 The expected value of sn/n ≈ 0.79295350640…

Julian D. A. Wiseman

Publication history: only here. Usual disclaimer and copyright terms apply.

Introduction

The problem, as described in Mathematical Constants, Steven R. Finch, Cambridge University Press, 362-363:

We observe a fair coin being tossed repeatedly and can stop observing at any time. When we stop, the payoff is the average number of heads observed. What is the best strategy to maximize the expected payoff? Chow & Robbins [21…] described a strategy that achieves an expected payoff > 0.79 = (0.59 + 1) /2.

 Y.S. Chow and H. Robbins, On optimal stopping rules for sn/n, Illinois J. Math. 9 (1965) 444-454; MR 31 #4134.

Here is described an algorithm which provides strong evidence that the expected value ≈ 0.79295350640….

Outline

Let the number of heads seen so far be h, and the number of tails t, so stopping at this time would have payoff h/(h+t). Consider a restricted strategy, in which one may choose to stop or continue whilst t<4; but once t reaches 4 one must either stop or toss forever (well, until h=t which happens eventually with probability 1).

The payoff from this restricted strategy can be calculated using the following ‘spreadsheet’:

 t=0 t=1 t=2 t=3 t=4 h=0 =Avg(↓,→) =Avg(↓,→) =Avg(↓,→) =Avg(↓,→) =Max( h/(h+t), 0.5 ) h=1 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=2 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=3 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=4 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=5 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=6 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=7 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=8 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=9 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=10 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=11 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=12 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=13 =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), Avg(↓,→) ) =Max( h/(h+t), 0.5 ) h=14 =Max( h/(h+t), 0.5 ) =Max( h/(h+t), 0.5 ) =Max( h/(h+t), 0.5 ) =Max( h/(h+t), 0.5 ) =Max( h/(h+t), 0.5 )

which evaluates to:

 t=0 t=1 t=2 t=3 t=4 h=0 599/768 215/384 349/672 113/224 1/2 h=1 1 269/448 359/672 57/112 1/2 h=2 1 2/3 47/84 29/56 1/2 h=3 1 3/4 101/168 15/28 1/2 h=4 1 4/5 2/3 4/7 1/2 h=5 1 5/6 5/7 5/8 5/9 h=6 1 6/7 3/4 2/3 3/5 h=7 1 7/8 7/9 7/10 7/11 h=8 1 8/9 4/5 8/11 2/3 h=9 1 9/10 9/11 3/4 9/13 h=10 1 10/11 5/6 10/13 5/7 h=11 1 11/12 11/13 11/14 11/15 h=12 1 12/13 6/7 4/5 3/4 h=13 1 13/14 13/15 13/16 13/17 h=14 1 14/15 7/8 14/17 7/9

where bolded numbers are stopping states in this restricted game. Already we know that the expected value of sn/n ≥ 599/768 ≈ 0.7799479166666.

• The interior of the spreadsheet contains formulae choosing the better of stopping (“h/(h+t)”), and continuing (“Avg(↓,→)” reflecting the ½ chance of a head or of a tail).
• On the lower edge (large h) and right edge (large t), the formulae choose between stopping immediately, and continuing until h=t.
• On the top edge, stopping with h=0 is pessimal, and not testing for whether to stop avoids the divide by zero when h=t=0.

Observe that the rows with h>4 are redundant: every element in the h=4 row is a stopping state, and the ‘spreadsheet’ below this can be truncated without changing the value at h=t=0.

Let EVn(h,t) be the expected value of this mini game with t ≤ n, assuming that h heads and t tails have already been tossed. We now describe the algorithm by which EVn(0,0) is calculated for large n, and when this is done we shall happily see that it appears to have a neat functional form, allowing ready approximation of the value of EV(0,0).

Algorithm

The algorithm can now proceed as follows.

