"English is the most ambiguous language in the world." ~ Yours Truly, 1969
So, is the probability of making a point on the first throw 1 - 7/72 - 2/72 = .875% ?? If so, how does one calculate the probability of ultimately winning at the start of the game?
Yeah, both of these probabilities you computed are correct. P(win on first throw) = 7/72
You are also correct that the probability of "his point" is 126/144 = 0.875.
What is P(win on second throw)?
What is P(win on third throw)?
What is P(win on fourth throw)?
You have to add all of these up...forever. Can you think of what kind of sequence you're going to end up with that has infinitely-many terms that add up to a finite sum?
Hint: P(win on second throw) = P(continue on 1st AND his point on second) = P(continue on 1st)*P(his point on second), where you take the product since these are independent events.
The Probability of winning on subsequent throws will consist of a similiar product of probabilities (continue n-1 times and get his point on the nth time). Except one more factor will be added on to that product in each subsequent term...
EDIT: This problem was a lot of fun! Slightly tricky until you see the pattern.
Since I'm not completely confident I wouldn't fudge one of the small calculations along the way (or even set them up properly), do you think there would be any way to calculate the win probability at the start of the game knowing the probability of making a point on the first throw (87.5%) and the probability of losing on any one of the subsequent throws (8.333333% per throw, 1/12, or the chances of getting 13 on each subsequent throw of the die)? I guess, by proxy, you know that the chances of winning on a subsequent throw or having to throw again would be 91.666666%).
On this or any subsequent throw the player loses if he gets the sum 13 and wins if he gets his point but must throw both dice again if any other sum occurs. The player continues until he either wins or loses. To the nearest percent, what is the probability at the start of any game that a dice thrower will win?
As I mentioned earlier, you can try to calculate the total probability of losing instead, and take 1 - P(lose). Either way, you have to set up "small calculations", one for each round. Just do it! Do it the way I suggested first, and post your work here. We can point out any mistakes and provide hints along the way. It just requires being a bit methodical. In the end, if computing the win probability directly, and excluding the first round, there are literally only two numbers that go into each term.
Last edited by j-astron; 02-19-2018 at 11:23 AM. Reason: typo
I must have made some mistakes in calculating the following outcomes because my numerator is more than 144 when you add them all up:
2 - 2/144
3 - 2/144
4 - 4/144
5 - 4/144
6 - 8/144
7 - 7/144
8 - 8/144
9 - 8/144
10 - 10/144
11 - 10/144
12 - 10/144
13 - 12/144
14 - 12/144
15 - 12/144
16 - 12/144
17 - 8/144
18 - 8/144
19 - 6/144
20 - 6/144
21 - 4/144
22 - 4/144
23 - 2/144
24 - 2/144
Whatever your point would be, I would guess you would take 1 - (probability of particular point) - 12/144 (lose on 13)
Last edited by bennyJ; 02-19-2018 at 10:21 PM.
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