[Probability] 150% bet or 1001% bets

Was just browsing a roulette thread and had a thought. If your goal is to double your money in the safest way possible. Which bet is better?

Putting the entire £100 on 50/50 odds so you get just 1 bet with the chance to win £100 giving you a total of £200.

Or placing 100 £1 bets, this way you may end up winning 1, 2 or more times and make double or triple your money.

I thought the 100 £1 bets would be the better opinion but this doesn't seem to be the case.

(.50)^1 = 1/2 obvious.

(.99)^1 = 1%
(.99)^2 = 1.99%
(.99)^3 = 2.97%
(.99)^4 = 3.94%
(.99)*100 = 36.66%

So after 100 bets your chances of winning once is actually 36.6% whereas the one 50% bet gives you a 50% chance.

I just can't get my head around it, why?! It's so strange...

Why is 1 50% so much better than 100 1% bets?

Reply 1

Vikingninja

1% bets you have a 50% chance of remaining where you are by losing then winning back and forwards. Makes sense to me.

Reply 2

EternalLight

Original post by Vikingninja

1% bets you have a 50% chance of remaining where you are by losing then winning back and forwards. Makes sense to me.

That certainly doesn't make sense to me.

Reply 3

Vikingninja

Original post by AishaGirl

That certainly doesn't make sense to me.

Forgotten what it was called but when you have a statistic with two possibilities it forms a bell curve which plots the possible number of routes to an outcome. The most numerous outcome will be the one that has an equal number of each possibility which is 50% wins since there are many routes to get that many wins, the least numerous outcome will be all wins or all losses since there is only one route to each one.

Reply 4

EternalLight

Original post by Vikingninja

Can you try to find this for me please?

Reply 5

nexttime

So 1% chance of winning a 100-fold return?

0.99^100 gives you the chance of something of 99% probability (losing) happening 100 times in a row. So after 100 bets your chance of losing all of them is 36.6%. Your chance of winning at least once is 1 - that.

Then you have to add the chance of winning twice etc, to show that the 'value' of 100 1% bets is about 1 to 1.

(edited 7 years ago)

Reply 6

NotNotBatman

It's a binomial distribution. Each Bernoulli trial is independent, so the probability of winning cumulative trials decreases as the number of trials increase.
If the losing once is

\frac{99}{100}

then on the second bet it's

\frac{99}{100}

so if you lose 99 times then it's

(\frac{99}{100})^{99}

The probability that you lose the second given that the first bet is lost is the product, so

(.99)^2

and carry this on 100 times for 100 trials and the probability gets smaller.

Reply 7

EternalLight

Original post by nexttime

So 1% chance of winning a 100-fold return?

You bet £1 and have 1% chance to win and your opponent bets £99 and has 99% chance to win. Or you both just bet £100 and each have 50% chance to win.

So after 100 bets your chance of losing all of them is 36.6%. Your chance of winning at least once is 1 - that.

And with the 50% bet your chance is losing is 50%. So clearly the 50% bet is better but it's a bit strange for me to understand

I know the maths holds but it's just weird for me to understand. You have 1% each time. (50)1% = 50% chance of losing.

but (.99)^50 - 60.5% chance of losing :s-smilie:

Reply 8

EternalLight

Original post by NotNotBatman

It's a binomial distribution. Each Bernoulli trial is independent, so the probability of winning cumulative trials decreases as the number of trials increase.
If the losing once is

\frac{99}{100}

then on the second bet it's

\frac{99}{100}

so if you lose 99 times then it's

(\frac{99}{100})^{99}

The probability that you lose the second given that the first bet is lost is the product, so

(.99)^2

and carry this on 100 times for 100 trials and the probability gets smaller.

Ah of course! Omg I'm so silly :tongue:

Wow... ok definitely time for bed :blushing:

Reply 9

Vikingninja

Original post by AishaGirl

Can you try to find this for me please?

http://www.statisticshowto.com/bell-curve/

(edited 7 years ago)

Reply 10

DotA2Player

I think expected value gives a better argument as to why it's better to place your money in the 50% scenario. Expected value of 50% scenario is simply 0*1/2 + 200*1/2=$100, seeing that you either lose all your money or double it, and chances are equivalent. Expected value of second scenario can be calculated by considering binomial distribution (seeing that events are independent, as pointed out by @NotNotBatman), and if you carry out the calculations (in the attached file), you get an expected value of $2, which is much worse.

Probability thingy.xlsx19.5KB

Reply 11

nexttime

Original post by AishaGirl

You bet £1 and have 1% chance to win and your opponent bets £99 and has 99% chance to win. Or you both just bet £100 and each have 50% chance to win.

Ok so not quite what i thought (you've a 1% chance of winning £99, not £100) but pretty close.

If you made it a 1% chance of winning £100 the bets would be of identical value, i think.

And with the 50% bet your chance is losing is 50%. So clearly the 50% bet is better but it's a bit strange for me to understand

It produces a more reliable return, but the amount it returns is limited.

I know the maths holds but it's just weird for me to understand. You have 1% each time. (50)1% = 50% chance of losing.

but (.99)^50 - 60.5% chance of losing

Yes, you have a lower chance of winning. But you've also got a chance of winning twice, three times... even 100 (extremely unlikely). If you add it all up the bet becomes almost of equal value (I think 99% as much?)

Original post by AishaGirl

Ah of course! Omg I'm so silly :tongue:

Wow... ok definitely time for bed :blushing:

That poster just did the exact same calculation you did in the OP (just giving the chance of you losing all 100 times). I'm confused as to why you've now come to some kind of realisation? :p:

Original post by DotA2Player

Your understanding of the scenario is different to mine and OP's.

You're saying its betting $1 for a 1% chance of winning $2. Any idiot can say that is a dumb bet with low returns.

OP is saying its a $1 bet with 1% chance of winning $99.

(edited 7 years ago)

Reply 12

DotA2Player

Bleh, didn't read the second bit from OP carefully. Expected value for that yields $99 with a few tweaks in my spreadsheet, surprisingly close actually. Seems like there may not be that much of a difference between the two

(edited 7 years ago)

Reply 13

EternalLight

Original post by nexttime

That poster just did the exact same calculation you did in the OP (just giving the chance of you losing all 100 times). I'm confused as to why you've now come to some kind of realisation? :p:

It was late and I was tired :redface:

Reply 14

nexttime

Original post by DotA2Player

Given that each iteration is independent, an easier way of calculating that would be:

Probability * return * number of iterations

0.01*99*100 = £99.

Your way looks very pretty though :p: