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Topic: World Series Betting (Read 3315 times) |
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Kozo Morimoto
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Posts: 114
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World Series Betting
« on: Oct 30th, 2002, 12:07am » |
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Ok, here's my attempt - I have a habit of embarrassing myself with probability riddles, but I'll give it a go...
for the 1st game:
(0-0)
bet $625
for the 2nd game:
(1-0, 0-1)
bet $625
for the 3rd game:
(2-0,1-1)
bet $750
(0-2)
bet $500
for the 4th game:
(3-0,0-3)
bet $250
(2-1,1-2)
bet $750
for the 5th game:
(3-1,1-3)
bet $500
(2-2)
bet $1000
for the 6th game:
(2-3,3-2)
bet $1000
for the 7th game:
(3-3)
bet $2000
I *think* there must be a more general solution that involves N = number of games and W = number of wins for your team as Win/Loss bet sizes seems to be symmetrical except for 2-0, 0-2...
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Garzahd
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Posts: 130
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Re: World Series Betting
« Reply #1 on: Oct 30th, 2002, 10:29am » |
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Interesting, but I think you need to explain where those numbers come from. I assume some clever working-backwards system would probably do the trick.
Under your system, the pattern of W W L L L W W won't leave you enough money; did you intend to have (2-0) be $500 instead?
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Kozo Morimoto
Junior Member
Posts: 114
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Re: World Series Betting
« Reply #2 on: Oct 30th, 2002, 3:41pm » |
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Hey, you were right, it should be $500 for (2-0). Now the symmetry is there and I believe more strongly that there gotta be a more general solution.
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ws1
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Posts: 1
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Re: World Series Betting
« Reply #3 on: Nov 12th, 2002, 5:50pm » |
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Can somebody provide a definitive answer? It's stumped me and some reasonably smart friends after much deliberation.
If the solution is not posted here, how can it be found? Am I missing the point?
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Garzahd
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Re: World Series Betting
« Reply #4 on: Nov 13th, 2002, 1:44pm » |
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Work backwards.
At 3-3 you clearly need to have $2000 in hand, and bet it all.
At 3-2 you want to be at $4000 if you win, $2000 if you lose. Therefore you need to have $3000 in hand and bet $1000.
At 2-3 you want to have $2000 in hand if you win, $0 if you lose. So $1000 in hand, bet it all.
At 2-2 you want to have $1000 in hand if you lose, and $3000 in hand if you win. So plan to have $2000 in hand and bet $1000.
At 3-1 you want to have $3000 in hand if you lose and $4000 if you win, so plan to be at $3500 and wager $500.
Etc.
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Allie
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OK, I did the working backwards thing and got all the bets that you'd make. Whoever said that there must be a formula is right though - the pattern is too neat (put it in a table and take a look). I know we're working with $125 increments ($2000 / 16 possible bets), and that f(x) = Amount remaining/125 - Bet/125, but I cannot figure out the function of wins and losses to make it work! Any hints??
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Kozo Morimoto
Junior Member
Posts: 114
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Re: World Series Betting
« Reply #6 on: Nov 25th, 2002, 4:27pm » |
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Using a standard binomial tree definition of naming nodes,
root = (0,0)
1st level = (1,1), (1,0)
2nd level = (2,2), (2,1), (2,0)
etc
we get
number of paths reaching node (k,i) = k! / [(k-i)!i!]
prob of getting to node (k,i) = k!(p^i)[(1-p)^(k-i)] / [(k-i)!i!]
(sorry for all the brackets...)
But I couldn't find a pattern to match the bet sizes...
Maybe I need to apply the formula backwards, but because the binomial tree chops off after 4 wins and does not continue, it gets a bit difficult.
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towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
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Re: World Series Betting
« Reply #7 on: Nov 25th, 2002, 11:10pm » |
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Well, in a sense the path to a node does matter.. You can optimize and take any extra winnings.. So a binary tree would perhaps make it easier, and more profitable..
Might be interesting to try that with an evolutionary algorithm.. since I'm too lazy to do it by hand and logic..
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Chronos
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Re: World Series Betting
« Reply #8 on: Dec 1st, 2002, 9:34pm » |
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I don't think that you can optimize this one in such a way as to pocket winnings (unless you're willing to take a risk, and know more about the true odds than the Vegas bookies). There's a well-known theorem that no combination of negative-expectation bets has a positive expectation, and this looks like it should be an application of it. If each team is equally likely to win any given game, then they're equally likely to win the whole series, which are the odds that your uncle wants to bet.
