One 50% bet is worse than fifty 1% bets?

[iDimension]

Imagine there is an empty pot with a max capacity of £1000. You are given £500 to bet with.

Option one is to bet the full £500 giving you a 50% stake in the pot. Other people make up the other £500. You have a 50% chance to win £500, simple enough.

The second option is to spread your bets, and instead you decide to make fifty £10 bets giving you a 1% chance to win, but you get fifty tries.

I favour the second option, despite the fact that when your £500 is gone, you would have had a 50% chance to win with both amounts but the 1% bets give you many chances to win. You may well win on your first bet, or several times for that matter.

With the 1% bets the absolute minimum you can win is £500.
With the 50% bet the absolute maximum you can win is £500.

So is it always better to bet smaller than larger where possible in a game like this? Of course I am not encouraging gambling, but this is just a thought experiment for a game I play.

Psychologically a lot of my friends prefer the 50% bet. Why?

 

[jedishrfu]

People prefer to win early and win big over winning small and spending a lot of time doing it.

There's a cool book on the history of this:

https://www.amazon.com/dp/0809045990/?tag=pfamazon01-20

 

[iDimension]

People prefer to win early and win big over winning small and spending a lot of time doing it.

There's a cool book on the history of this:

https://www.amazon.com/dp/0809045990/?tag=pfamazon01-20

So it's purely a greed factor? Even though the fifty 1% bets offers them the chance to win much more money? It just takes a bit longer.

 

[jedishrfu]

I don't think its greed but rather impatience and fear. Imagine you place your bet and win on the first try that's the dream of the first time bettor. They'd feel like they have a gift. The more seasoned bettor realizes that they're not always going to win but are compelled to try again.

Where things differ is in how a master bettor plays the game, weighing the odds and betting accordingly not unlike how expert traders play the stock market.

The beginning bettor starting with say $16 might bet $1 and when they lose try to recoup their losses by betting $2 in the next round ... and in the end lose everything. In contrast, a preferred betting strategy is to bet half of what you have so bet $8 of the $16 and when you lose bet $4 then $2 then $1 ... In this way, you don't bet more than you have.

From my experience with $1 slot machines, I noticed I started with $10 in coins lost the first few times and then won a pot putting my total back to $9 then losing some more and winning another pot putting my total back to $8 ... So you lose several times then win a bit then lose some more then win... its a gentle rollercoaster to the bottom and to me was an introduction into the business of entertainment gambling.

 

[iDimension]

Yes of course Jedishrfu but I'm thinking of this purely as the better strategy. You could bet all you have and ultimately lose it but for my thought experiment if you were to bet, it's almost certainly better to bet lots of small bets than one large bet. Remember the goal is the bet your £500 in a way which maximises your probability and amount won. Of course you could take this even further and bet five hundred £1 bets or one thousands £0.50 bets etc etc.

In fact betting the full £500 in one go seems like a terrible terrible idea.

 

[jbriggs444]

Yes of course Jedishrfu but I'm thinking of this purely as the better strategy. You could bet all you have and ultimately lose it but for my thought experiment if you were to bet, it's almost certainly better to bet lots of small bets than one large bet. Remember the goal is the bet your £500 in a way which maximises your probability and amount won. Of course you could take this even further and bet five hundred £1 bets or one thousands £0.50 bets etc etc.

In fact betting the full £500 in one go seems like a terrible terrible idea.

It is clearly a zero sum game and every play has a zero expected value. No scheme that carries on for a bounded number of plays can obtain anything but a result with a zero expected value. The things you can manipulate are your odds of winning big, your odds of going broke and your chance of getting done quickly.

Betting the full amount in one go sounds like a wonderful idea to me. Finish the game and go on to do something worthwhile instead.

 

[iDimension]

It is clearly a zero sum game and every play has a zero expected value. No scheme that carries on for a bounded number of plays can obtain anything but a result with a zero expected value. The things you can manipulate are your odds of winning big, your odds of going broke and your chance of getting done quickly.

Betting the full amount in one go sounds like a wonderful idea to me. Finish the game and go on to do something worthwhile instead.

It might be a wonderful idea if you're rushed and doing it to make a quick profit (hopefully) but I still think fifty 1% bets is without question the better option. Both give a total of 50% chance but with the 1% bets you have the added chance of multiple wins with a minimum take home of £500... Which is the maximum the 50% bet can win.

