SOLE objective of leaving with MORE money than you left. The doubling idea leaves you with specifically one dollar more. You can give yourself more chance by being happy with 1 cent more. Bet the following sequence instead:
1, 1.01, 2.02, 4.04, 8.08, 16.16, 32.31
and you get one more bet, and so your chances of walking away with more money is now 94%, as opposed to the 91% you had with the doubling system.
With the doubling system, you were going to get a dollar though. This way, you get a cent, but the main thing is you don't lose a whole $100 once in every ten times.
Originally posted by uzlessDepends on the ODDS you are giving me!
If i said you have a one chance in three to win, do you want to try once or three times?
If you give even money I only what to take that bet once (or not at all!)
If you are giving me 10 to 1 then I'll take it all day long.
Using your logic the more often you go to the bookies the liklier you are to come out ahead!!! Unfortunately there are people like this; gambling addicts.
Wolfgang is correct if you're looking to make the least loss. Of course, if you don't want to lose anything, you should not bet at all, but if you want to bet and not lose much, the correct thing to do if the odds are against you is to minimise your bet number.
However, in this case, you want good odds of increasing your money. The method I outlined increases your money by 1 cent 94% of the time, and decreases your money by about $50 6% of the time. It's the best method for the stipulation given.
Originally posted by doodinthemoodAgreed. But I think the question assumed whole dollar bets.
Wolfgang is correct if you're looking to make the least loss. Of course, if you don't want to lose anything, you should not bet at all, but if you want to bet and not lose much, the correct thing to do if the odds are against you is to minimise your bet number.
However, in this case, you want good odds of increasing your money. The method I outlined ...[text shortened]... eases your money by about $50 6% of the time. It's the best method for the stipulation given.
Originally posted by doodinthemoodI'm guessing your 32.31 should be 32.32.
SOLE objective of leaving with MORE money than you left. The doubling idea leaves you with specifically one dollar more. You can give yourself more chance by being happy with 1 cent more. Bet the following sequence instead:
1, 1.01, 2.02, 4.04, 8.08, 16.16, 32.31
and you get one more bet, and so your chances of walking away with more money is now 9 ...[text shortened]... way, you get a cent, but the main thing is you don't lose a whole $100 once in every ten times.
I like your strategy. Good work, thanks for the improvement.
If you were to lose all seven initial rounds, what would you do with your remaining 35.37 dollars?
Originally posted by wolfgang59Look, the question wants to know the best strategy to increase your money.
Depends on the ODDS you are giving me!
If you give even money I only what to take that bet once (or not at all!)
If you are giving me 10 to 1 then I'll take it all day long.
Using your logic the more often you go to the bookies the liklier you are to come out ahead!!! Unfortunately there are people like this; gambling addicts.
The odds of winning are 1 in 3.
So, therefore you want a strategy that will give you as many chances as possible. 3 chances would be better than 1 chance. 5 chances would be better than 3 chances.
So, the best strategy would maximize the amount of times you can bet and still recover all the money you have previously lost.
Originally posted by LemonJelloAs we're looking to increase our money, it may be the case that the question assumes losing 1 cent is equivalent to losing all 100$
I'm guessing your 32.31 should be 32.32.
I like your strategy. Good work, thanks for the improvement.
If you were to lose all seven initial rounds, what would you do with your remaining 35.37 dollars?
Once you reach only $35.37 left, you want to get back to the point where you can reach a bet capable of pushing you up to $100.01
The first time this is possible is $50.01 with a bet of $50, so you have to make $14.64. As wolfgang says, the best odds of doing this are by going there in one. So you bet $14.64 at this stage, then $50 if you win.
If you lose, you want to get back into a situation where you can bet up to the $50 dollar mark, or in other words, you want to get to $25.01 so you bet up to that in one. Then you bet up to $12.51 and so on, until you lose everything, then you leave.
This isn't a new strategy to the main one, though it may look like it. In the first one, you start at an odd amount, and are constantly trying to bet up to $100.01. Then whenever that isn't possible, you half the value you're trying to bet up to.
Originally posted by uzlessI don't think you are this stupid so I will assume this is a wind-up.
Look, the question wants to know the best strategy to increase your money.
