Elimination Dice Roll Game

Question

What are the odds and or probability of winning this game?

How it works: Roll two 6 sided dice for 11 rounds. You are eliminated if you roll over the highest number for the round. Round one # is 12, round 2 is 11, round 3 is 10 and so on. So to win, in the 11th round you must roll 2 (snake eyes).

Answer 1

All 11 rounds are independent in that no round affects the probability of the next round. Consequently, calculate the probability of each round and multiply them all together.

For a single roll of two dice, there are 36 possible outcomes:

1-1 (total of 2), 1-2 (total of 3), 1-3 (total of 4), 1-4 (total of 5) etc, all the way up to 6-4 (total of 10), 6-5 (total of 11), 6-6 (total of 12)

Round 1 - Roll 12 or less = 1/1

Round 2 - Roll 11 or less = 35/36

Round 3 - Roll 10 or less = 33/36 = 11/12

Round 4 - Roll 9 or less = 30/36 = 5/6

Round 5 - Roll 8 or less = 26/36 = 13/18

Round 6 - Roll 7 or less = 21/36 = 7/12

Round 7 - Roll 6 or less = 15/36 = 5/12

Round 8 - Roll 5 or less = 10/36 = 5/18

Round 9 - Roll 4 or less = 6/36 = 1/6

Round 10 - Roll 3 or less = 3/36 = 1/12

Round 11 - Roll 2 or less = 1/36

The total probability of winning all 11 rounds is: (1/1)(35/36)(11/12)(5/6)(13/18)(7/12)(5/12)(5/18)(1/6)(1/12)(1/36) = 4379375/(3.13457x10¹¹) = 1.397x10^-5 or 0.0000139712 or 0.00139712%

The odds are slightly different. To calculate odds, you divide successes by failures (where probability is successes divided by total) so, since there are 4379375 and 3.13457x10¹¹ total then there are (3.13457x10¹¹ - 4379375) failures which is 3.13452x10¹¹ failures meaning the odds of winning are 4379375/(3.13452x10¹¹) = 0.0000139714 or 0.00139714%

Bottom line, bet money on not winning.