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  1. #1
    Eric's Avatar
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    Default the hot hand debate

    First off, I believe sports can be streaky and a basketball player can get in the zone and have a hot hand. However, I'm confused about what they're saying regarding coin tosses in this http://www.wsj.com/articles/the-hot-...ead-1443465711 article:
    The new paper, written by Joshua Miller and Adam Sanjurjo, begins with a riddle. Toss a coin four times. Write down what happened. Repeat that process one million times. What percentage of the flips after heads also came up heads?


    The obvious answer is 50%. That answer is also wrong. The real answer is 40%—and the authors say their correction should alter years of thinking about the hot hand.
    I thought past coin tosses have nothing to do with a current coin toss. How can the percentage of heads go down from 50% to 40% based on the past?
  2. #2
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    So to clarify what they're talking about, considering the following. Here are the 14 possibilities for tossing a fair coin four times that have a heads in the first three flips. Each of them will happen with the same frequency. Beside of these, I've listed how many times a heads came after a heads along with how many times a tails came after a heads.

    1. HHHH 3-0
    2. HHHT 2-1
    3. HHTH 1-1
    4. HHTT 1-1
    5. HTHH 1-1
    6. HTHT 0-2
    7. HTTH 0-1
    8. HTTT 0-1
    9. THHH 2-0
    10. THHT 1-1
    11. THTH 0-1
    12. THTT 0-1
    13. TTHH 1-0
    14. TTHT 1-0

    There are 56 total tosses in this sample. A total of 33 of them are heads, and the remaining 23 are tails. That's about 60% heads, which falls in line with their "finding."

    Basically, the coin example is a fluke based on the approach that they're using to measure it.
  3. #3
    Something doesn't add up there spoon. For a start, I only count 31 heads, and there's 25 tails.

    TTTH and TTTT are omitted, and when they are added, there are 32 heads and 32 tails.

    Furthermore, it doesn't make sense to use all the flips for the total when the first flip of each set of four is not relevant to the stats.

    I can't see where they're getting 40% from. I'm not saying they didn't, but your explanation spoon doesn't seem to cut it.
    Last edited by OngBonga; 09-29-2015 at 12:38 PM.
    Quote Originally Posted by wufwugy View Post
    ongies gonna ong
  4. #4
    1. HHHH 3-0
    2. HHHT 2-1
    3. HHTH 1-1
    4. HHTT 1-1
    5. HTHH 1-1
    6. HTHT 0-2
    7. HTTH 0-1
    8. HTTT 0-1
    9. THHH 2-0
    10. THHT 1-1
    11. THTH 0-1
    12. THTT 0-1
    13. TTHH 1-0
    14. TTHT 1-0
    15. TTTH 0-0
    16. TTTT 0-0

    Now there's 16 x 3 = 48 flips that come after another flip (ie, not the first). Of those 48, 24 are heads and 24 are tails. But how many heads come after heads compared to tails after heads? That's what we're looking at.

    There's 24 times where heads comes up before the last one, and it's the flip after these 24 that we need to look at. 13 are heads, and 11 tails. I make it roughly 54% in favour of heads, not the 40% they got.

    I dunno what I'm doing wrong, but their methods are definitely skewing the results. There's a lot of flips they aren't considering, and if they're getting 40% heads in their sample, I'm gonna assume the missing head flips are in the omitted sample.
    Quote Originally Posted by wufwugy View Post
    ongies gonna ong
  5. #5
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    They aren't counting the TTTH and TTTT because the whole premise is whether or not you can get "hot". In this case, they are saying being hot = getting a heads.
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  6. #6
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    They concluded that if you throw a heads within the first three throws, that you will get another heads right after that heads throw only 40% of the time. So they aren't counting how many heads versus tails total. They are counting how many times you get a heads AFTER you throw a heads.

    Here is their intro:

    "Jack takes a coin from his pocket and decides that he will
    flip it 4 times in a row, writing down
    the outcome of eachflip on a scrap of paper. After he is done
    flipping, he will look at the flips that immediately followed an outcome of heads, and compute the relative frequency of heads on those flips. Because the coin is fair, Jack of course expects this empirical probability of heads to be equal
    to the true probability of flipping a heads: 0.5. Shockingly, Jack is wrong. If he were to sample one
    million fair coins and flip each coin 4 times, observing the conditional relative frequency for each
    coin, on average the relative frequency would be approximately 0.4."
    Last edited by chardrian; 09-29-2015 at 03:34 PM.
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  7. #7
    Eric's Avatar
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    Trying to get my head around this.