The odds of winning are 1 in 3.
So, therefore you want a strategy that will give you as many chances as possible. 3 chances would be better than 1 chance. 5 chances would be better than 3 chances.
So, the best strategy would maximize the amount of times you can bet and still recover all the money you have previously lost.
But if you are serious (?!?!?!). Try it experimentally.
OR think of situation where chance of winning is 1 in 1,000 and you are getting even money. Do you seriously want to have lots of bets?
What about the bookie? Are you suggesting he should minimise the number of bets?
I don't think this post will help answer the question, but I did note some interesting behaviour.
n = number of bets
W = number of wins
L = number of losses
kW = amount won per win
kL = amount lost per loss (positive number to keep things simple)
V = value of bets to the bettor
V = kW*W - kL*L
In order to be ahead, V > 0. So we have:
V = kW*W - kL*L > 0
kW*W > kL*L
W > (kL/kW)*L
Substituting n = W + L, we get:
W > (kL / (kW + kL)) * n
This is the minimum number of wins required in "n" trials to end up ahead, "Wmin". The probability of ending up ahead in "n" trials is the probability of getting at least "Wmin" wins in "n" trials, and is given by:
P(ahead) = SUM{i=Wmin...n} CHOOSE(n,i) * P(win)^i * P(lose) ^(n-i)
This expression arises from choosing "i" bets out of "n" to be winning. Ever one to make computers do work for me, I put this into *gasp!!* an Excel spreadsheet! 😲 I did get some interesting results, though.
From the problem statement:
P(win) = 1/3
P(lose) = 2/3
kW = 1
kW = 1
The spreadsheet makes all bet sizes equal, so setting n = 1 makes the bet size $100 (one-shot deal), which gives a probability of 1/3 as expected. Setting n = 2 makes the bet size $50, which gives a probability of 1/9, also as expected. However, plugging in n = 3 gives a probability of 0.259259... which is 259/999 and UNexpected! Similarly, n = 4 gives 1/9, again a little unexpected, while n = 5 gives approximately 0.209877, again unexpected.
Basically, the overall behaviour of the P(ahead) function us decreasing, but with spikes at all odd values for "n". The reason for this is that Wmin is the same for both an even number and the odd number following it (i.e. 3 wins out of 4 will put you ahead, and so will 3 wins out of 5), so you get one freebie on an odd number of bets. Changing kW from 1 to 2 changes the picture completely. Now, n = 1 gives you a probability of 1/3, but n = 2 gives 5/9!!
So the general rule of thumb that minimizing the number of bets will maximize your chance of coming out ahead seems to be true, but there can be islands of higher probability depending on the kW/kL ratio.
Originally posted by wolfgang59I think you're forgetting that you quit betting once you win. You win once and then you quit. you don't keep betting.
I don't think you are this stupid so I will assume this is a wind-up.
But if you are serious (?!?!?!). Try it experimentally.
OR think of situation where chance of winning is 1 in 1,000 and you are getting even money. Do you seriously want to have lots of bets?
What about the bookie? Are you suggesting he should minimise the number of bets?
Originally posted by PBE6Not sure i follow the text. Did you try calculating what happens if you increase your bet each time you lose? And, did you calculate that you will stop betting after you win once?
I don't think this post will help answer the question, but I did note some interesting behaviour.
n = number of bets
W = number of wins
L = number of losses
kW = amount won per win
kL = amount lost per loss (positive number to keep things simple)
V = value of bets to the bettor
V = kW*W - kL*L
In order to be ahead, V > 0. So we have:
V = kW*W ...[text shortened]... is a factor when deciding which exact large number will maximize your chances.
If so, then ok, i'll give it up.
Originally posted by uzlessYou're right, this analysis only answers the question "a person made n bets, each of equal size, with the following probabilities and payouts...what is the probability that this person is ahead?".
I think you're forgetting that you quit betting once you win. You win once and then you quit. you don't keep betting.
Originally posted by uzlessNope, that could be stupidly complicated using my original method. I think the other method presented by doodinthemood looks promising...
Not sure i follow the text. Did you try calculating what happens if you increase your bet each time you lose? And, did you calculate that you will stop betting after you win once?
If so, then ok, i'll give it up.