aa r X i v : . [ phy s i c s . pop - ph ] O c t A Fresh Look at the “Hot Hand” Paradox
S. Redner
Santa Fe Institute, 1399 Hyde Park Road, Santa Fe, New Mexico, 87501
We discuss the “hot hand” paradox within the framework of the backward Kolmogorov equation.We use this approach to understand the apparently paradoxical features of the statistics of fixed-length sequences of heads and tails upon repeated fair coin flips. In particular, we compute theaverage waiting time for the appearance of specific sequences. For sequences of length 2, the averagetime until the appearance of the sequence HH (heads-heads) equals 6, while the waiting time forthe sequence HT (heads-tails) equals 4. These results require a few simple calculational steps bythe Kolmogorov approach. We also give complete results for sequences of lengths 3, 4, and 5; theextension to longer sequences is straightforward (albeit more tedious). Finally, we compute thewaiting times T n H for an arbitrary length sequences of all heads and T n (HT) for the sequence ofalternating heads and tails. For large n , T n H ∼ T n (HT) . I. INTRODUCTION
In a repeated flips of a fair coin, the outcomes H (heads) or T (tails) occur with 50% probability. Thus in a longstring of N coin flips, the number of heads and tails, N H and N T , will be nearly equal, with | N H − N T | of the order of √ N . Given that H and T appear equiprobably, a naive expectation might be that the average frequencies of specificfixed-length sequences of H’s and T’s should be the same; that is, the sequence HHTH should occur with the samefrequency as HTTH. As a corollary of this expectation, the waiting time before encountering either of these sequencesshould be the same. Surprisingly, this expectation is false.The paradoxical nature of this so-called “hot hand” paradox has spawned considerable discussion and literaturethat has ultimately resolved this intriguing issue, see, e.g., Refs. [1–10]. However, the approaches given in thesereferences are complicated and the simplicity of the mechanism that underlies the apparent paradox can be lost incalculational details. Here we give an alternative route to understand the hot hand paradox that is based on thebackward Kolmogorov equations [11, 12]. This formulation has proved to be extremely useful in a variety of first-passage processes. We will use this approach to compute the waiting time for specific sequences of H’s and T’s oflength up to 5, and it may be straightforwardly extended to longer sequences. We give an intuitive reason whydifferent sequences of the same length do not occur with the same frequency. We also derive the waiting time forparticularly simple sequences of arbitrary length, namely, the sequence of n consecutive H’s and the sequence of n consecutive (HT)’s. We find that T n H ∼ T n (HT) , so that 2 n heads in a row is three times less frequent than n (HT)’sin a row.The idea underlying the backward Kolmogorov equation is quite simple. Consider a Markov process that is currentlyin a particular state S . We want to compute the average time T S → F until the process reaches a specified final state F . Suppose that there are two possible outcomes at each stage of the process that occur with equal probability. Thatis, from state S , the process transitions either to state S ′ or to S ′′ , each with probability . Suppose further that thetime required for each transition equals 1. Since the Markov process has no memory, when either of the states S ′ or S ′′ are reached, the the process starts anew. Consequently, the hitting time from S is just the average of the hittingtimes starting from either S ′ or S ′′ plus the time spent in the transition itself. That is T S → F = ( T S ′ → F + 1) + ( T S ′′ → F + 1) . (1)We will use this basic equation to compute the waiting time for specific sequences of H’s and T’s of a given length asa result of repeated flips of a fair coin. II. DOUBLET SEQUENCES