 1 Let max_h be n. 2 Create a one-dimensional array of maximum-precision real numbers EV[], the array being of length n+1. 3 Fill the array with values 0.5. EV[] is now the rightmost column in which t=n. 4 Loop from t=n−1 to 0. 5 Set h = max_h, and EV[h] = h/(h+t). The last row, in which h=max_h. 6 Loop over h downwards to 1. …so going up the rows from large h to small h. 7 Calculate ( EV[h] + EV[h+1] ) / 2. EV[h] contains the expected value of the game at h,t+1. Hence the value from tossing again is ( EV[h] + EV[h+1] ) / 2. 8 Set EV[h] to the greater of this and h/(h+t). 9 Next h. 10 EV = ( EV + EV ) / 2. This avoids the possible divide-by-zero when h=0. 11 Next t. 12 Output EV. This replicates the ‘spreadsheet’ in memory linear in n. There are some improvements to this, which between them reduce the n² run-time by a factor of something like √n. The areas of calculation eliminated by these improvements are shown in the diagram on the right for n=19683; proportionately larger areas are eliminated for larger n.

• It is trivial to see that if (h,t) is a stopping state, then so are all states with the same h and smaller t. So when a stopping state is identified, max_h can be reduced to h+1, the additional unit being required because of the boundary. This removes the area below the red line in the diagram.
• Once a keep-playing state (h,t) has been identified, all states with smaller h and the same t are also keep-playing states. So we no longer need to test these, and can implement the EV[h] = ( EV[h] + EV[h+1] ) / 2. This speeds, though does not eliminate, all the states above the red line.
• Assume that we have a measure of machine precision (in C, LDBL_EPSILON). Once we reach an EV[h] ≤ 0.5+LDBL_EPSILON, then EV[h] will be between this value and 0.5 for all smaller h. Yes, continuing the h loop might marginally increase some of the EV[h], but by almost zero. So the h loop can be terminated, reducing the computed EVn(0,0) by, at most, one machine precision. This removes the area above the grey line in the diagram.

A recent version of the code that implements this algorithm with improvements is available at www.jdawiseman.com/papers/easymath/coin-stopping.c. those desiring the latest version should contact the author.

It is possible to rewrite the program such that the memory usage is linear in the widest gap between the red and grey lines (rather than being linear in n), albeit at the price of a loss of readability. This recoding has not yet been attempted.

Results

The following table shows results for n = 2i (calculations run on an Intel Pentium IV processor under Windows XP using cygwin gcc C compiler):

i n = 2^i EV Difference Ratio Estimated limit =
EV + Diff
/ (1/Ratio − 1)
1 2 0.770833333333333
2 4 0.779947916666666 0.009114583333
3 8 0.785247657412574 0.005299740746 0.58145727 0.79261028151309
4 16 0.788190898971545 0.002943241559 0.55535576 0.79186697529802
5 32 0.790023224424476 0.001832325453 0.62255354 0.79304542974422
6 64 0.791196842160634 0.001173617736 0.64050725 0.79328787366908
7 128 0.791897537079360 0.000700694919 0.59703845 0.79293570515041
8 256 0.792318224291505 0.000420687212 0.60038570 0.79295027021847
9 512 0.792570438324007 0.000252214033 0.59952864 0.79294801722604
10 1,024 0.792722493847599 0.000152055524 0.60288289 0.79295333676356
11 2,048 0.792814414356913 0.000091920509 0.60451937 0.79295492118365
12 4,096 0.792869752675300 0.000055338318 0.60202363 0.79295346361236
13 8,192 0.792903056616781 0.000033303941 0.60182424 0.79295339398384
14 16,384 0.792923109615955 0.000020052999 0.60212090 0.79295345634652
15 32,768 0.792935195956992 0.000012086341 0.60271987 0.79295353233307
16 65,536 0.792942476379465 0.000007280422 0.60236778 0.79295350539518
17 131,072 0.792946862428220 0.000004386049 0.60244426 0.79295350891743
18 262,144 0.792949504079600 0.000002641651 0.60228500 0.79295350449952
19 524,288 0.792951095314407 0.000001591235 0.60236367 0.79295350581352
20 1,048,576 0.792952053932079 0.000000958618 0.60243634 0.79295350654503
21 2,097,152 0.792952631391198 0.000000577459 0.60238731 0.79295350624769
22 4,194,304 0.792952979294206 0.000000347903 0.60247210 0.79295350655746
23 8,388,608 0.792953188872027 0.000000209578 0.60240302 0.79295350640540
24 16,777,216 0.792953315120851 0.000000126249 0.60239592 0.79295350639599
25 33,554,432 0.792953391174430 0.000000076054 0.60241019 0.79295350640739
26 67,108,864 0.792953436989929 0.000000045815 0.60241083 0.79295350640770
1. The first column contains i,
2. and the second n = 2i;
3. The third contains the computed value of EVn(0,0), to fifteen decimal places. These increase with i, as expected.
4. The fourth column shows the increase between adjacent rows, these increases being positive, but decreasing.
5. The fifth column has the ratio between successive increases. This appears to be tending to a constant limit.
6. If this ratio does indeed reach a constant limit, then the limiting value can be estimated as the current bound, plus the increase divided by one less than the reciprocal of the ratio. This appears to be tending to 0.79295350640….