    2 tosses have 4 outcomes:
    HH
    HT
    TH
    TT
    Here it would be 50% because HH and HT are equally likely.


    3 tosses have 8 outcomes. I like to think of it as putting H in front of the 4 above and then putting T in front of the 4 above.
    Still seems like 50%. We have the 50% from within the above twice (once with H in front and once with T in front). We also have 50% from the 4 times where we put H in front of the 4 above and measure the first 2 tosses.


    4 tosses have 16 outcomes. Seems like we can think of it as putting H in front of the 8 and then 8 in front of the T. Should still be 50%.


    I don't understand how they get 40%.
  8. #8
    Well I'm still getting 54.1666666667% heads, which is wrong on both counts. Outstanding.
    Quote Originally Posted by wufwugy View Post
    ongies gonna ong
  9. #9
    chardrian's Avatar
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    meh - I tried to copy and paste their table but it came out horribly. So just take a look at their actual paper and it shows the table.
    Last edited by chardrian; 09-29-2015 at 04:15 PM.
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  10. #10
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    Their choice of the number of tosses in the group is what's doing this. I apologize to anyone who missed that in my original reply.
  11. #11
    Apology accepted.
    Quote Originally Posted by wufwugy View Post
    ongies gonna ong
  12. #12
    Eric's Avatar
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    Ok, I think I see how they get .40 but it looks like they incorrectly gave each sequence/line equal weight.


    Here's their table:


    ----------
    0: TTTT: -
    ----------
    1: TTTH: -
    1: TTHT: 0
    1: THTT: 0
    1: HTTT: 0
    ----------
    2: TTHH: 1
    2: THTH: 0
    2: THHT: 1/2
    2: HTTH: 0
    2: HTHT: 0
    2: HHTT: 1/2
    ----------
    3: THHH: 1
    3: HTHH: 1/2
    3: HHTH: 1/2
    3: HHHT: 2/3
    ----------
    4: HHHH 1
    ----------


    The first column is the number of heads.
    The second column is the sequence.
    The third column is the probability of heads following tails.


    The sum of the probabilities in column 3 is 5.66. Dividing this by the 14 sequnces, we get .40.


    It doesn't seem right for them to be able to add the probabilites this way. For example, the "1" sequences are not all equal. The last line, HHHH, has a "1" but it is a case where H followed another H 3 different times. This is not the same as the TTHH line which also has a "1" where H followed another H just 1 time. In other words, the sequences/lines do not have equal weight.
  13. #13
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    Also, the 0 lines/sequences are not equal weight either. HTHT and HTTT both have 0 on the right but HTHT represents two failures and HTTT represents just one failure.
  14. #14
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    Another way to see that their 40% number can't be right is to take their table and use a lowercase letter immediately after each heads toss. We see that there are 12 lowercase tails and 12 lowercase heads:
    ----------
    0: TTTT: -
    ----------
    1: TTTH: -
    1: TTHt: 1t, 0h
    1: THtT: 1t, 0h
    1: HtTT: 1t, 0h
    ----------
    2: TTHh: 0t, 1h
    2: THtH: 1t, 0h
    2: THht: 1t, 1h
    2: HtTH: 1t, 0h
    2: HtHt: 2t, 0h
    2: HhtT: 1t, 1h
    ----------
    3: THhh: 0t, 2h
    3: HtHh: 1t, 1h
    3: HhtH: 1t, 1h
    3: Hhht: 1t, 2h
    ----------
    4: Hhhh: 0t, 3h
    ----------


    12 lowercase heads
    12 lowercase tails
    ==> 50%
  15. #15

    Default Hot Hand Reply

    I see it now, having worked on this at length (3+ hours), did the same exact thing you did Eric 3x, and having read both papers. It's quite subtle.

    To be formal, we're sampling length-4 sequences and computing the expectation of p(H|H) from our sample.

    It's easier to visualize if we talk about P(seq) where seq is a particular sequence and P(H|H,seq) is the conditional probability of H following H given a particular sequence. We estimate P(H|H,seq) as C(H follow H)|C(H) in seq. Then P^(H|H) is our estimation of conditional probability of H following H in any sequence, and is equal to SUM over all sequences, P(H|H,SEQ) * P(SEQ). The sum over equiprobable sequences (SUM P(SEQ)) is where your intuition that the sequences need to equally count comes in, but it isn't the whole story.

