Roll It Twice
May 31st, 2013 at 11:49:58 AM
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"Roll It Twice" is a dice based, "roulette" game where the dealer, then the player, each roll two six sided die. One of the die is red, and the other black. A player continues to roll until one of his die match the dealer die color for color and spot for spot.
What are the odds of this occurring? I assume it is (1/6)+(1/6) = 2/6 = 1 in 3. Does this mean the average number of rolls per player is only ~3?
What are the odds of this occurring? I assume it is (1/6)+(1/6) = 2/6 = 1 in 3. Does this mean the average number of rolls per player is only ~3?
Simplicity is the ultimate sophistication - Leonardo da Vinci
May 31st, 2013 at 1:24:34 PM
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Nope, not quite, because you are double counting when both dice match the dealer.
Each die has a 5/6ths chance of NOT matching the corresponding dealer die.
So the probability of you continuing to roll is:
(5/6)*(5/6) = 25/36 = 0.694444
Probability of number of rolls:
1 roll: 1 - 25/36 = 11/36 = 0.30555556
2 rolls: (11/36)*(25/36) = 0.212191358
3 rolls: (11/36)*(25/36)^2 = 0.1473551097
4 rolls: (11/36)*(25/36)^3 = 0.1023299373
and so on until numbers get small enough for us to quit
Then expected number of rolls becomes:
1*(11/36) + 2*(11/36)*(25/36) + 3*(11/36)*(25/36)^2 + 4*(11/36)*(25/36)^3 + ... = 36/11 = 3.27272727
Each die has a 5/6ths chance of NOT matching the corresponding dealer die.
So the probability of you continuing to roll is:
(5/6)*(5/6) = 25/36 = 0.694444
Probability of number of rolls:
1 roll: 1 - 25/36 = 11/36 = 0.30555556
2 rolls: (11/36)*(25/36) = 0.212191358
3 rolls: (11/36)*(25/36)^2 = 0.1473551097
4 rolls: (11/36)*(25/36)^3 = 0.1023299373
and so on until numbers get small enough for us to quit
Then expected number of rolls becomes:
1*(11/36) + 2*(11/36)*(25/36) + 3*(11/36)*(25/36)^2 + 4*(11/36)*(25/36)^3 + ... = 36/11 = 3.27272727