The same table is next shown for powers of 3:

i n = 3^i EV Difference Ratio Estimated limit =
EV + Diff
/ (1/Ratio − 1)
1 3 0.777083333333333
2 9 0.785866262932785 0.008782929599
3 27 0.789626479634224 0.003760216701 0.42812784 0.79244153793616
4 81 0.791476153990993 0.001849674357 0.49190632 0.79326689955370
5 243 0.792293464103993 0.000817310113 0.44186703 0.79294051849865
6 729 0.792657276357000 0.000363812253 0.44513367 0.79294913959852
7 2,187 0.792820943435251 0.000163667078 0.44986687 0.79295478084870
8 6,561 0.792894165592400 0.000073222157 0.44738476 0.79295344459041
9 19,683 0.792926925277956 0.000032759686 0.44740126 0.79295344854644
10 59,049 0.792941602603975 0.000014677326 0.44803013 0.79295351608787
11 177,147 0.792948175932809 0.000006573329 0.44785602 0.79295350770317
12 531,441 0.792951119086408 0.000002943154 0.44774173 0.79295350523929
13 1,594,323 0.792952437204528 0.000001318118 0.44785910 0.79295350637219
14 4,782,969 0.792953027559343 0.000000590355 0.44787702 0.79295350644967
15 14,348,907 0.792953291954919 0.000000264396 0.44785876 0.79295350641431
16 43,046,721 0.792953410364442 0.000000118410 0.44784986 0.79295350640660

Other geometrically increasing sequences of n behave likewise, including 2×3i−1; 3×2i−1; 5×2i−1; 7×2i−1; 9×2i−1; 11×2i−1; 13×2i−1; 15×2i−1; 17×2i−1; 19×2i−1; 21×2i−1; 23×2i−1; and 25×2i−1. For each of these, only the last four rows are shown here:

i n EV Difference Ratio Estimated limit =
EV + Diff
/ (1/Ratio − 1)

n = 2×3i−1
13 1,062,882 0.792952068251885 0.000001772987 0.44783045 0.79295350621149
14 3,188,646 0.792952862302889 0.000000794051 0.44786062 0.79295350638697
15 9,565,938 0.792953217946628 0.000000355644 0.44788526 0.79295350645113
16 28,697,814 0.792953377218624 0.000000159272 0.44784142 0.79295350639998

n = 3×2i−1
22 6,291,456 0.792953114533501 0.000000258644 0.60243695 0.79295350646377
23 12,582,912 0.792953270339568 0.000000155806 0.60239469 0.79295350639463
24 25,165,824 0.792953364197547 0.000000093858 0.60240260 0.79295350640243
25 50,331,648 0.792953420738771 0.000000056541 0.60241254 0.79295350640833

