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Given two dices, we roll them and add the result to a sum (initialised to 0) till sum is $\ge 100.$ The resultant sum can be any number in [100 111]. Which among them have the highest probability of being the resultant sum.

miracle173
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KSHITIZ KUMAR
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    I think you mean >100 –  Aug 04 '17 at 17:35
  • We keep rolling the 2 dice till we have sum less than 100 and stop as soon as sum reaches 100 or more. – KSHITIZ KUMAR Aug 04 '17 at 17:38
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    I think you mean as long as the sum is < 100. – peterwhy Aug 04 '17 at 17:40
  • By simulation with 10 million iterations: it's either 100 or 101, possibly a tie, but if not, I'd bet on 101. No idea how to get an elegant exact analytic solution. – BruceET Aug 05 '17 at 04:29
  • @Gigaboggie: As simulations go, you did a fine job simulating in Excel. – BruceET Aug 05 '17 at 04:52
  • @BruceET Reiner Martin's answer shows you that 101 is slightly more likely than 100. He wrote the probabilities in decimal form, but a computer can give you the exact fractions. –  Aug 05 '17 at 15:48
  • @ByronSchmuland: I do not regularly use Mathematica and am not sure I want to trust its 10th decimal place of accuracy after an intricate chain of computations. Certainly my simulation with $10^7$ iterations is not that accurate. Would welcome authoritative views on accuracy of Mathematica. – BruceET Aug 05 '17 at 18:18
  • @BruceET I don't use Mathematica either, but Maple tells me that the chance of ending at state 100 is $${93331231822982469878637815679368814800320167759280160765014068179459505217069\over 653318623500070906096690267158057820537143710472954871543071966369497141477376}$$ while the chance of ending at 101 is $${23332807968278042259341308510960904430945596638576435328945697702687096283165\over 163329655875017726524172566789514455134285927618238717885767991592374285369344}$$ –  Aug 05 '17 at 20:02
  • @ByronSchmuland. Thanks. That does seem very precise indeed. // I retract my commented speculation that 101 may be more likely than 100. Not because it is incorrect, but because my simulation doesn't really provide the evidence to support that guess. – BruceET Aug 06 '17 at 20:32

2 Answers2

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The probabilities for the possible final sums are: $$ \begin{array}{cc} 100 & 0.1428571427 \\ 101 & \textbf{0.1428571428} \\ 102 & 0.1388888889 \\ 103 & 0.1309523810 \\ 104 & 0.1190476191 \\ 105 & 0.1031746032 \\ 106 & 0.08333333337 \\ 107 & 0.05952380953 \\ 108 & 0.03968253967 \\ 109 & 0.02380952379 \\ 110 & 0.01190476189 \\ 111 & 0.003968253964 \\ \end{array} $$ Thus, 101 has the highest probability by a very small margin.

So how is this computed? Let $a_k$ be the probability that the number $k$ appears as a sum at some point in the game. Then $a_0=1, a_k=0$ for $k<0,$ and $$ a_k = \frac{1}{36} \sum_{i=1}^6 \sum_{j=1}^6 a_{k-i-j} \cdot \chi_{n-i-j<100}, $$ where $\chi_s$ is 1 if $s$ is true, and 0 otherwise.

I have computed this using Mathematica:

a[0] := 1

a[n_] := 0 /; n < 0

a[n_] := a[n] = Sum[If[n - i - j < 100, a[n - i - j], 0], {i, 1, 6}, {j, 1, 6}]/36

Also, the expected sum (this has not been asked) is about 103.41666666745046042. This can be computed from a similar recursion.

Reiner Martin
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5

The single die version of this question is analyzed here: Probability of dice sum just greater than 100

In my paper Linear recurrences via probability, American Mathematical Monthly 122, 386-389 (April 2015), I consider the general version of this game. In particular, equation (5) gives the asymptotic hitting probabilities as the lower limit (100 in this problem) gets large.

For the 2 dice game, the asymptotics say that $$\mathbb{P}(\mbox{games ends at state }99+k)\approx{\mathbb{P}(\xi\geq k)\over\mathbb{E}(\xi)}, \mbox{ for }k=1,2,\dots, 12,$$ where $\xi$ is the sum on a pair of dice. These probabilities are $$1/7, 1/7, 5/36, 11/84, 5/42, 13/126, 1/12, 5/84, 5/126, 1/42, 1/84, 1/252,$$ or in decimal form $$.1429, .1429, .1389, .1310, .1190, .1032, .08333, .05952, .03968, .02381, .01190, .003968.$$

  • This contradicts @Reiner Martin's answer. Is this due to the inaccuracy of floating point numbers? – Henry Aug 04 '17 at 20:16
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    Of course rounding will cause some discrepancies. Bear in mind that he has calculated the exact results, while I only give approximate results. –  Aug 04 '17 at 20:19
  • Gotcha, thanks. – Henry Aug 04 '17 at 20:19