In-Game "Gambling" - Looking for chance at winning
Thread Rating:
1. A random number 1-99 is picked by the computer and told to you. At the same time, another random (secret) number that is NOT that number is picked but kept from you.
2. You are given one number and told to guess whether the other (secret) number is above or below that number. They cannot be the same number. Let us assume that you always go with the mathematically correct answer.
3. You are allowed to be wrong three times and must be correct six times in order to "win". In reality, you can only be wrong twice because #3 means you lose the item and experience so please analyze for both.
What is the probability for winning - or would this require a computer simulation? If it's the latter no need just state it please.
Example:
Above or below 60.
Guess below.
Answer is 53 - win 5 more to go.
If you are allowed to be wrong three times, I get 0.83896.
I used 149/198 as the probability of getting any one guess right.
(These are my revised results.)
I think the probability of winning against a number N is (N-1)/98 when N > 50 and (99-N)/98 when N < 50.
The probability of getting 6 guesses right with at most two mistakes is (I think) p^6+6*(1-p)p^6+21*(1-p)^2*p^6, where p is the probability of the correct guess for the given number. The overall probability of a win is an average of all those individual probabilities for each n (because the probability of a particular n is 1/99).
I am getting 0.6455 by this method ...
Quote: weaselmanWhere does 149/198 come from?
149/198 is p in your formula. There are 99 numbers for the given number and 98 for the unknown number. This gives us 99*98 = 9702 total combinations. If the given number is 1 then we have 98 unknown numbers that will give us a win. For 2 its 97,.... 49 its 50, 50 its 49 , 51 its 50, 52 its 51 ..... 98 its 97 , 99 its 98. Add all the second numbers up, we get 7301. So we win with a probibility of 7301/9702. Simplified its 149/198.
I agree with ChesterDog's numbers.
Quote: miplet149/198 is p in your formula. There are 99 numbers for the given number and 98 for the unknown number. This gives us 99*98 = 9702 total combinations. If the given number is 1 then we have 98 unknown numbers that will give us a win. For 2 its 97,.... 49 its 50, 50 its 49 , 51 its 50, 52 its 51 ..... 98 its 97 , 99 its 98. Add all the second numbers up, we get 7301. So we win with a probibility of 7301/9702. Simplified its 149/198.
I agree with ChesterDog's numbers.
I see. Indeed, the sum of the p's over 99 is 149/198s, but don't you think you sum them too early? I think, the probability you are agreeing with is the probability of six wins (with a most two losses) of a single guess - each guess going against a different number, and the way I read the question is that one number is picked, and then you can keep guessing if the next one is below or above until you are wrong three times.
I guess, it could be interpreted your way just as well though ...
Quote: weaselmanI see. Indeed, the sum of the p's over 99 is 149/198s, but don't you think you sum them too early? I think, the probability you are agreeing with is the probability of six wins (with a most two losses) of a single guess - each guess going against a different number, and the way I read the question is that one number is picked, and then you can keep guessing if the next one is below or above until you are wrong three times.
I guess, it could be interpreted your way just as well though ...
OK, I see what you did. You are using the unknown number as the new given number after each guess. Back to the speadsheet I go. :+)
Quote: mipletOK, I see what you did. You are using the unknown number as the new given number after each guess.
No, not quite. I always use the same given number until you either win or lose.
To continue the OP's example:
Example:
Above or below 60.
Guess below.
Answer is 53 - win 5 more to go.
Guess below
Answer is 65 - lose 1 more left
Guess below
Answer is 7 - win 4 more to go
Guess below
Answer is 45 - win 3 more to go
etc.
All guesses going against the original 60.
Quote: weaselmanNo, not quite. I always use the same given number until you either win or lose.
To continue the OP's example:
Example:
Above or below 60.
Guess below.
Answer is 53 - win 5 more to go.
Guess below
Answer is 65 - lose 1 more left
Guess below
Answer is 7 - win 4 more to go
Guess below
Answer is 45 - win 3 more to go
etc.
All guesses going against the original 60.
Ok using that method I get the same as you and .7548 if you can have 3 wrong guesses.