n = 5×2i−1
22 10,485,760 0.792953236675307 0.000000178026 0.60239696 0.79295350639723
23 20,971,520 0.792953343917754 0.000000107242 0.60239801 0.79295350639842
24 41,943,040 0.792953408522001 0.000000064604 0.60241303 0.79295350640861
25 83,886,080 0.792953447440291 0.000000038918 0.60241071 0.79295350640766

n = 7×2i−1
21 7,340,032 0.792953156304784 0.000000231073 0.60241671 0.79295350642634
22 14,680,064 0.792953295502404 0.000000139198 0.60239569 0.79295350639561
23 29,360,128 0.792953379356017 0.000000083854 0.60240694 0.79295350640552
24 58,720,256 0.792953429870388 0.000000050514 0.60241139 0.79295350640787

n = 9×2i−1
21 9,437,184 0.792953215074249 0.000000192283 0.60239934 0.79295350640012
22 18,874,368 0.792953330905141 0.000000115831 0.60239692 0.79295350639717
23 37,748,736 0.792953400683046 0.000000069778 0.60241188 0.79295350640813
24 75,497,472 0.792953442718023 0.000000042035 0.60241099 0.79295350640774

n = 11×2i−1
20 5,767,168 0.792953088791376 0.000000275635 0.60244710 0.79295350648587
21 11,534,336 0.792953254832655 0.000000166041 0.60239494 0.79295350639491
22 23,068,672 0.792953354855966 0.000000100023 0.60240027 0.79295350640051
23 46,137,344 0.792953415111305 0.000000060255 0.60241296 0.79295350640854

n = 13×2i−1
20 6,815,744 0.792953136810324 0.000000243941 0.60242671 0.79295350644375
21 13,631,488 0.792953283759023 0.000000146949 0.60239524 0.79295350639518
22 27,262,976 0.792953372281616 0.000000088523 0.60240474 0.79295350640401
23 54,525,952 0.792953425608702 0.000000053327 0.60241216 0.79295350640817

n = 15×2i−1
20 7,864,320 0.792953173528347 0.000000219705 0.60240697 0.79295350641128
21 15,728,640 0.792953305877818 0.000000132349 0.60239574 0.79295350639569
22 31,457,280 0.792953385606316 0.000000079728 0.60240889 0.79295350640670
23 62,914,560 0.792953433635636 0.000000048029 0.60241094 0.79295350640773

n = 17×2i−1
20 8,912,896 0.792953202640363 0.000000200490 0.60240124 0.79295350640269
21 17,825,792 0.792953323414915 0.000000120775 0.60239637 0.79295350639652
22 35,651,584 0.792953396170865 0.000000072756 0.60241127 0.79295350640790
23 71,303,168 0.792953439999837 0.000000043829 0.60241082 0.79295350640770

n = 19×2i−1
19 4,980,736 0.792953041537121 0.000000306823 0.60246218 0.79295350652302
20 9,961,472 0.792953226366925 0.000000184830 0.60239806 0.79295350639854
21 19,922,944 0.792953337707907 0.000000111341 0.60239734 0.79295350639770
22 39,845,888 0.792953404781104 0.000000067073 0.60241248 0.79295350640837

n = 21×2i−1
19 5,505,024 0.792953074341674 0.000000285172 0.60245230 0.79295350649762
20 11,010,048 0.792953246128268 0.000000171787 0.60239617 0.79295350639635
21 22,020,096 0.792953349612308 0.000000103484 0.60239881 0.79295350639922
22 44,040,192 0.792953411952475 0.000000062340 0.60241334 0.79295350640873

n = 23×2i−1
19 6,029,312 0.792953102146911 0.000000266820 0.60244184 0.79295350647418
20 12,058,624 0.792953262877942 0.000000160731 0.60239471 0.79295350639462
21 24,117,248 0.792953359702563 0.000000096825 0.60240155 0.79295350640158
22 48,234,496 0.792953418030943 0.000000058328 0.60241269 0.79295350640840

n = 25×2i−1
19 6,553,600 0.792953126057606 0.000000251038 0.60243188 0.79295350645355
20 13,107,200 0.792953277281640 0.000000151224 0.60239496 0.79295350639492
21 26,214,400 0.792953368379547 0.000000091098 0.60240363 0.79295350640322
22 52,428,800 0.792953423258054 0.000000054879 0.60241238 0.79295350640826