We start with the simplest example of length-2 sequences. The possible sequences are HH, HT, TH, and HH.Because the coin is fair, we obtain the same statistics by the substitution H ↔ T , so that the waiting time for thesequences TT and TH is the same as that for HH and HT. Consequently, we only need to consider the first twosequences. How long does one have to wait before encountering each of these sequences in a long string of fair coinflips?Using Eq. (1) as our starting point, we first compute the waiting time T HH to encounter an HH sequence. For thispurpose, we introduce the auxiliary restricted times: • A , the average waiting time for the sequence HH starting with an H. • B , the average waiting time for the sequence HH starting with a T.These two times obey the backward equations A = × (1 + B ) B = (1 + B ) + (1 + A ) . (2a)These two equations express the waiting times A and B as the average time to reach the desired final state afterone coin flip, plus the time for a single coin flip itself. Thus in the equation for A , the first term accounts for thenext coin flip being H (which occurs with probability ) after which the sequence HH has been generated. The factor2 arises because it takes two coin flips to generate the sequence HH; the fact that the starting state is H does notcount as time step. The second term accounts for the next coin flip being T. Again, the probability for this eventis . Once a T appears, the waiting time to generate an HH sequence is B by definition. Consequently, the factor(1 + B ) accounts for the time spent in making a single coin flip plus the waiting time when the sequence string startswith T. Solving these two equations gives A = 5, B = 7. Since H and T appear equiprobably, on average, in a longseries of fair coin flips, we have T HH = ( A + B ) = 6.For T HT , we introduce • A , the average waiting time for the sequence HT starting with H. • B , the average waiting time for the sequence HT starting with T.Using the same reasoning as given above, these two times obey the backward equations A = × (1 + A ) B = (1 + A ) + (1 + B ) . (2b)The solution to (2b) are ( A, B ) = (3 , T HT = ( A + B ) = 4.Why are these two times different? The key lies in the second term of first lines of Eqs. (2a) and (2b). These termsaccount for a “mistake”. For example, in Eq. (2a), if the next coin flip is T, one has to “start over” to generate HH.The soonest that the next HH can happen immediately after a T is after two more coin flips. In contrast, in Eq. (2b),if the next coin flip is H (again a mistake), the process “starts over”. Now, however, the soonest that the next HTsequence can appear is after only one more coin flip. III. TRIPLET SEQUENCES
Let’s now generalize to triplet sequences. The 2 = 8 distinct triplet sequences are HHH, HHT, HTH, and THHand their counterparts that are obtained by the substitution H ↔ T. By left/right symmetry, the triplets HHT andTHH have identical statistics, so the only distinct sequences are HHH, HHT, and HTH.We define T HHH as the waiting time to encounter the sequence with three consecutive H’s. To compute this time,we define the auxiliary restricted times: • A , the waiting time for HHH when the current state is H; • B , the waiting time for HHH when the current state is HH; • C , the waiting time for HHH when the current state is T.Following the same reasoning that led to Eqs. (2a), the above times satisfy A = (1 + B ) + (1 + C ) B = (2) + (1 + C ) C = (1 + A ) + (1 + C ) . (3)The first term in the equation for B merits explanation. From the state HH, the desired sequence HHH is obtainedwith probability , while the time for this event is 2. Here the time is measured starting before the second H has beenadded to the sequence. The solution to (3) is ( A, B, C ) = (13 , , T HHH = ( A + C ) = 14.Similarly, let T HHT be the waiting time to encounter the sequence HHT. Here, we introduce the auxiliary times: • A , the waiting time for HHT when the current state is H; • B , the waiting time for HHT when the current state is HH; • C , the waiting time for HHT when the current state is T.These times satisfy A = (1 + B ) + (1 + C ) B = (2) + (1 + B ) C = (1 + A ) + (1 + C ) , (4)There is a subtlety in the second equation that needs explanation. If the initial state is HH, then after adding an H,the current state is still HH, so that that the second term involves B . This feature that the initial state consists of asubsequence of length greater than one plays an increasing role for longer sequences (see Appendices A and B). Thesolution to (4) is ( A, B, C ) = (7 , , T HHT = ( A + C ) = 8.Finally, let T HTH be the waiting time to encounter the sequence HTH. We introduce the auxiliary times: • A , the waiting time for HTH when the current state is H; • B , the waiting time for HTH when the current state is HT; • C , the waiting time for HHT when the current state is T.These times satisfy A = (1 + A ) + (1 + B ) B = (2) + (1 + C ) C = (1 + A ) + (1 + C ) , (5)with solutions ( A, B, C ) = (9 , , T HTH = ( A + C ) = 10.To summarize, T HHH = 14, T HHT = 8, and T HTH = 10, in agreement with the times quoted in, for example, Ref. [6].
IV. QUARTET AND QUINTET SEQUENCES
The six distinct quartets are HHHH, HHHT, HHTH, HHTT, HTHT, and HTTH. To not inflict even more tediousalgebra upon the casual reader, all the calculational details are given in Appendix A. Here we merely quote the waitingtimes in reverse time order: T = 30 T HTHT = 20 T HHTH = T HTTH = 18 T HHHT = T HHTT = 16 . (6)There are nine distinct quintets: HHHHH, HHHHT, HHHTH, HHTHH, HHHTT, HHTHT, HTHHT, HTHTH,and HTTHH. There are additional non-independent sequences that are obtained by either the interchange H ↔ T orby reading the above sequences in reverse order. The waiting times are (see Appendix B for details): T = 62 T HTHTH = 42 T HHTHH = 38 T HTHHT = 36 T HHHTH = T HTTHH = 34 T HHHTH = T HHTHT = 32 . (7)All these results agree with those given in [6]. V. SIMPLE ARBITRARY LENGTH SEQUENCESA. n Consecutive H’s
While the calculational details for longer sequences are straightforward, they become progressively more tediousas the sequence length is increased. However, for the sequence of n consecutive H’s, the equations for the restrictedtimes are sufficiently systematic in character that they can be solved. To this end, we first define the following set ofrestricted times: • A k , the waiting time for n H when the current state consists of k consecutive H’s; • B , the waiting time for n H when the current state is T.These times satisfy A = (1 + A ) + (1 + B ) A = (1 + A ) + (1 + B )... A n − = (1 + A N − ) + (1 + B ) A n − = (2) + (1 + B ) B = (1 + A ) + (1 + B ) . (8a)From the last equation, we have B = 2 + A , while from the penultimate equation we can replace the factor (1 + B )everywhere with A n − −
1. Thus Eqs. (8a) become A = (1 + A ) + A n − − A = (1 + A ) + A n − − A n − = (1 + A n − ) + A n − − A n − = (1 + A n − ) + A n − − A n − can be written in terms of A n − only. Similarly, the equation for A n − can be writtenin terms of A n − only. Continuing this procedure, we find A n − k = − k − − k − + 2 k − k − A n − . (9a)In particular A = − n − − n − + 2 n − − n − A n − . (9b)From the original equation for A n − , we eliminate B in favor of A and obtain A n − = + A . Using this in (9b),we solve for A and find, after some straightforward steps, A = 2 n +1 −
3. Finally, T nH is the average of the waitingtimes starting from H and starting from T. That is, T nH = 12 ( A + B ) = 2 n +1 − . (10) B. n Consecutive (HT)’s
A similar calculation can be carried out for the sequence of n consecutive (HT)’s. Here, we first define the followingset of restricted times: • A k − , the waiting time for n (HT) when the current state is ( k − • A k , the waiting time for n (HT) when the current state is k (HT); • B , the waiting time for n (HT) when the current state is T.These times satisfy A = (1 + A ) + (1 + A ) A = (1 + A ) + (1 + B ) A = (1 + A ) + (1 + A ) A = (1 + A ) + (1 + B )... A n − = (1 + A n − ) + (1 + A ) A n − = (1 + A n − ) + (1 + B ) A n − = (2) + (1 + A ) B = (1 + A ) + (1 + B ) . (11)From the last two equations, we obtain (1 + A ) = A n − (1 + B ) = A n − . (12)We now eliminate A and B from Eqs. (11). With this substitution, we can then recursively solve for A n − , A n − , . . . in terms of A n − and obtain A n − k = − S k k − + 2 k − k − A n − . (13)where S k = ( k − / X j =0 j . We now use k = 2 n − A in terms of A n − , and then use the first of (12) to eliminate A n − infavor of A and ultimately solve for A . After straightforward algebra, the final result is A = − S n − + 3(2 · n − −
