequence.
    Then I'd walk up to my brother and say: "Wanna bet 10 euro's that this coin toss gives head?"
    The odds would be greatly in my favour.
     
  16. Jan 19, 2005 #15
  17. Jan 19, 2005 #16
    A Computer Simulation Program

    I never trust mathematics so I had to prove to myself that Peng was wrong by writing a simulation program. :biggrin:

    Years ago, in 1996 I had a bit of a problem believing this one:

    The Monty Hall Dilemma

    After writing a BASIC program to simulate the doors I could see that Marilyn was right. Of course, that problem also loans itself to intuitive insight by simply imagining that there are thousands of doors.

    In any case, I know that coin tosses must always be 50/50 because there's no way that history can change how a coin behaves. None the less, the Monty Python Dilemma does seem to suggest that since we have more information this might somehow have an effect on the probabilities. So I wrote the following Visual Basic program to toss the coin and keep track of the results.

    The program pasted below is hard-wired to only tally up the heads or tails that come up after 4 heads in a row. It also only does this for 1000 tosses. I actually ran the program to toss millions of times, and played with various numbers of heads in a row. The results were always 50/50 after millions of tosses.

    Anyhow, for anyone who's interested in BASIC programming here's the program code. It uses a form with a Flip Coin button called cmdFlip, and has three text boxes named, txtTallyHeads, txtTallyTails, and txtPaper where I print out the actual results of tossing a coin 1000 times. I place underscores after every occurrence of N-heads in a row so they can easily be found in the print out. I've also attached a small picture of the output. I had to clip it down to fit into PF's 400 pixel limit for GIFs.

    It seems that Peng's friend is right. Which I figured had to be the case since coins have to be 50/50 no matter what. But in light of the Monty Python Dilemma I thought I'd just simulate it to be sure. Sorry Peng, you were wrong. :yuck:

    Here's the code for the VB program: (be sure to scroll down to see the last part)
    A clipped image of the output of 1000 tosses is also attached.



    Code (Text):

    Option Explicit
    Dim Toss As Double: Rem For the coin tossing loop.
    Dim Flip As Double: Rem Just a random flip of the coin
    Dim Coin As String: Rem Keeps track of how the coin landed
    Dim N As Integer: Rem The number of heads in a row to test.
    Dim CountHeads As Integer: Rem Counts heads in a row.
    Dim TallyHeads As Integer: Rem Tallys # of heads after N-heads
    Dim TallyTails As Integer: Rem Tallys # of tails after N-heads

    Private Sub cmdFlip_Click()
    Rem This routine actually flips the coins
    TallyHeads = 0
    TallyTails = 0
    N = 4
    txtPaper.Text = "": Rem clear the paper
    For Toss = 1 To 1000
    Flip = Rnd(1)
    If Flip < 0.5 Then Coin = "T" Else Coin = "H"
    txtPaper.Text = txtPaper.Text + Coin
    If CountHeads = N Then Call Tally_Toss
    Call Count_Heads
    Next Toss
    End Sub

    Public Sub Count_Heads()
    Rem This routine counts N heads in a row
    If Coin = "H" Then CountHeads = CountHeads + 1
    If Coin = "T" Then CountHeads = 0
    End Sub

    Public Sub Tally_Toss()
    Rem This routine tallies the heads or tails after N-heads in a row.
        If Coin = "H" Then TallyHeads = TallyHeads + 1
        If Coin = "T" Then TallyTails = TallyTails + 1
        txtTallyHeads = Str(TallyHeads)
        txtTallyTails = Str(TallyTails)
        txtPaper.Text = txtPaper.Text + "_____": Rem Just a marker
        CountHeads = 0
    End Sub
     

    Attached Files:

    • Coin.GIF
      Coin.GIF
      File size:
      12.5 KB
      Views:
      406
    Last edited: Jan 20, 2005
  18. Jan 20, 2005 #17

    cepheid

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    It's Monty HALL...the host of the game show Let's Make a Deal. And I don't understand how it cast into doubt your understanding of whether coin tosses are 50/50 or not. The situations don't seem related in any way...why bring it up?
     
  19. Jan 20, 2005 #18

    matt grime

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    And why say marilyn was right when she got it wrong because she didn't know anything about conditional probablity? I trust you built the basic program to allow for the fact that the door they open is KNOWN to not have the car behind it?
     
  20. Jan 20, 2005 #19

    HallsofIvy

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    A good reason to say Marilyn vos Savant was right is that she was!

    I have no idea whether she did or did not know anything about conditional probability but her answer was correct. The question was "Suppose you choose one of three doors, knowing that there is a car behind one of them, equally likely to behind any one of the three doors. The game show host (Monty Hall) who knows which door the car is behind opens a door showing the car is not there. Would the person improve his/her chances of winning the car by changing his/her choice?

    Marilyn vos Savant did not use conditional probability to answer: you did what mathematicians often do to think about a problem intially- look at an extreme case. She said "suppose there were 1000 door, with a car behind one of them. You choose one, the game show host opens 998 of the doors showing no car behind them. In other words, there are now two doors, your choice and one other, one having the car behind it. Would you change? You bet you would! Monty Hall has completely changed the odds, using his superior knowledge.
    Of course, one can apply conditional probability to show that same result. I remember seeing this problem about 15 years ago as an exercise, in chapter one of an introductory probability book.

    It is an interesting exercise to see what happens if Monty Hall does NOT know which door the car is behind but opens doors at random. You can use conditional probability to show that, in that case, given that the door he opens HAPPENS not to have the car behind it, there is no advantage to changing.
     
  21. Jan 20, 2005 #20

    Hurkyl

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    IIRC, Marylin first gave the wrong answer, then later gave the right answer.
     
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