This lacks a proof that the ratio tends to a constant limit. But even if the ratio merely changes slowly, then the estimated limit should still be good. Hence the claim that this algorithm “provides strong evidence that the expected value ≈ 0.79295350640…”, with the next digit probably being not more than 1 different from 7.

This also lacks a proof that the limit of the restricted strategy is indeed the value of the whole game. But small changes to the algorithm (present but deactivated in the code) allow calculation of the probability of reaching the rightmost boundary at t=max_t, and this probability appears to be proportional to n−0.231…. As n tends to +∞ this probability tends to 0, suggesting that the boundary becomes irrelevant in the limit. Which is encouraging.

For these sequences the raw output data is available:

 Scores n = 3i; 2×3i−1; 2i; 3×2i−1; 5×2i−1; 7×2i−1; 9×2i−1; 11×2i−1; 13×2i−1; 17×2i−1; 19×2i−1; 21×2i−1; 23×2i−1; 25×2i−1 Strategy t≤2048, n = 3i; 2×3i−1; 2i; 3×2i−1; 5×2i−1; 7×2i−1; 9×2i−1; 11×2i−1; 13×2i−1; 17×2i−1; 19×2i−1; 21×2i−1; 23×2i−1; 25×2i−1 Precision Some t, n = 3i; 2×3i−1; 2i; 3×2i−1; 5×2i−1; 7×2i−1; 9×2i−1; 11×2i−1; 13×2i−1; 17×2i−1; 19×2i−1; 21×2i−1; 23×2i−1; 25×2i−1 Scores n = 226; Strategy t≤26(216−2), t≡0 mod 64, n = 226; Precision Some t, n = 226

Strategy  The above process estimates the value of the game as a limit, without specifying the strategy that would achieve that limit. Specifying the strategy is not easy: if n=1,048,576 then h=177 and t=162 is a stopping point. But with n twice as big it becomes profitable to keep playing: with 162 tails one would need at least 178 heads to stop. Let h[t] be, for any value of t, the largest number of heads with which one would toss again. The algorithm does not provide a direct estimate of h[t], but we can guess that with t<<n, the algorithm’s estimate is close.

The upper chart on the right, with the dark blue line, shows ( h[t] − t ) / √t. This is not quite a horizontal line, though over a huge t range does not vary much. We might guess that ( h[t] − t ) is, in the limit, ≤O(√t). The lower chart, with the dark green line, replaces “√t” with “√( h[t] + t )”, and this make no difference to the conclusion: stopping states require that the number of heads exceed the number of tails by something of the order of something no greater than the square root of the number of tails, or the square root of the number of tosses. The author has failed to find a better-behaving more-plausible combination of √t, Ln(t), and Ln(Ln(t)).

Next Steps

For large increases in CPU time (and perhaps memory), this algorithm could calculate a few more decimal places of accuracy. Readers with CPU time and memory to spare are invited to cope.

But an algorithmic improvement is needed to calculate many more decimal places, which might work by better estimating the value of the game on the boundary, and then using backward induction from this improved estimate. The above algorithm estimates the boundary values as Max( h/(h+t), ½). A better estimate might come from solving a continuous version of the game, and using that value on the boundary. A solution to such a continuous version might perhaps have the following properties.

• ∃ function EV[h,t], with ½≤EV[h,t]≤1 ∀h≥0, t≥0, ht>0, representing the value of the continuous game (the rules of which have not been specified) after h heads and t tails.