 

[nuuskur]

It might be a wonderful idea if you're rushed and doing it to make a quick profit
We refer to them as donkeys in the poker world.

You will definitely see more rounds on average when you bet 1% at a time, but there is no actual way of manipulating the odds (poker strategy), therefore just bet the 50% in one go and be done with it.

 

[micromass]

The problem is that betting is rarely fair. That is, the house always has more chance of winning than you. So even if the probability of you winning is 0.49, then making only 1 large bet is a vast improvement over betting 50 times (and thus letting yourself be exposed to unfair probabilities 50 times).

And it turns out that it can be mathematically proven that bold play (betting as much money as possible) is superior to timid play. In fact, it can be proven that bold play is optimal! (it is not the only optimal way to play though).

I refer you to the nice book by Dubbins and Savage, Inequalities for Stochastic Processes (How to Gamble if You Must).

 

[russ_watters]

Both give a total of 50% chance but with the 1% bets you have the added chance of multiple wins with a minimum take home of £500... Which is the maximum the 50% bet can win.

No, the minimum you can win in both cases is zero: you could bet 50 times and never win.

Also, the odds in the two situations are not equivalent because in the second case you are not required to bet the same amount of money: you could bet once and then walk away if you win. The knowledge of the previous bet's outcome affects both your behavior and therefore the cumulative odds of all the bets you choose to make.

 

[Hornbein]

Imagine there is an empty pot with a max capacity of £1000. You are given £500 to bet with.

Option one is to bet the full £500 giving you a 50% stake in the pot. Other people make up the other £500. You have a 50% chance to win £500, simple enough.

The second option is to spread your bets, and instead you decide to make fifty £10 bets giving you a 1% chance to win, but you get fifty tries.

I favour the second option, despite the fact that when your £500 is gone, you would have had a 50% chance to win with both amounts but the 1% bets give you many chances to win. You may well win on your first bet, or several times for that matter.

With the 1% bets the absolute minimum you can win is £500.
With the 50% bet the absolute maximum you can win is £500.

So is it always better to bet smaller than larger where possible in a game like this? Of course I am not encouraging gambling, but this is just a thought experiment for a game I play.

Psychologically a lot of my friends prefer the 50% bet. Why?

This isn't correct. Your chance of winning nothing with the 50 1% bets is .605.

Proof: (.99)^50 = .605

The expectation of the two betting systems is exactly the same, zero. With the 1% you have the possibility of winning more -- you might win 25000 -- but the price is the increased chance of winning nothing.

Not only that, let's say you win exactly one of those 50 1% bets. Your net profit is £10.

If you continue to play this game indefinitely then you will go broke with probability one.

Indeed, even if you play the 50-50 bet indefinitely then you will go broke with probability one.

------

If you read books on winning poker, quite often the strategy is to make bets with LOWER expectation in order to reduce variance.

 

[gill1109]

This isn't correct. Your chance of winning nothing with the 50 1% bets is .605.

Proof: (.99)^50 = .605

The expectation of the two betting systems is exactly the same, zero. With the 1% you have the possibility of winning more -- you might win 25000 -- but the price is the increased chance of winning nothing.

Not only that, let's say you win exactly one of those 50 1% bets. Your net profit is £10.

If you continue to play this game indefinitely then you will go broke with probability one.

Indeed, even if you play the 50-50 bet indefinitely then you will go broke with probability one.

------

If you read books on winning poker, quite often the strategy is to make bets with LOWER expectation in order to reduce variance.

If you are playing a game with a systematic advantage to the house, e.g. roulette, then there are quite a few mathematical results saying that "bold play" is optimal. You go to the casino with say 100 dollars and you desperately need 10 000. Your best strategy is to bet all 100 on just one number, then if you win, bet your whole capital on one number. And so on. The intuition is that every time you play you lose on average a fixed percentage of what you had. Better to play as few times as possible, therefore. See the movie "Run Lola Run" for an example of success using the bold play strategy.

 

[mfb]

This isn't correct. Your chance of winning nothing with the 50 1% bets is .605.

Proof: (.99)^50 = .605

That is the important point.

