1 in 100, with a twist

Originally posted by mtthw
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...

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 luskin
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?

No, 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.

I knew I asked the right guys!! Great stuff, mttw. I also posed this question to the Wizard of Odds, hopefully he'll post his analysis in his next column. I'll keep ya's posted.

Originally posted by mtthw
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. 🙂

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?

Originally posted by uzless
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?

Ah 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.

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 uzless
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?

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.

Originally posted by mtthw
No, 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.

Wait; is this the Ultimate Question to Life, the Universe, and Everything?!

Originally posted by AThousandYoung
Wait; is this the Ultimate Question to Life, the Universe, and Everything?!

42 is the answer. I just can't remember the question.

( Sorry, deleted because of ill formatting )

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?

Does anyone know how to find the Nash equilibrium for this game?

Originally posted by mtthw
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.

how bout you give us a number assuming your opponent picks 42 since he has figured out it's the optimum number

Originally posted by uzless
how bout you give us a number assuming your opponent picks 42 since he has figured out it's the optimum number

Well, that one's easy, innit!

Originally posted by PBE6
I knew I asked the right guys!! Great stuff, mttw. I also posed this question to the Wizard of Odds, hopefully he'll post his analysis in his next column. I'll keep ya's posted.

He did analyze it!

Question #3 - http://wizardofodds.com/askthewizard/205

Originally posted by PBE6
He did analyze it!

Question #3 - http://wizardofodds.com/askthewizard/205

i went to his mathinfo website to see the answer but it just gives the answer without showing how he got it.