1) = − (4 n − −
1) + 3(2 · n − − . (14)The waiting time T n (HT) is given by ( A + B ) = A + 1, and after some algebra, we obtain T n (HT) = (4 n − . (15)It is intriguing to compare the times T n H and T n (HT) . The fair comparison is between T n H and T n (HT) ; i.e, betweenstrings of the same length. Asymptotically, Eq. (10) gives T n H ∼ n +1 , while (15) gives T n (HT) ∼ · n +1 . Onehas to wait three times as long, on average, to encounter a sequence of 2 n H’s in a row compared to a sequence of n (HT)’s in a row. VI. CONCLUDING COMMENTS
While most of the results given here are already known, the backward Kolmogorov approach provides a fresh andpowerful perspective on how to calculate waiting times for specific sequences of H’s and T’s in a long string of repeatedflips of a fair coin. Once one understands the idea that underlies the Kolmogorov approach, the computation of thewaiting times for specific sequences is straightforward and direct. Another important aspect of this approach is thatit is not limited to mean waiting times. This same method can be applied to compute any functional of the waitingtime, such as higher moments or even the characteristic function, h exp( − sT ) i .A surprising conclusion of repeated fair coin flips is that the waiting times, or equivalently, the occurrence frequen-cies, for specific sequences of H’s and T’s of the same length are different. The effect is quite pronounced for a longstring of 2 n H’s in a row compared to the string of n (HT)’s in row. For large n , one has to wait three times longerto encounter the former sequence compared to the latter.Although our approach unambiguously illustrates the different waiting times/frequencies of fixed-length sequences,this seemingly paradoxical phenomenon simple requires careful thought to appreciate intuitively.I thank David Atkinson and Porter Johnson for helpful suggestions while this manuscript was written, and MichaelMauboussin for his encouragement. I also gratefully acknowledge financial support from NSF Grant DMR-1608211. Appendix A: Calculational Details for Quartet Sequences
The calculation T HHHH was given in Sec. V A and we start with T HHHT . To compute T HHHT , we define A , B , C ,and D as the waiting time for HHHT when the current state is H, HH, HHH, and T, respectively. These times satisfy A = (1 + B ) + (1 + D ) B = (1 + C ) + (1 + D ) C = (2) + (1 + C ) D = (1 + A ) + (1 + D ) , (A1)whose solutions are ( A, B, C, D ) = (15 , , , T HHHT = ( A + D ) = 16.To compute T HHTH , we define A , B , C , and D as the waiting time for HHTH when the current state is H, HH,HHT, and T, respectively. These times satisfy A = (1 + B ) + (1 + D ) B = (1 + B ) + (1 + C ) C = (2) + (1 + D ) D = (1 + A ) + (1 + D ) , (A2)whose solutions are ( A, B, C, D ) = (17 , , , T HHTH = ( A + D ) = 18.To compute T HHTT , we define A , B , C , D as the waiting time for HHTT when the current state is H, HH, HHT,and T, respectively. These times satisfy A = (1 + B ) + (1 + D ) B = (1 + B ) + (1 + C ) C = (2) + (1 + A ) D = (1 + A ) + (1 + D ) , (A3)whose solutions are ( A, B, C, D ) = (15 , , , T HHTT = ( A + D ) = 16.To compute T HTHT , we define A , B , C , D as the waiting time for HTHT when the current state is H, HT, HTH,and T, respectively. These times satisfy A = (1 + A ) + (1 + B ) B = (1 + C ) + (1 + D ) C = (2) + (1 + A ) D = (1 + A ) + (1 + D ) , (A4)whose solutions are ( A, B, C, D ) = (19 , , , T HTHT = ( A + D ) = 20.Finally, to compute T HTTH , we define A , B , C , D as the waiting time for HTTH when the current state is H, HT,HTT, and T, respectively. These times satisfy A = (1 + A ) + (1 + B ) B = (1 + A ) + (1 + C ) C = (2) + (1 + D ) D = (1 + A ) + (1 + D ) , (A5)whose solutions are ( A, B, C, D ) = (17 , , , T HTTH = ( A + D ) = 18. Appendix B: Calculational Details for Quintet Sequences
