Originally posted by mtthwAnother question: is 42 the only number that would give you a winning edge in the original version of the problem, or just the best one?
Of course, the next question is what your strategy is if your opponent has had time to think about it! That's where the game theory really comes in...
Originally posted by luskinNo, as it happens any number above 1 and below 75 gives you an edge. 42 is the best, but it's a pretty flat distribution.
Another question: is 42 the only number that would give you a winning edge in the original version of the problem, or just the best one?
Originally posted by mtthwthe intent of the question should be to beat your opponent...not pick the optimum number. If your opponent had the same strategy as you, and you therefore knew he'd pick 42, what number would you pick to beat him? 43? 41?
When you take this into account, an optimal solution will take the form of a probability distribution - you need to pick randomly, but with a bias towards the more effective numbers (so with a maximum probability somewhere around 42, probably). You then assume your opponent has an identical strategy.
So now all those 'simple' probabilities take the form of l probably have some nasty integral equation that I have no intention of trying to solve. 🙂
Originally posted by uzlessAh yes, but we assumed our opponent was moving randomly when we chose 42, so he would have to assume that we were moving randomly for him to want to choose 42.
the intent of the question should be to beat your opponent...not pick the optimum number. If your opponent had the same strategy as you, and you therefore knew he'd pick 42, what number would you pick to beat him? 43? 41?
So how do we choose what to play when both players are optimising against each other? We utilise the concept of Nash Equilibrium, and try to find a solution where, for each player, they are playing their optimal strategy given what the other player is playing.
Originally posted by uzlessAs I said, the optimum strategy (if your opponent is also playing the optimum strategy) is not to pick a particular number. It's to pick from a random distribution that is biased towards particular numbers.
the intent of the question should be to beat your opponent...not pick the optimum number. If your opponent had the same strategy as you, and you therefore knew he'd pick 42, what number would you pick to beat him? 43? 41?
I want to answer the original question, what is the best number
when both the referee and the opponent randomly choose their number.
Let a be my choice, p the opponents choice and r the referee's choice
There are 6 disjoint distinctions:
#1 p>r and a > r (will be repeated)
#2 p>r and a < = r (I win immediately)
#3 p < = r and a > r (opponent wins immediately)
#4 p < =r and a > p and a < =r (I win immediately )
#5 p < =r and a=p (will be repeated)
#6 p < =r and a < p (opponent wins immediately )
The likelihood to win for a choosen a is:
P(#2)+P(#4) / ( P(#2)+P(#3)+P(#4)+P(#6) )
The problem is to find the best a.
Ok, I was not able to calc the conditional probabilities for
the 6 cases, but I counted them for all p and r and a
( just 1 mio cases to count, did only take some millisecs ).
The shortend result table is:
1: 0,5000000
2: 0,5049990
...
26: 0,6000000
...
36: 0,6164190
37: 0,6169666
38: 0,6172911
39: 0,6173862
40: 0,6172459
41: 0,6168639
42: 0,6162340
...
50: 0,6015301
...
75: 0,4204524
...
99: 0,0390176
100: 0,0196078
So the optimum is at 39.
Can anyone calc the conditional probabilities?
Originally posted by mtthwhow bout you give us a number assuming your opponent picks 42 since he has figured out it's the optimum number
As I said, the optimum strategy (if your opponent is also playing the optimum strategy) is not to pick a particular number. It's to pick from a random distribution that is biased towards particular numbers.