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?