Yes, you can win more (up to $990), but you are more likely to lose the whole $500. The expectation value is zero with both approaches.

 

[iDimension]

Guys I'm not talking about playing against the house or temptation or anything like that.

The game is simple. You have £500 to play with and the

That is the important point.

Yes, you can win more (up to $990), but you are more likely to lose the whole $500. The expectation value is zero with both approaches.

In both examples you may lose it all sure... but both bets give you a total of 50% chance right?

I just made a python program to run the simulation and I run it 100 times to simulate 100 plays at 1% each.

Sometimes I'll get 0 wins, others I'll get 1 or 2 wins... occasionally I'll get 3 wins.

So to illustrate my point I know if you play the games longer and longer it will balance out to 1 win, law of probability right?

But if you have one sum of money, I see no reason why you would not split it up into 1% bets and have 50 tries, after all you only need to win once which might happen after the 25th bet, which means you've still got a free 25 bets remaining to win more money.

I'm talking purely from a statistical point of view here and not actually going to the casino and playing. It's just a thought that's all.

 

[micromass]

What is the probability of winning?

 

[mfb]

Wait, do you continue to make 1%-bets even if you win? The first post didn't seem to suggest this.
If you do, your expectation value to win something is 0.5 in both cases. You are still more likely to lose $500 with the many smaller bets. This is balanced by the small probability to win more than once.

But if you have one sum of money, I see no reason why you would not split it up into 1% bets and have 50 tries, after all you only need to win once which might happen after the 25th bet, which means you've still got a free 25 bets remaining to win more money.

As I said, the probability that you win nothing is larger that way. 60% instead of 50%.

 

[Hornbein]

What is the probability of winning?

You have to define what "winning" is. Will you quit once you've made X dollars? Will you stop after Y bets?

 

[iDimension]

As I said, the probability that you win nothing is larger that way. 60% instead of 50%.

So 50(1%) is < 1(50%) how??

So what you're saying is with 50 bets you're chances of winning is only actually 40% after all your 50 bets?

You have to define what "winning" is. Will you quit once you've made X dollars? Will you stop after Y bets?

It's upto you, you can quit after 1 win or bet the remaining amount left over excluding of course the money you win. Personally I would favour quitting after winning once.

 

[mfb]

So 50(1%) is < 1(50%) how??

See post 11.

It becomes obvious if you take more extreme values:
A) one attempt, 100% chance to win
B) two attempts, 50% chance to win each time

With (A), you win each time, with (B), you do not (you do not win with 25% probability, you win once with 50% probability, you win twice with 25% probability).

So what you're saying is with 50 bets you're chances of winning is only actually 40% after all your 50 bets?

Your chances of winning at least once, yes.

 

[WWGD]

Problem with considering expectation is that expectation only kicks in in the long run , after a very large number of plays, and it may fluctuate pretty wildly before converging.

 

[Hornbein]

So 50(1%) is < 1(50%) how??

Personally I would favour quitting after winning once.[/QUOTE]

Chance of losing that 1% bet is 0.99. The chance of losing 50 times in a row, thus going broke, is 0.99 times 0.99 fifty times, which is 0.99 to the fiftieth power. That's equal to .605

(.99)^50 = .605

So you have a 60.5% chance of going broke without having won even once.

Personally I would favour quitting after winning once.

With that strategy and 1% bets then your expectation is going to be about -375. You're better off making 50 bets.

 

[jbriggs444]

With that strategy [quit after the first win] and 1% bets then your expectation is going to be about -375. You're better off making 50 bets.

Your expectation of what, exactly? It is clear that the expected value of your net winnings is zero.

 

[Hornbein]

Your expectation of what, exactly? It is clear that the expected value of your net winnings is zero.

No so, if you quit the first time you win a 1% bet.

In such a case, the chance of never winning a bet and losing all your money is .605.

The most you can possibly make with the 100-to-one bet is $500. This happens when your first bet wins. If you win on a subsequent bet you make less than 500 because of those prior losing bets, each of which lost $10. If you win on the 50th bet, you net only $10.

Expectation = -500(.605) + (1 - .605)[$500-$10*(0+1+2+3+...49)/50]

= -500(.605) + (1 - .605)[$500-$10*(50.5*24)/50]

which is about -$395.

