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 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)

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 re(2): Different numbers | Comment 7 of 8 |
(In reply to re: Different numbers by armando)

(I posted before in a hurry without reviewing. There are some faults, so I do it again).

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 one of this sequencies:

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

The possibility of having a 9 in the first column is 1/9. Similarly for 9+1+5 (with 9 in the first column + 1 in other column and 5 in a third one is), possibility is 1/9*8*7.

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

The second player only wins in his third ball (the sixth one) when he obteins the winner sequence and the first player failed to obtein it in his fifth ball. The possibility for both players obteining the 15 in six balls can be deduced 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 would be able to have two of the eight other sequencies (8+3+4; 7+6+2). As the order of digits doesn't count, there are 12 possibilities for him (six for each number). Considering that the first player can obtein the sequence 9+1+5 in different order, the possibilities are 72, and adding the other sequencies (different of 9+1+5) that allow player 2 to obtein himself also a 15, it totalize 72*6 = 432. It means there are 432 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 432/9*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, if he has 15 but not with one of this numbers. His possibility of winning is then 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 12:03:52

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