    The problem with your math, and what I also did, lies in cases with multiple H. We both naively summed up H and T from that to all the other cases, and got 0.5. Take for instance THht, where we both got one head and one tail. P(THht) = 1/16 (or 1/14 counting just the eligible sequences) just like every other sequence. The problem with summing down H and T and ignoring P(seq) is that the denominator C(H) here is 2 but in THtT the denominator C(H) is 1, so summing just the counts over the rows misses the fact that they have different denominators.

    I'll derive your example according to the sum over products I outlined. I'm "shoving" the sequence probability (1/14) into each outcome,so instead of the average of conditional probabilities, the total probablility is just the sum of these products..comes out the same.

    0: TTTT: - X -- conditioned out
    ----------
    1: TTTH: - X -- conditioned out
    1: TTHt: 1t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 0/1 = 0/14 = 0
    1: THtT: 1t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 0/1 = 0/14 = 0
    1: HtTT: 1t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 0/1 = 0/14 = 0
    ----------
    2: TTHh: 0t, 1h = P(SEQ) * P(H|H,SEQ) = 1/14 * 1/1 = 1/14 = 0.0714
    2: THtH: 1t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 1/1 = 0/14 =0
    2: THht: 1t, 1h = P(SEQ) * P(H|H,SEQ) = 1/14 * 1/2 = 1/28 = 0.0357
    2: HtTH: 1t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 0/1 = 0/14 = 0
    2: HtHt: 2t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 0/1 = 0/14 = 0
    2: HhtT: 1t, 1h = P(SEQ) * P(H|H,SEQ) = 1/14 * 1/2 = 1/28 = 0.0357
    ----------
    3: THhh: 0t, 2h= P(SEQ) * P(H|H,SEQ) = 1/14 * 2/2 = 2/28 = 0.0714
    3: HtHh: 1t, 1h= P(SEQ) * P(H|H,SEQ) = 1/14 * 1/2 = 1/28 = 0.0357
    3: HhtH: 1t, 1h= P(SEQ) * P(H|H,SEQ) = 1/14 * 1/1 = 1/28 = 0.0357
    3: Hhht: 1t, 2h= P(SEQ) * P(H|H,SEQ) = 1/14 * 2/3 = 2/42 = 0.0476
    ----------
    4: Hhhh: 0t, 3h= P(SEQ) * P(H|H,SEQ) = 1/14 * 3/3 = 3/42 = 0.0714

    0.0714 * 3 + 0.0357 * 4 + 0.0476 = 0.4046

    Couple of things are going on here...you're conditioning out two sequences that are heavy with tails, which means that the remaining sequences are more likely to have multiple heads. Now, you'd think that would make P(H|H) go up, but each of those sequences is still weighted by the sequence probability, so you're just shoving the cases where H follows H into the same sequences, such that the individual counts of H all "share" the same 1/14 sequence probability (bc P(SEQ) * C(H follow H) / C(H)), while the cases where T follows H are more likely to occur "by themselves".

    Not sure this translates into the Hot Hand effect (my own take--Steph Curry has some psychological/physical factor that violates independence to make him streakier when the game is on the line)< but their statistical argument bears out.
    Last edited by kyleg; 09-30-2015 at 12:01 AM. Reason: formatting
  16. #16
    Quote Originally Posted by Eric View Post
    Another way to see that their 40% number can't be right is to take their table and use a lowercase letter immediately after each heads toss. We see that there are 12 lowercase tails and 12 lowercase heads:
    ----------
    0: TTTT: -
    ----------
    1: TTTH: -
    1: TTHt: 1t, 0h
    1: THtT: 1t, 0h
    1: HtTT: 1t, 0h
    ----------
    2: TTHh: 0t, 1h
    2: THtH: 1t, 0h
    2: THht: 1t, 1h
    2: HtTH: 1t, 0h
    2: HtHt: 2t, 0h
    2: HhtT: 1t, 1h
    ----------
    3: THhh: 0t, 2h
    3: HtHh: 1t, 1h
    3: HhtH: 1t, 1h
    3: Hhht: 1t, 2h
    ----------
    4: Hhhh: 0t, 3h
    ----------


    12 lowercase heads
    12 lowercase tails
    ==> 50%
    Well done, I miscounted and got 13-11, hence my 54%. Actually it was spoon who miscounted, I added up his numbers because I'm lazy and assumed he got that bit right.
    Quote Originally Posted by wufwugy View Post
    ongies gonna ong
  17. #17
    Eric's Avatar
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    Thanks, OngBonga.


    kyleg,


    I don't understand what you're saying. My last post shows that their own table tells us that we get heads 12/24 times or 50% of the time.