So it's even worse than I thought.

 

[iDimension]

So betting big is the absolutely the better option it would seem. I never would have thought that to be honest. Always good to ask on things like this.

Thanks.

 

[Hornbein]

So betting big is the absolutely the better option it would seem. I never would have thought that to be honest. Always good to ask on things like this.

Thanks.

Yes, I think that betting the whole shot is the only reasonable way you can get a zero expectation. Everything else is worse, I think, but I'm not going to try to prove it.

I too am surprised that the "bet 100-to-one and quit on the first win" strategy is so poor. No wonder people do so badly in casinos.

 

[mfb]

No...
@Hornbein: if you win the first 1% bet and stop you win $990, if you win the last 1% bet of 50 you still get $500.
The expectation value is zero because each bet has zero expectation value.
I don't know where your numbers come from but they are wrong.

So betting big is the absolutely the better option it would seem.

It is not, all bets have expectation value of zero. It's up to your personal preference which distribution you like more, but mathematically in terms of expected money there is no advantage of one over the other (or over not playing at all).

 

[Hornbein]

No...
@Hornbein: if you win the first 1% bet and stop you win $990, if you win the last 1% bet of 50 you still get $500.
The expectation value is zero because each bet has zero expectation value.
I don't know where your numbers come from but they are wrong.
It is not, all bets have expectation value of zero. It's up to your personal preference which distribution you like more, but mathematically in terms of expected money there is no advantage of one over the other (or over not playing at all).

'Tis a puzzlement. The original post is a bit confused. It seems to say that 1% of $500 is $10, and that the payoff for winning such a bet would be $500. I didn't question this. It appears that we have both filled in the blanks differently. Surely he meant to make 50 50-to-one bets, which is what I did. So each time one is betting 2% of one's money, not 1%.

You are calculating that the bets pay off 100-to-one, that $10 earns $1000. But that's not the question the original poster asked.

 

[mfb]

The whole thread, including multiple posts from iDimension, clearly shows that the amount of money to win is $1000 in all bets (not including the money used for the bet).
The other interpretation would make the discussion pointless as it would make one option clearly superior.

 

[iDimension]

The whole thread, including multiple posts from iDimension, clearly shows that the amount of money to win is $1000 in all bets (not including the money used for the bet).
The other interpretation would make the discussion pointless as it would make one option clearly superior.

The total amount the pot will contain after your bet is $1000.

With the 50% bet you get one bet of $500 obviously, with the 1% bets you get 50 bets of $10.

With both bets the minimum you can walk away with is $0
With the 50% bet the most you can win is $500
With the 1% bets the minimum (except $0) you can win is $500

 

[mfb]

The point that was unclear in the last few posts is the amount of money you get if you win - the $1000.

With the 1% bets the minimum (except $0) you can win is $500

Yes, but you have a higher probability to not win, as shown before.

 

[iDimension]

The point that was unclear in the last few posts is the amount of money you get if you win - the $1000.
Yes, but you have a higher probability to not win, as shown before.

Yeh I was just clearing up any confusion caused by my previous posts. So if you ran a computer simulation 10million times. the 1% bets would end up about 40% win and 60% lose whereas the 50/50 bets would end up at win 50% and lose 50% ?

 

[mfb]

The 1% bets (assuming we play all 50 and have a chance to win $1000 each time) would end up:
- 60.5% to not win
- 30.6% win once
- 7.6% win twice
- 1.2% win three times
- 0.1% win four times
negligible chance to win more often

Your expectation value is then given by
-500 + 1000*(0.605*0 + 0.306*1 + 0.076*2 + 0.012*3 + 0.001*4) = 0 (the numbers here would give -2 but that is a rounding error)For the 50%-bet, there are just two options:
50% chance to not win
50% chance to win
Expectation value:
-500 + 1000*(0.5*0 + 0.5*1) = 0If you play with the 1%-bets and stop as soon as you win:
1% chance to win with the first attempt
0.99*1% chance to win with the second attempt (0.99 accounts for the 1% probability that we stopped before)
0.99*.99*1% chance to win with the third attempt
...
0.99⁴⁹*1% chance to win with the last attempt
0.99⁵⁰=60.5% chance to not win at all
The expectation value is zero again.
0.01*990 + 0.99*0.01*980 + ... + 0.99⁴⁹*500 + 0.99⁵⁰*(-500) = 0

 

[iDimension]

OK so just one last question.

