Problem 232

[sfabriz]

After having read Daniel's stunning explanation for 233 I was wondering if anyone of you mathematicians would like to do the same thing for this one. It's been really hard to solve for me and still I can't understand many of the methods used by those who posted on the problem's thread.
So, if anyone would like to do it, I'll be a very interested reader. Probability theory is really one of my worst weaknesses...
Cheers,
sfabriz

[hk]

But please do it not here in this forum....

[sfabriz]

Of course, I meant in the problem's thread...

[hk]

I was afraid that some helpful guy would start something here we would've had to delete immediately.

[sfabriz]

Oh yes, of course. I didn't specify that 'cause i gave for granted that we can't write anything here that could spoil the problem for others.
Cheers,
Sfabriz

[zwuupeape]

Just to find out if my solution is at least approximately true.

If the target is 66 points, Player 2 wins with probability 0.811849718826
If the target is 50 points, Player 2 wins with probability 0.78973645941617376
If the target is 10 points, Player 2 wins with probability 0.6999731081357714

[daniel.is.fischer]

Close, but not correct

[sivakd]

Is

Expand

0.66057914

correct for 10?

Edit: Never mind, I solved this problem.

[DDgeva]

I still don't get this problem. My algorithm is pretty straight-forward and very fast, but the answer isn't correct.
This happens to me way too often..

Am I missing something here?
Am I correct stating the the probability of player 2 winning when they're playing upto 1 is 1/3?

My program's outputs:
Playing up to 2: 0.52
up to 4: 0.601777777777777777777777777778
up to 10: 0.667026820050150500796231363509
up to 30: 0.761458127957947605300352303773
up to 80: 0.823494959687249598060259595678

[jaap]

Am I correct stating the the probability of player 2 winning when they're playing upto 1 is 1/3?

Yes, but your other answers are incorrect.
When playing up to 2 the answer should be 0.5259259259259259...
For 4 it is 0.603753086419753...

[DDgeva]

Am I correct stating the the probability of player 2 winning when they're playing upto 1 is 1/3?

Yes, but your other answers are incorrect.
When playing up to 2 the answer should be 0.5259259259259259...
For 4 it is 0.603753086419753...

I'm getting 0.52 even when calculating it by hand, using the same approach that produced a correct result for a game up to 1.
I'm guessing my method is probably incorrect. I'll try to figure out what I'm missing..

[Waldovski]

Question: When you say "On each turn Player 2 selects the number, T, of coin tosses that maximises the probability of her winning.", do you mean maximising the probability of reaching 100 in one turn? (in other words, am I right in saying that the value of T will depend on Player Two's current score only?)

[harryh]

Question: When you say "On each turn Player 2 selects the number, T, of coin tosses that maximises the probability of her winning.", do you mean maximising the probability of reaching 100 in one turn?

Your question has already been answered; see this previous post.

[Hippunky]

Hi All,

I've tried calculating T (the number of throws player 2 chooses) using a very simple formula based on "P2 score" - "P1 score". Should I instead be looking to a more dynamic means of calculating T on each turn (based on more that just the difference in scores) as I can't convince my current approach to return a probability >= 0.6 for the 'first to four points' variant.

Thanks,

Will.

[ffff0]

I'm stuck here.

Can someone check my values please:
10 - 0.6700226472266161789570328463236506093
20 - 0.7289790525667717256992685609669337712
50 - 0.7943581957799124887668550310834422535

And i'd like to PM my value for 100 to someone.

Edit:
Finally solved this one! I thought that I calculate T values for all possible parameters, but I totally forget about another one.

Correct values will be:
10 - 0.670064571682367
20 - 0.729009756737729
50 - 0.794385833126634

[MuthuVeerappanR]

I solved this problem yesterday and was playing around with the values... I'm noticing a slightly counter-intuitive result which may be wrong or may be hard to believe...

I tried to find the probability when both the players are presented with the option of choosing the number of tosses before their turn. I thought, now that the game is symmetric except for the first mover advantage (or disadvantage), the B's winning probability should be somewhere around 1/3 because in the classic case of one toss allowed for both players, the first player wins with probability 2/3.

But when I modified my program, it says no matter what the final limit is, when A starts the game, B will win with probability 1. And when the limit is high, no matter who starts B will win with probability 1. Am I missing something? or is there a reason for this?


PS: Though the probability of my program being wrong is very low , it still is possible. If thats the case, can someone determine the value of B's winning probability when A starts the game and both players have the option of choosing T before their turn?

[papadil]

I know I'm a little late to this problem. Just noting that the phrase "will always selects the number" does not seem correct. Should be one of: "will always select the number" (no 's' in select) or, "always selects the number" (no 'will'). I know...nitpicking....

[RobertStanforth]

I know I'm a little late to this problem. Just noting that the phrase "will always selects the number" does not seem correct. Should be one of: "will always select the number" (no 's' in select) or, "always selects the number" (no 'will'). I know...nitpicking....

Thank you for taking the time to flag this! I have corrected it to "will always select the number".