    Cheers,
    Eric
  18. #18
    I think if you take 20% of the heads flips and throw them in the bin, then the H/T ratio that remains is 40/60.

    That's the gist of it as far as I can tell.
    Quote Originally Posted by wufwugy View Post
    ongies gonna ong
  19. #19

    Default Hot Hand Followup

    Eric, I'm working backwards from the table they have in the paper, and the specific thing that I am doing differently than you is I am weighting (multiplying) each of the relative probabilities (e.g. 0.5 in this sequence, 0.7 elsewhere) by the associated sequence probabilities (1/14 in all cases). According to that logic, adding Hs and Ts across the sequence outcomes and dividing by the denominator is invalid because the counts that give you the denominators for the relative probabilities are not the same; the relative number of heads across different sequences therefore make different contributions to the probability.

    This logic, albeit unintuitive, (the whole problem is highly unintuitive so intuition isn't going to give us good answers, and you know how slippery statistical intuition can be), falls directly out of the definition of conditional expectation/Bayes Rule: P(B) = P(B|A) * P(A) / P(B). For a set of outcomes A_N for N = 1 .. 14 (our 14 sequences), P(B) = Sum for A_N 1 to 14 P(B|A_N) * P(A_N). Now, for all 14 sequences, P(A_N) the sequence probability is always 1/14, but the P(B|A_N) varies depending on the number of H in that sequence. The key is that for each sequence, we need to compute each P(A_N) separately from each set of counts, then and there, and its the probabilities that we need to do a (weighted) sum over, not the numerator and denominator counts.

    Why would this give a different (and I believe, correct) result, in an intuitive way? Consider a variant of this game where in each of the sequences we first roll a 20 sided dice to determine how many flips you do in that sequence, maybe another to determine how many sequences. Consider one outcome of this game, with three sequences A, B, C: Hhht, HhhtHhhtHhht, and TTHh. It should be intuitive that a) A and B both convey the same long-run expectation of p(T|H), and b) that the raw count of H and T from B can't matter more overall than the raw counts from A, the probabilities you get from each do. But all I did was allow the sequence lengths (and therefore probabilities) to vary, so my correct procedure for computing expectation of p(T|H) hasn't changed from the above case.
    Last edited by kyleg; 10-01-2015 at 02:21 AM. Reason: typo
  20. #20
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    Quote Originally Posted by kyleg View Post
    But all I did was allow the sequence lengths (and therefore probabilities) to vary, so my correct procedure for computing expectation of p(T|H) hasn't changed from the above case.
    If their paper is right then why aren't there casinos that have a coin tossing game where gamblers are only allowed to bet on heads when the casino knows the chances are just 40% based on the recent past? If the paper is right then the casino could payout 6:5 and make money if the true odds based on the recent past are closer to 3:2. There is no type of casino game like this and there never will be - the chances are always 50% regardless of what happened in the recent past.
  21. #21

    Default Response

    Before I respond to that, which I'm more than willing to do (I'm enjoying this), is it out of place to suggest you first identify a flaw in the formal argument I laid out, not a link to an intuition that could or could not be misleading? All the interesting problems in statistics are subtle, and it's burdensome to some extent to lay out this math to chase down each intuition with the amount of rigor I'm trying to provide. The debate doesn't progress if one side is conversing in math and the other is conversing in intuitions.

    Not to be cantankerous, but I think that's the only way this will get figured out.
  22. #22
    chardrian's Avatar
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    Quote Originally Posted by Eric View Post
    If their paper is right then why aren't there casinos that have a coin tossing game where gamblers are only allowed to bet on heads when the casino knows the chances are just 40% based on the recent past?
    I think there are a couple of reasons for this: 1) no one knew about this until recently - the paper was just published; 2) the conditions required to make the game work are too cumbersome - I still haven't totally wrapped my head around this problem but I know that the key words to this problem are "conditional relative frequency" - we aren't talking about what the authors call the "true probability" of flipping a coin (which they agree is 50%). So I think part of the problem is that we don't really understand what they mean by "conditional relative frequency."