If 50% bets are better than 1% bets. Is 90% bets better than 50% bets?

Betting 90% ($900) to win $100 each time.

 

[jbriggs444]

OK so just one last question.
If 50% bets are better than 1% bets. Is 90% bets better than 50% bets?
Betting 90% ($900) to win $100 each time.

A 90% bet has a 90% chance to win. Winning yields a net profit of 10% of the pot.
A 90% bet has a 10% chance to lose. Losing yields a net loss of the players outlay, i.e. 90% of the pot.

The expected value is 0.90 * $100 - 0.10 * $900 = $0

No matter how much or how little you bet, your expected net on a single play is $0.00. No matter what strategy or pattern of bets that are employed, the expected net on any finite sequence of plays is also $0.00.

 

[iDimension]

What do you mean expected net? Do you mean that when it all balances out over a large number of games, the money you spent might be $20million and the amount you win is $10million and the amount you lose is $10million meaning you've made 0% profit?

 

[jbriggs444]

What do you mean expected net? Do you mean that when it all balances out over a large number of games, the money you spent might be $20million and the amount you win is $10million and the amount you lose is $10million meaning you've made 0% profit?

By expected net, I mean the "expected value" of your net winnings -- (i.e. winnings minus losses). "Expected value" is a term used in probability and statistics when you have a probability distribution for a set of numeric outcomes. You multiply the probability of each outcome by the numeric value of that outcome and add up those products. The total is the expected value. It is also called the mean of the distribution.

Yes, over a sufficiently large number of bets, it is overwhelmingly likely that the total earnings will be approximately equal to the expected value per bet multiplied by the number of bets. This is called the "law of large numbers".

 

[Hornbein]

OK so just one last question.

If 50% bets are better than 1% bets. Is 90% bets better than 50% bets?

Betting 90% ($900) to win $100 each time.

You have to define what "better" means.

 

[Hornbein]

You have to define what "better" means to you.

 

[mfb]

If 50% bets are better than 1% bets.

It is not, and I'm getting tired of repeating this.

 

[StoneBurner]

I'm confused about the initial set-up, and I think other people may be too. The one bet of $500, paying $500 50% of the time is clear enough. Net expectation $250, right?
It's the 50 small bets of $10. What are the odds there? 100:1 or 1:1? I think people are confusing the fraction of the pot that you're betting, with the odds on the small bets, which aren't stated explicitly. Either case is easy to analyse:
If the $10 bet is 100:1 against, each bet has a net expectation of 10 cents. Fifty of these makes a grand total expectation of $5.00. Not so good.
If the $10 bets has 1:1 odds, paying $10, the expectation of each bet is $5. Fifty of these makes your net expectation $250 - same as with the single big bet, as it should be with 50:50 bets. BUT WITH THIS DIFFERENCE: with the 50 small bets you are virtually guaranteed of getting a significant amount of money! $250 on the average, but the chance of getting $0 or $500 is about a quadrillion to one against. So you can take a guaranteed (almost) win of a few hundred, or take a 50-50 chance of $500. This will depend on your own situation. Me, I'd take the found money.

 

[xman720]

Based on what other people have said, it looks like your expected value numbers are wrong.

If you put $500 into the pot, you have a 50% of winning $500 and a 50% chance of losing $500 giving an expected value of $0.00, not $250.00

If you put $10 into the pot, you have a 1% chance of winning $990 and a 99% chance of losing $10. Again, you can multiply this out and get an expected value of $0.00

In the long run, neither strategy is dominant. Both strategies lead to what everybody was thinking from the beginning- you end up with however much money you started with.

Your chance of winning is equal to x/1000 where x is how much money you put in.

How much money you win is equal to 1000 - x

So positive expected value is always (x/1000) * (1000 - x) which is (x - x^2/1000)

But you also have a chance to lose all the money you put in the pot. This is equal to (1 - x)/1000 and you lose x. This is expected value of (-x + x^2)/1000

Total expected value is the two added together: (x - x^2/1000) + (-x + x^2)/1000

No matter how much money you put into the pot and no matter how many bets you make, the expected value is always 0.

