All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Fateful Fickle Fifteen (Posted on 2005-03-24)
There is a bag with balls numbered 1 to 9. Two players take turns at randomly taking a ball from the bag. If a player gets three balls that sum fifteen, he wins. Getting fifteen with more or less balls doesn't count.

What are the odds of the first player winning? Of a draw?

 See The Solution Submitted by Old Original Oskar! Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Different numbers | Comment 6 of 8 |
(In reply to Different numbers by e.g.)

I've done it "manually" obteining Charlie's values. So it seems they are ok.

For the first player winning in his third ball he has to obtein:

9+1+5; 9+2+4; 8+1+6; 8+2+5; 8+3+4; 7+6+2; 7+3+5; 6+4+5 in which order doesn't count. So there are 8*6=48 possibilities of having 15 adding three balls.

The possibility of having a 9 fixed in the first column is: all the numbers with the nine digits/9 = (9!/9). Similarly of 9+1+5 (with 9 in the first column + 1 in other column and 5 in a third one is): 1/9*8*7.

So, the possibility for player 1 of winning in his third ball is: 48/9*8*7 = 0.0952380.

The second player only wins in his third ball (the sixth one) when he obteins the winner sequence and the first player failed in his fifth one. The possibility for both players obteining the 15 in six balls can be obtein from the above numbers. You can see that if the first player obtein a 6+4+5 there is no possibility for the second of obteining any of the other sequencies. Instead if he obteins a 9+1+5, the second can obtein two of the eight other sequencies (8+3+4; 7+6+2). As the order doesn't count, there are 12 possibilities for him. Considering the sequence 9+1+5 can be obtein in different order the possibilities are 72, and adding the other sequencies that allow player 2 to obtein also a 15 himself, you totalized 72*6 = 426. It means there are 426 numbers with 6 digits and with the sum of 15 in three columns and 15 in the other three.

The possibility of obteining one of these numbers is 426/9*8*7*6*5*4 = 0,0071428 But player 2 only wins in his sixth ball is he has a 15 in his balls but player 1 hasn't. So his possibility of winning is 0.0952380 - 0,0071428 = 0,0880952, which is Charlies value.

I stop here; I intended only say that Charlies' values seems ok to me.

 Posted by armando on 2005-03-28 10:44:16

 Search: Search body:
Forums (0)