    I think in your mind the rules of the casino game (and for this problem) would be: person walks up and is allowed to flip a coin until he gets a heads, as soon as he does then the casino will pay him even money (or anything above the 40% that they are now expecting after this problem) if he flips another heads in the next toss. But I don't think that those are the actual parameters of this problem.

    Maybe kyleg could explain what the rules of your proposed casino game would actually be if they were taken from this problem?
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  23. #23
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    Do the same thing with three coins:

    HHH 2-0
    HHT 1-1
    HTH 0-1
    HTT 0-1
    THH 1-0
    THT 0-1
    TTH 0-0
    TTT 0-0

    And the score is 4-4
  24. #24
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    Quote Originally Posted by spoonitnow View Post
    Do the same thing with three coins:


    And the score is 4-4
    Applying kyleg's demonstrated method:

    HHH 2-0 = P(SEQ) * P(H|H,SEQ) = 1/6 * 2/2 = 1/6
    HHT 1-1 = P(SEQ) * P(H|H,SEQ) = 1/6 * 1/2 = 1/12
    HTH 0-1 = P(SEQ) * P(H|H,SEQ) = 1/6 * 0/1 = 0/6 = 0
    HTT 0-1 = P(SEQ) * P(H|H,SEQ) = 1/6 * 0/1 = 0/6 = 0
    THH 1-0 = P(SEQ) * P(H|H,SEQ) = 1/6 * 1/1 = 1/6
    THT 0-1 = P(SEQ) * P(H|H,SEQ) = 1/6 * 0/1 = 0/6 = 0
    TTH 0-0 - X - conditioned out
    TTT 0-0 - X - conditioned out

    Total = 1/6 + 1/12 + 1/6 = 5/12 ~= 42%
  25. #25

    Default Hot Hand Followup

    @Chardrian -- great post

    "I know that the key words to this problem are "conditional relative frequency" - we aren't talking about what the authors call the "true probability" of flipping a coin (which they agree is 50%)".

    I'm still vague on how the upshot of this experiment gives you the hot hand effect, but I can tell you that "relative frequency" formally refers to an estimator (which they claim is biased) and not the true probability. If I had to guess their larger point (and I'll go take another look at the paper), I think their claim is that "hot hand" represents an observational or sample bias of attending to one particular player in one game conditional on an outcome. Why that should be different than the true conditional probability I'm still unclear on.
  26. #26
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    Quote Originally Posted by kyleg View Post
    Before I respond to that, which I'm more than willing to do (I'm enjoying this), is it out of place to suggest you first identify a flaw in the formal argument I laid out, not a link to an intuition that could or could not be misleading?
    I'm thinking someone will do this soon. I guess it hasn't happened yet because the paper is so new.
  27. #27
    I think I more-or-less know what's going on here.

    So let's say there's a casino that has a game where they flip 4 coins, and they only allow you to bet after a Heads is flipped. Well, if you're only allowed to play the game once, the best possible outcome you can get is to win two bets with the *exact* outcome of HTHT. In other words: H (you're allowed to bet) T (you win!) H (you're allowed to bet) T (you win!).

    You can, however, lose thrice: H (you're allowed to bet) H (you lose, but at least you're allowed to bet again) H (you lose again, but at least you're allowed to bet once more) H (you lose, fyl).

    But let's say you're allowed to play 6 flips: you can now get HTHTHT to win three times in a row. The only problem is that that exact combination only happens 1/64 times, whereas HHHH (wherein you lose three times in a row) happens 1/16th of the time.

    However, if you're given infinite flips, you find that there are FAR more ways to win 3 straight times: it could be TTHTTTTTTTTHTTTTTTTTHT, for example. In fact, if you're given infinite flips, then there are infinite permutations where you would win 3 times in a row, and I suspect the sum of all those permutations approaches 1/16 as your sample approaches infinity.

    In other words, the parameters for what samples we look at and how we analyze them is fucking with all the possible outcomes to contribute to all the various ways that bad results can follow good results.

    This paper is useful in proving why certain forms of statistical analysis will give the *allusion* of the hot-hand effect. Basically, whenever we look at any finite example that necessarily excludes any case where the first several shots are misses, then we're gonna end up with more makes than misses. However, it doesn't (as far as I can tell), prove that the probability of any one flip changes based on what the previous flip was.
    Last edited by surviva316; 10-05-2015 at 08:55 PM.

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