Here is a way to help understand this.

Imagine there are 10 people with $100. They put all their money into a pot and one guy leaves with $1000

The same 10 people decide to put another $100 into the pot. Another person leaves with $1000. People have spent $2000 and $2000 has left the pot.

If this process continues forever, it's obvious that however much money goes into the pot is how much money leaves the pot.

Imagine that this process is repeated 10 times. All 10 people have won the pot once and have $1000. But they realize that that is exactly how much money all of them have put into the pot- 10 bets of $100 10 times, or $10,000.

There is no way anybody comes out of this system with any sort of advantage, and there is no way more money will ever leave the pot than what went into the pot.

In order to believe that there is a dominant strategy, you have to believe that money comes from nowhere. The pot does not make any money, it only redistributes the money that people put in in an even manner.

 

[iDimension]

...

Yeah xman I agree with this and I can see why this is now the case. But what other people have said here is that with the 1% bets, your chances of losing is greater than your chances of winning. Thus if you played a large number of games, your wins would be less than your loses which is clearly worse than 50/50 bets.

I don't know why some people are struggling to understand what I'm saying, maybe I'm not great at explaining things. The key factor here is we don't have infinite money. You have just $500... when it's gone it's gone that's it.

Let's just use a $100 pot.

Method One: You put your full $50 into the pot giving you 1(50%) chance to win the other $50. Whether you win or lose you only ever play one game.

Method Two: You split your $50 into $1 bets.
Game 1: You bet $1 giving you a 1% chance to win the $100 pot. If you win you walk away $149 and never play again. If you lose, you play again.
Game 2: You bet another $1 giving you a 1% chance to win $99. If you win you walk away with $148 and never play again. If you lose, you play again.
...
...
...
Game 50: You finally win with your last dollar. You walk away with the $100 pot bringing you to $100 which is the same amount that Method One could have won but you could have won more money with method two making it the "better" option.

But then I see people here saying that the 1% bets actually give you a 60% chance to lose and a 40% chance to win which is why I thought that the 1(50%) bet is the "better" option

 

[zinq]

Apologies, but I don't understand the original question. (I have taught probability at Stanford, so the theory is no problem.)

What is the role of the "other people" who "make up the other £500 ?

And where it says "The second option is to spread your bets, and instead you decide to make fifty £10 bets giving you a 1% chance to win, but you get fifty tries," the fifty £10 bets give you a 1% chance to win what, exactly?

And are you saying the 50 bets each give you a 1% chance of winning [whatever], or do they give you a 1% chance of winning [whatever] when taken together?

Is there any chance that someone can express the question clearly and unambiguously? In my fairly extensive experience, probability questions are often asked unclearly and ambiguously, and that is the main thing that makes them hard.

 

[mfb]

Game 50: You finally win with your last dollar. You walk away with the $100 pot bringing you to $100 which is the same amount that Method One could have won but you could have won more money with method two making it the "better" option.

But then I see people here saying that the 1% bets actually give you a 60% chance to lose and a 40% chance to win which is why I thought that the 1(50%) bet is the "better" option

Okay, one last time, and I think this thread went in circles for way too long so I'm out after this post: those two effects exactly cancel each other. Both options have exactly the same expectation value of zero.

 

[pbuk]

I think I can help clear up the confusion. The game is defined as follows:

On each play the gambler can choose to make exactly 1 bet from a choice of 2:

• Bet50: pay $50 for a 50% chance of a payout of $100
  
• Bet1: pay £1 for a 1% chance of a payout of $100

The Expected Values for each bet are easily calculated as zero (remembering to subtract the cost of the bet):

• E(Bet50) = $100 x 0.5 - $50 = $50 - $50 = $0
• E(Bet1) = $100 x 0.01 - $1 = $1 - $1 = $0

The outcome of Method One is clear: you have a 50% chance of walking away with $100 and a 50% chance of walking away with nothing.

