Probability of iindependent event ocurring with dice thrown in matches
Pairs of d10 dice are thrown by two players in matches, with let's say one player trying to get 2 and the other trying to get 10. The probability of getting a 2 is 1/25 or 4%, and the probability of getting a 10 is 19/25 or 76%
Questions:
1. What is the probability that one player or the other, but not both, will obtain their number?
This one stumps me.
2. What is the probability that neither player will obtain their number?
I believe the answer to this is to multiply the improbability of getting one number by the improbability of the other number. so .96 x .24 = .23 or 23%
3. What is the probability that both will obtain their number?
I believe you multiply the two probabilities, so the answer would be .04 x .76 = .0304 or 3%
4. What is the probability that one player or the other or both will get their number?
Perhaps multiply the probability found for question 1 by the probability found for question 3?
Thanks in advance.
Probability of getting a 2 is 1 out of 36 roll. Prob of a 10 is 3 out of 36. What is a d10 dice?
Let us assume that getting a 2 is 4% and 10 is 76%
Question 1
prob of 2 rolling but not 10 is (.04)(1-.76)=.0096
prob of 10 rolling but not 2 is (.76)(1-.04)=.7296
prob of both 2 and 10 is rolled (.04)(.76)=.0304
prob of neither 2 or 10 is (1-.04)(1-.76)=.2304
P(t)= 1
1-P(neither)-P(both)=1-.2304-.0304=.7392
which is same as P(only 2)+P(only 10) =.7392
Question 4
P(neither)=.2304
1-P(neither)=.7696
Quote: morphix
1. What is the probability that one player or the other, but not both, will obtain their number?
p2 * (1 - p10) + (1 - p2) * p10 = 462 / 625 = 0.7392
Quote:
2. What is the probability that neither player will obtain their number?
(1 - p2) * (1 - p10) = 144 / 625 = 0.2304
Quote:
3. What is the probability that both will obtain their number?
p2 * p10 = 19/625 = 0.0304
Quote:
4. What is the probability that one player or the other or both will get their number?
1 - p2 * p10 = 606 / 625 = 0.9696
Quote: MangoJp2 is the probability, that a throw will evaluate as "2", same for p10 evaluating to "10". You gave some numbers, but I don't know if they are correct. Let's asume so to give some numbers. If they turn out to be different, just plug them in into the formula and evaluate again (you can do that, Right?)
Yes, no problem. Maybe some background will explain my questions.
The examples I gave were correct for probability, but the actual game uses d10 dice numbered 1-5 twice. There are challenges in the game using such dice thrown by two players in matches to resolve things like who gets to hold a position and who must retreat, whether one will proceed forward or be blocked by an opponent's barrier, etc.
Players must use sets of opposing numbers corresponding to their current position on the noard (1 vs 10, 2 vs 9, 3 vs 8, 4 vs 7, and 5 vs 6) The lower number in each set can be obtained either on one die or by adding the dies together. In each match, the challenger gets to pick either number and the defender takes the other number. The lower number always has the better odds, those odds being greatest for numbers farthest apart and poorest when closest together. For example, 1 vs 10 being 9/25:1/25 or 9:1 for 90% to 10%, and 5 vs 6 being 13/25:5/25 or 13:5 for 72% to 28%. The odds for the other challenge number combinations distribute somewhere in between.
My recent design goal has been to shift the odds so that the one seeking the lower number does not have such an advantage. One solution is to have the challenger seeking the lower number throw the dice once but the defender seeking the higher number throw twice. This changes the odds for 1 vs 10 to 82% to 18%,, and 5 vs 6 to 59% to 41%, and similarly shifted for 2 vs9, 3 vs 8, etc.
My four questions pertain to deriving the probability for any one of these challenges to be settled, and ultimately the probability of settling over two or three matches, as I don't want it to take more than a few matches on average to settle any challenge. After I play around with the formulas you provided I should be able to see how it all plays out and will post if any here are interested. May have more questions too!
Really appreciate it:)