• ∃ continuous stopping function h[t]≥t, monotonically increasing in tt>0, such that h≥h[t] ⇒ EV[h,t]=h/(h+t).

• t and h such that 0<t and 0<h<h[t], EV[h,t] is continuous and once partially differentiable in h and t. EV[h,t] is also continuous on the h=h[t] boundary.

• h<h[t] ⇒ ∂EV[h,t]/∂h = −∂EV[h,t]/∂t. This is the analogue of the “Avg(↓,→)” from the discrete game.

• ∀ fixed h, as t tends to +∞ the limit of EV[h,t] is ½. This defines the rightmost ‘boundary’.

• Then h[t] is chosen to maximise EV[0,t′], for some t′>0. (It is suspected that replacing t′ with a smaller positive value would leave h[t] unchanged for t>t′. Hopefully this would be clear from the functional form of ht[t].)

A solution would be an estimate of the value of the discrete game for large h and t. An analytical solution could be used to generate the starting values of the backward induction, substantially improving the estimate of sn/n. Mathematicians skilled in partial differential equations are invited to help. All, whether or not skilled mathematicians, are invited to invite skilled mathematicians to www.jdawiseman.com/papers/easymath/coin-stopping.html. Please, somebody help. (This request for help is echoed in Open Mathematical Questions at jdawiseman.com.)

Unimportance

Pretend for a moment that this puzzle has an expected value that is some combination of elementary mathematical constants and functions. If this were true, the game would, in some way, illuminate a property of those constants and functions. There would be a connection between this puzzle and the rest of mathematics.

But the Inverse Symbolic Calculator at the Centre for Experimental and Constructive Mathematics does not know of any such combination for even the first eight digits of the limit (“0.79295350”). So it seems likely that this problem is of no importance whatsoever, even if it has given the author a little entertainment.

Julian D. A. Wiseman
October and November 2005

Afterword

In June 2009 this paper was cited: Luis A. Medina & Doron Zeilberger, ‘An Experimental Mathematics Perspective on the Old, and still Open, Question of When To Stop?’, arXiv:0907.0032.

September 2010: Medina and Zeilberger asked whether one should stop at h=5 t=3. This is answered in the affirmative by The Chow & Robbins Problem: Stop at h=5 t=3.

January 2012: this paper was cited by Olle Häggström and Johan Wästlund in ‘Rigorous computer analysis of the Chow-Robbins game’, arXiv:1201.0626.

In March 2012 this paper was cited in Stochastic Games and Dynamic Programming, by Professor Henk Tijms, published in the July 2012 issue of the Asia Pacific Math Newsletter.

August 2012: in the above analysis, estimates of the limiting value were obtained from by considering three values of the restricted game, always from n’s in geometric progression. But it cannot be optimal to take just three data points at a time. Surely one could derive a better estimate by considering all the data simultaneously. Presumably the fitting must be done with a greater weight given to larger n’s, and hopefully in a manner that still allowed an estimate of the uncertainty about the limiting value. Help from a competent statistician would be greatly welcomed.

March 2013: help was sought on MathOverflow.net: The Chow & Robbins game ≈ 0.79295350640: improvements could come from simple statistics, or from a continuous version of the game. And help came: hurray!

September 2013: the problem had previously been discussed with friend Phil Wakely, who has extended the calculation and its precision. His paper agrees with “0.7929535064”, but not necessarily the next 0, and hence has even less support for “the next digit probably being not more than 1 different from 7”. So this author’s last 1½ decimal places should be considered speculative rather than known.

The results suggest EV(0,0)=0.7929535064… Note that this is one less digit of precision than in the conjecture from  ("the expected value ≈0.79295350640…, with the next digit probably being not more than 1 different from 7") as despite the increased precision and larger n computation, the ratio of deltas between successive EV(0,0) estimates with increasing n does not as yet seem to be converging sufficiently.

(Also note that our definitions of n differ by a factor of 2, as is apparent from the tables of results.)