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SWF
Uberpuzzler
Posts: 879
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Re: World Series Betting
« Reply #9 on: Dec 7th, 2002, 7:36pm » |
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There may not be a neat formula. The best I can find, that seems to work is:
Let Vi,j be the value of the bankroll when there are i wins and j losses of the team being bet on. Bi,j is the amount to bet when there are i wins and j losses. Let p be the number of wins needed to win or lose the series (for example, p=4 in the case of seven game series). Assume V0,0=1, and this can be scaled by whatever the starting bankroll is ($2000 in the original example).
The value of Vi,j is obviously 2 if i=p (i.e. you win the series and double your starting value) and 0 if j=p (i.e. you lose the series and all the starting funds). If neither i nor j equals p, then:
Vi,j=Sum(k=0 to p-j-1) of (0.5)p-i+k-1*(p-i+k-1)!/(p-i-1)!/k!
The value to bet in each case is given by:
Bi,j=(Vi+1,j+Vi,j+1)/2
I don't see a way to simplify that except for special cases, like when i=p-1.
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alan
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Re: World Series Betting
« Reply #10 on: May 27th, 2004, 3:34pm » |
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Like Kozo, I worked it backwards and filled in a grid
I choose to bet on Team A.
The first column represents the current score between the teams. The next 2 columns are what I now have if team A wins or loses. The next 2 are the bankroll I have left and how much I bet to ensure I end up with the values in columns 2 or 3.
so...
The starting point is game 7
score w L have bet
3-3 4000 0 2000 2000
therefore if the score is 3-2 we need to end up with 4000 if we win and 2000 if we lose
3-2 4000 2000
so we see that we need to have 3000 and bet 1000 of it
and so we go on filling in the last 2 columns for preceding games and then calculating what we need to have and need to bet from those columns. The final answer is...
score W L have bet
3-3 4000 0 2000 2000
3-2 4000 2000 3000 1000
2-3 2000 0 1000 1000
3-1 4000 3000 3500 3500
1-3 1000 0 500 500
2-2 3000 1000 2000 1000
2-1 3500 2000 2750 750
1-2 2000 500 1250 750
3-0 4000 3500 3750 250
0-3 500 0 250 0
2-0 3750 2750 3250 500
0-2 1250 250 750 500
1-1 2750 1250 2000 750
1-0 3250 2000 2625 625
0-1 2000 750 1375 625
0-0 2625 1375 2000 625
or more neatly....
A0 A1 A2 A3
B0 625 625 500 250
B1 625 750 750 500
B2 500 750 1000 1000
B3 250 500 1000 2000
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Vivek
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Re: World Series Betting
« Reply #11 on: Jan 2nd, 2006, 11:28pm » |
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This one works, but not sure if this is optimal:
Without loss of generality lets assume A wins the world series. Now find all ways in which A can win 4 matches:
AAAA
BAAAA
ABAAA
AABAA
AAABA
BBAAAA
BABAAA
BAABAA
BAAABA
ABBAAA
ABABAA
ABAABA
AABBAA
AABABA
AAABBA
BBBAAAA
BAAABBA
…
now one needs to bet:
2000*1/15 on the first game,
2000*2/15 on the second game,
2000*4/15 on the third game,
2000*8/15 on the fourth game,
2000*16/15 on the fifth game .. so on till 7th.
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mingus33
Newbie
Posts: 5
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Re: World Series Betting
« Reply #12 on: Jan 15th, 2007, 2:39pm » |
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 i'm afraid i must concede that the 1000 *1/15 on game 1 1000*2/15 on game 2; 1000* 4/15 on game 3 etc. is the correct solution if you have enough money to be losing $3200 (assuming you lose game 4 and 5 and 6)
there is another mathematical solution which is based on determining an algorithm that does not create more than $1000 of exposure
gk
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mingus33
Newbie
Posts: 5
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envelope gamble 1
« Reply #13 on: Jan 15th, 2007, 2:41pm » |
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can someone help me with the formal solution of why it is correct to change envelopes.
thanks,
tony the simpleton
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towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
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Re: envelope gamble 1
« Reply #14 on: Jan 15th, 2007, 2:49pm » |
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on Jan 15th, 2007, 2:41pm, mingus33 wrote:| can someone help me with the formal solution of why it is correct to change envelopes. |
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Wrong thread; and it's not (well, actually it doesn't matter).
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