Math for interesting dice game

Hi, I am playing a dice game, quickly explained as follows:

2 die(?) are rolled. first dice determines winning number and second dice determines gain ammount (roll+2). so the minimum gain is 3 times your bet, and max is 8 times your bet.
you can bet on as many numbers as you want.

my question is, if I always bet on 4 numbers and have a real 2/3 chance of winning (not like roulette), what are the odds that I lose x times in a row? (formula?)

same if I bet on 2 numbers for 1/3rd chance of winning. odds of losing 8 in a row for example?

thanks!

vk

With 4 numbers, the odds of losing on any roll are 2/6 = 1/3. To see the odds of losing x times in a row, just raise 1/3 to the power of X. E.g. the odds of losing 5 times in a row are 1/3^5 = 1/3 * 1/3 * 1/3 * 1/3 * 1/3 = 0.00411522634 or 4.12%. You can easily calculate this with Google, just put (1/3)^x in google where x is the number of rolls.

For betting 2 numbers, the chances of losing are 4/6 = 2/3, and the math is (2/3)^x. In the case of 8 rolls, the probability of losing all 8 times is 0.0390184423.

A question about the rules of the game posed here:

Are the winnings paid as "(x+2) to 1", or "(x+2) for 1"?
If the original bet is returned with the winnings on a successful roll, this game seems to have positive expectation for the player.

Average winnings per roll, 5.5 units
Betting all 6 numbers gives a loss of 5 units for the five losing numbers, and a gain of 5.5 on average for the winner, if the winning number returns the original bet.