2048 Tile Game probability
I calculated the probability to obtain (with the perfect game) the max score in the original configuration of 2048. The value is 2.205919*10^(-6012). Here is why. The max number of moves is: we consider first the maximum tile on the board which is 131971 and the numbers that appear all 2 except for at least 15 "four" that are necessary to preceed to the bitter end, we obtain: 131072-2-(1*15)-3=131052. Concluding with 8 and a tile of 2 or 4 is the meaning of the "-3" at the end, so 3 moves are lacking to even the score in (131072*2)/2. The maximum number (that I use for the probability) of numbers that appear is greater than two units to the maximum number of moves (131052), since the beginning of the game we already have 2 numbers on the board. But the real influential numbers are 131051 because the the last can't be combined with other tiles, no matter if 2 or 4. So, the max possibile score is 12+44+124+316+764+1788+4092+9212+20476+45052+98300+212988+458748+983036+2097148=3932100. And the probability to obtain it (with a perfect game) is 10^(-15)*(0.9)^131053=2.205919*10^(-6012) and the most likely score in the is 3932100-2097148=1835008. The scores are defined by tiles of 2^k, so the score=(k-1)*2^k. With a perfect game and a game finished with tiles from 131072, 65536,.., 64, 32, 16, 8, Y (happens once in a million billion games), the expected score would be 3932156-(131053/10)4 = 3879734.8 (probability of having a 4 on the board in the individual event is 1/10 and assuming that in 15 cases this happens by default, otherwise you do not continue the game.
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