Well then, I'm only about four years late with this reply, but here it is!

From the 5488 rolls collected, the breakdown was as follows:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 2832 1332 766 269 179 67 25 10 4 1 2 1 0 0 0 0

No rolls above 13 occurred, which is to be expected when dealing with a sample size this small. This resulted in the following actual percentage, which I'll use for extrapolating the hypothetical probabilities:

1 2 3 4 5 6 7 8 9 10 11 12 51.6035% 24.2711% 13.9577% 4.9016% 3.2617% 1.2208% 0.4555% 0.1822% 0.0729% 0.0182% 0.0364% 0.0182%

Eyeballing this, the game likely uses the simplest infinite series possible. The odds we're looking for 2^{-n} would be:

1 2 3 4 5 6 7 8 50.0000% 25.0000% 12.5000% 6.2500% 3.1250% 1.5625% 0.7813% 0.3906% 9 10 11 12 13 14 15 16 0.1953% 0.0977% 0.0488% 0.0244% 0.0122% 0.0061% 0.0031% 0.0015%

Our deviations from the pattern are all due to sampling error, which is a byproduct of using a finite sized dataset. Emulating these results have a wide range of outcomes that show how much variation occurs purely due to chance:

> table(sample(1:16,5488,replace=T,prob=prob)) 1 2 3 4 5 6 7 8 9 10 11 14 2777 1363 652 345 180 90 42 23 10 4 1 1 > table(sample(1:16,5488,replace=T,prob=prob)) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 2764 1346 671 349 179 85 51 16 12 6 4 2 2 1 > table(sample(1:16,5488,replace=T,prob=prob)) 1 2 3 4 5 6 7 8 9 10 11 12 13 2723 1398 689 340 159 84 46 19 16 9 2 1 2 > table(sample(1:16,5488,replace=T,prob=prob)) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 16 2767 1374 656 340 167 92 43 17 15 8 2 3 1 2 1 > table(sample(1:16,5488,replace=T,prob=prob)) 1 2 3 4 5 6 7 8 9 10 11 12 2752 1356 711 338 172 77 41 24 13 1 1 2 > table(sample(1:16,5488,replace=T,prob=prob)) 1 2 3 4 5 6 7 8 9 10 11 12 14 2807 1299 710 318 178 94 42 10 15 8 2 4 1 > table(sample(1:16,5488,replace=T,prob=prob)) 1 2 3 4 5 6 7 8 9 10 11 13 2766 1343 694 329 194 83 46 17 10 3 2 1

Chi-Square Test:

> prob<-2^(-(1:16)) > chisq.test(tab/length(cc),prob) Pearson's Chi-squared test data: tab/length(cc) and prob X-squared = 176, df = 165, p-value = 0.2646 Warning message: In chisq.test(tab/length(cc), prob) : Chi-squared approximation may be incorrect

A p-value lower than a cut-off of 1 standard deviation (p=0.05) would suggest the game does not follow this payout pattern.

If you paid 250 million to play Bagatelle one million times, and each jackpot ambitiously gave you 500,000 NP, you would only win back between 160 and 165 million NP. With a jackpot payout of 1,000,000 NP, this goes up to a whopping 161 to 169 million NP.

**TL;DR : Bagatelle is a pretty predictable game when you play it enough, and pays out 2/3 of the money you put in.**