Again, the calculation T HHHHH was given in Sec. V A and we start with T HHHHT . To compute T HHHHT , we define A , B , C , D , E as the waiting time for HHHHT when the current state is H, HH, HHH, HHHH, and T, respectively.These times satisfy A = (1 + B ) + (1 + E ) B = (1 + C ) + (1 + E ) C = (1 + D ) + (1 + E ) D = (2) + (1 + D ) E = (1 + A ) + (1 + E ) (B1)whose solutions are ( A, B, C, D, E ) = (31 , , , ,
33) and we obtain T HHHHT = ( A + E ) = 32.To compute T HHHTH , we define A , B , C , D , E as the waiting time for HHHTH when the current state is H, HH,HHH, HHHT, and T, respectively. These times satisfy A = (1 + B ) + (1 + E ) B = (1 + C ) + (1 + E ) C = (1 + C ) + (1 + D ) D = (2) + (1 + E ) E = (1 + A ) + (1 + E ) (B2)whose solutions are ( A, B, C, D, E ) = (33 , , , ,
35) and we obtain T HHHTH = ( A + E ) = 34.To compute T HHTHH , we define A , B , C , D , E as the waiting time for HHTHH when the current state is H, HH,HHT, HHTH, and T, respectively. These times satisfy A = (1 + B ) + (1 + E ) B = (1 + B ) + (1 + C ) C = (1 + D ) + (1 + E ) D = (2) + (1 + E ) E = (1 + A ) + (1 + E ) (B3)whose solutions are ( A, B, C, D, E ) = (37 , , , ,
39) and we obtain T HHTHH = ( A + E ) = 38.To compute T HHHTT , we define A , B , C , D , E as the waiting time for HHHTT when the current state is H, HH,HHH, HHHT, and T, respectively. These times satisfy A = (1 + A ) + (1 + E ) B = (1 + C ) + (1 + E ) C = (1 + C ) + (1 + D ) D = (2) + (1 + A ) E = (1 + A ) + (1 + E ) (B4)whose solutions are ( A, B, C, D, E ) = (31 , , , ,
33) and we obtain T HHHTT = ( A + E ) = 32.To compute T HHTHT , we define A , B , C , D , E as the waiting time for HHTHT when the current state is H, HH,HHT, HHTH, and T, respectively. These times satisfy A = (1 + B ) + (1 + E ) B = (1 + B ) + (1 + C ) C = (1 + D ) + (1 + E ) D = (2) + (1 + B ) E = (1 + A ) + (1 + E ) (B5)whose solutions are ( A, B, C, D, E ) = (31 , , , ,
33) and we obtain T HHTHT = ( A + E ) = 32.To compute T HTHHT , we define A , B , C , D , E as the waiting time for HTHHT when the current state is H, HT,HTH, HTHH, and T, respectively. These times satisfy A = (1 + A ) + (1 + B ) B = (1 + C ) + (1 + E ) C = (1 + D ) + (1 + B ) D = (2) + (1 + A ) E = (1 + A ) + (1 + E ) (B6)whose solutions are ( A, B, C, D, E ) = (35 , , , ,
37) and we obtain T HTHHT = ( A + E ) = 36.To compute T HTHTH , we define A , B , C , D , E as the waiting time for HTHTH when the current state is H, HT,HTH, HTHT, and T, respectively. These times satisfy A = (1 + A ) + (1 + B ) B = (1 + C ) + (1 + E ) C = (1 + A ) + (1 + D ) D = (2) + (1 + E ) E = (1 + A ) + (1 + E ) (B7)whose solutions are ( A, B, C, D, E ) = (41 , , , ,
43) and we obtain T HTHTH = ( A + E ) = 42.To compute T HTTHH , we define A , B , C , D , E as the waiting time for HTTHH when the current state is H, HT,HTT, HTTH, and T, respectively. These times satisfy A = (1 + A ) + (1 + B ) B = (1 + A ) + (1 + C ) C = (1 + D ) + (1 + E ) D = (2) + (1 + B ) E = (1 + A ) + (1 + E ) (B8)whose solutions are ( A, B, C, D, E ) = (33 , , , ,
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