maths again
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19 Nov 2007, 23:20
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Journals
Hi, i have one task for u :
(my english sucks so i will try my best):
2 players ar playing a game, and the player, who will win 6 rounds/make 6 wins as first will take the whole money. As the game was 5:3, they stopped to play. How should they share the money? (how much should get every player?)
(my english sucks so i will try my best):
2 players ar playing a game, and the player, who will win 6 rounds/make 6 wins as first will take the whole money. As the game was 5:3, they stopped to play. How should they share the money? (how much should get every player?)
(x\8)*5=first player
(x\8)*3=second
only other option is judging by probability of winning next round but who said that its more fair?
if the task mentioned strong 50\50 chances it would force the one and only solution, but it didnt
the guy with 5 wins wouldnt agree to stop playing to get only 5/8.
The only way the first player could lose the game is losing 3 next rounds and the probability of that is (1/2)^3 = 1/8.
there is no rule how it should be done, there could be many solutions based on different logic
the probability theory is a theory based on probability, it has a random factor included in it, wich makes it non solid ;)
Ofcourse it's a way to solve the problem, but since the numbers generated by the theory are completely random, the other solutions are correct too :)
Let's assume they have equal probabilities to win subsequent rounds. Thus, for the second guy to win, he has to win 3 consecutive rounds. That has a probability of (1/2)^3 = 1/8. That means the first guy has a probability of 7/8 of winning the game. This is the ratios of the money I would give them.
BUT. It could be done differently. The probability could be assumed to be 3/(5+3) = 3/8 for the second guy to win a round. This makes for a probability of getting the whole money of 27/512. The first guy would win with a probability of 485/512. This is another way of splitting the money.
i was just about to write an example with betting systems on sport matches
but if till now A won 5 and B won 3, that can also mean that A got bigger chances not only to win w whole match but also a next round as till now he won more rounds.
hmm btw, 50/50 is good at start of game. but when you see that A won 5 and B won 3 you cant rly say they are of the same strenght.
nono rly, i get your point. gn8, give some more logic tommorow or smth :)
mm you prolly mean "tought". but cant it be "learned" as well?
e..ooOps..nice€ skillz here..
gun=natürliche auslese?\?\\?+\++?
or... tommorow :D now im going to sleep =)
Rize and I are going to bring more cookies :)
I'll be up for the one tomorrow :)
i see smarter people have taken it as 50/50 :]
if first guy wins a round it's a, if 2nd guy wins it's b
ways for A to win: a = 0,5 ba = 0,5*0,5 = 0,25 bba = 0,5*0,5*0,5 = 0,125 -> P(A wins) = 0,5 + 0,25 + 0,125 = 7/8
ways for B to win: bbb = 0,5*0,5*0,5 = 0,125 = 1/8
Keep it simple