The outcome of Method Two needs a bit of work:

• 1% chance of winning the first play and walking away with $149
• 99% x 1% chance of losing the first play, winning the second and walking away with $148
• 99% x 99% x 1% chance of losing the first two plays...
• ...
• 99%⁴⁹ x 1% chance of losing 49 plays, winning the 50th and walking away with $100
• 99%⁵⁰ ≈ 60.5% chance of losing 50 plays and walking away with nothing

So although there is a chance with Method Two to walk away with up to $49 more than with Method One, you are more likely to walk away with nothing.

There is also Method Three where you play 50 times whether you win or not. The chances of losing 50 times and walking away with nothing are again approximately 60.5%, but now you have a chance to win even more - you could even walk away with $5,000! (compare the probability of this with the number of sub-atomic particles in the observable universe).

 

[iDimension]

The outcome of Method One is clear: you have a 50% chance of walking away with $100 and a 50% chance of walking away with nothing.

• 99%⁵⁰ ≈ 60.5% chance of losing 50 plays and walking away with nothing

So although there is a chance with Method Two to walk away with up to $49 more than with Method One, you are more likely to walk away with nothing.

Method One: Probability to win 0 = 50%
Method Two: Probability to win 0 = 60.5%

So betting once for 50% is the better option. Which is basically what I've been asking this entire thread. Apologies if my explanation of the game was terrible.

Thanks to all who posted and had the patience to help me lol.

 

[jbriggs444]

Method One: Probability to win 0 = 50%
Method Two: Probability to win 0 = 60.5%

So betting once for 50% is the better option.

The probability to lose everything is not the only possible measure of success. If you measure success by expected value, all options are equally poor.

 

[pbuk]

So betting once for 50% is the better option.

I didn't say that. Consider this:

You believe that Method One is the better option, but there is also another gambler Two who chooses Method Two. After playing your games you compare notes.

• There is a 30.25% chance you both end up with $0
• There is a 0.31% chance you both end up with exactly $100
• There is a 30.25% chance you end up with $100 and Two leaves with nothing
• There is a 19.60% chance that you end up with $0 and Two ends up with more than $100
• There is a 19.60% chance that you end up with $100 and Two ends up with more than $100

So the approximate chances are 31% that you both walk away with the same, 30% you walk away with more than Two and 39% the player of Method Two walks away with more than the player of Method One. Which do you think is the best method now?

 

[pbuk]

"What kind of crazy maths is this?" you might ask, "I ask a simple question and get three different answers!"

This is the maths of decision theory. It doesn't have a lot of significance for the natural sciences, but in the field of economics it is very important. Key to decision theory is the concept of a utility function: this is a rigorous way of evaluating different outcomes to decide what is "best". The reason we have three different answers in the game posed by the OP is that we have used three different utility functions:

• maximising the expected value of the winnings (note that this is rarely used as a utility function because expected values depend on a large number of trials, in decision theory we generally only get to make a decision once) which is $0 for both Methods and so they are equal
• maximising the probability of (winnings > $0) which is higher for Method One
• minimising the probability of (regret > $0) which is higher for Method Two (regret is an important concept within decision theory)

 

[votingmachine]

I still think the question is a bit vague.

Bet #1. $500. Odds 50%. Win pays $500 (plus the original $500, net $1000)
Bet #2 $10. Odds 1%. Win pays $990 (plus the original $10, net $1000)

The choice is between making 1 of Bet #1 or 50 of Bet #2? For convenience assume they are not 50 sequential bets. Say you can buy 1 of type Bet #1 or 50 of type Bet #2. There is no time difference or difficulty difference ... it is purely a matter of preference.

I see 50 of Bet #2 as better. You have a chance of winning 50-times. The downside is the same: you lose all your money. The upside on 50 bets is higher.

A lot of people point out that the game of chance proposed is unrealistic. I agree. In a casino, you always have worse odds than the payout. But if it was just a matter of putting the same $500 down, either into a box marked Bet#1, or a box marked Bet#2, and then getting the outcome, I would choose the box marked Bet#2.

I've only casino gambled once, and my goal was to play the game with the fairest odds (craps), and to make the smallest bets. Doing that, some people lose and some people win. I won a small amount. But the house ALWAYS wins in the long run.

Perhaps I misunderstand the problem, but the time element seemed to be a side issue that should be subtracted out. I thought the problem was not how to sequentially gamble, but whether to take 50 betting positions vs 1 betting position, with the odds and payout as described. I think the 50 betting positions is superior.