Probability question

[Malky]

If I bet $100 on the toss of a coin, I have a 50% chance of doubling my money to $200.

If the rules of the game are that I can only bet $50, and I still want to double my money to $200, is the probability 25%?

[dngf]

I think the answer is 12.5% because in your example you could only bet $50 per and you needed to win three times in a row. The $50 you didn't bet, the $50 you win first time - 50/50 the $50 you win the second toss 25% and the $50 you win on the third toss 12.5%.

[rfr3sh]

Assuming you can only bet 50$ at a time and you want to win 3 times in a row to get to 200$ the probability is 12.5%

[Malky]

Hang on....

I start with $100 balance

First bet wins, that takes me to $150 balance..

Second bet wins, that takes me to $200 balance...

that's just two bets in a row, not three.

Just want to make sure my question is clear before I start throwing some other stuff into the mix.....

I perhaps didn't make it clear that my betting bank starts at $100, not $50 (I agree with you though, if you were only starting with $50 then yes, the probability is 12.5%)

[the woodpile]

I would say it is still 50/50. All you need to do is win 2 times more than you lose before you lose 2 time more than you win. 50/50 proposition either way.

[benjy]

Are you limited to only two wagers? Or would you keep betting until you either went broke or doubled your money?

[dngf]

Sorry, I misread the starting point, it is early here. I agree it is 25% it is two tosses as you describe in post number 4

[Malky]

Are you limited to only two wagers? Or would you keep betting until you either went broke or doubled your money?

Keep betting until went broke or doubled money...

[Malky]

Sorry, I misread the starting point, it is early here. I agree it is 25% it is two tosses as you describe in post number 4

OK yeah, we're now singing from the same hymn sheet....

I think it is 25% also.

An example of this would be W, W, = 25% chance.

But it could also be acheived by W, L W, W........... now, I still think that represents a 25% chance of the result occuring, because the loss is cancelled out by a win, do you think I have that correct?

[captrobey]

What if it lands on its side

[konck]

If you flipped a coin 100 times and it came up heads every time...what would the pct be of heads on the next flip....it's still 50% therefore your 25% is correct

[Malky]

What if it lands on its side

I refer you to my terms and conditions;

3.) If a coin lands on it's side then we flip again until such time that the result is heads or tails.

[dngf]

I believe that you are correct to say that after the second toss, if you are back to even, then there is a 25% probability of hitting the next two tosses in order to double your money. However, the pattern shown of W, L, W, W is not a 25% probability.

[Malky]

I believe that you are correct to say that after the second toss, if you are back to even, then there is a 25% probability of hitting the next two tosses in order to double your money. However, the pattern shown of W, L, W, W is not a 25% probability.

Yeah this is what is confusing me, the chance of that specific pattern, W, L, W , W occuring is 6.25%.

However, sureley the probability of hitting the target result is 25% regardless of what combination gets you to it? (all of this assuming we are making our calculations before we start flipping.

[benjy]

There are 3 ways of doubling your money:

W, W = 25%
W, L, W, W = 6.25%
L, W, W, W = 6.25%

so the chances are 37.5%

[Malky]

There are 3 ways of doubling your money:

W, W = 25%
W, L, W, W = 6.25%
L, W, W, W = 6.25%

so the chances are 37.5%

OK so you have combined those probabilities by addition - what's that principle called so I can look it up?

Thanks

[benjy]

OK so you have combined those probabilities by addition - what's that principle called so I can look it up?

Thanks

Not sure of the name. It's the winning side of a probability tree (where all outcomes must sum to 100%)

The rest of the tree (going bankrupt) starts:

L, L = 25%
L, W, L, L, L = 3.125%
L, W, L, W, L, L = 1.6125%
W, L, L, L = 6.25%
W, L, W, L, L = 3.125%
etc.

[benjy]

Hmm the more I draw out the tree there are other winning ways:

W, L, W, L, W, W
W, L, W, L, W, L, W, W
etc.

so my initial answer was incorrect. you'd have to sum all these too.

[Malky]

Hmm the more I draw out the tree there are other winning ways:

W, L, W, L, W, W
W, L, W, L, W, L, W, W
etc.

so my initial answer was incorrect. you'd have to sum all these too.

Is there no "shortcut" way to do it without listing all the different possible ways of doing it?

[benjy]

There's got to be a better was to do this but here's a logical argument:

After two tosses you have:
W, W = 25% (win)
L, L, = 25% (lose)
or
W,L = 25% (even)
L, W = 25% (even)

The last two outcomes are equivalent to starting over. You need two straight tosses of either head or tails to win or lose. Both are equally likely. Thus the outcome of the bet is 50/50.

[Justin7]

If you start at "2" (how many big bets you have)... and you want to know how likely you are to get to 4 before 0 taking steps of 1, this is a "random walk" problem.

You are equally likely to go 2 forward, as 2 back. The odds of busting out are the same as doubling up. p = 0.50.

[csm506]

Wow this should be a calculus class, move on

[crazyfied]

it's actually statistics not calculus

[Justin7]

it's actually statistics not calculus

This sounds more like a Disco (Discrete and Combinatorial) problem.

[Malky]

If you start at "2" (how many big bets you have)... and you want to know how likely you are to get to 4 before 0 taking steps of 1, this is a "random walk" problem.

You are equally likely to go 2 forward, as 2 back. The odds of busting out are the same as doubling up. p = 0.50.

Obviously I "respect your authoritah" (http://www.youtube.com/watch?v=4rtwb34Pd1k)

but I am suprised the p = 0.50. I guess these things are counter intuitive though.

[dngf]

There must be a forumla that would calculate this, but I sure do not know what that is. It seems intuitive to me that there would be potentially an infinite number of possibilities to win or loose, which is best evidenced by benjy's post 17, in that after an initial win you can suffer three losses before failure, and often you can suffer two. Since the next toss is always 50/50 it seems from this perspective this could go on forever. It also seems impossible that the answer could be infinite because over any decent period, say 100, or 1,000 tosses the likelihood of more than two or three losses in a row occuring has to approach 100%.

[forsberg21]

The answer is 50%.

Trial 1:

1st flip loses 50% of the time.
If it goes to 2nd flip, you have a 50% chance of winning.
So, 0.5 x 0.5 = 0.25, or 25% chance of reaching $200 from your initial $50.

Now if your first flip loses, you take the remaining $50 and repeat with the probability outcomes being the same as those listed above.
So again, you have a 25% chance of turning your $50 into $200.

25% + 25% = 50% probability that you turn your $100 into $200 by wagering $50 at a time on a coin flip.

[Cookie Monster]

I found the general formula a good while ago, when studying the probability of busting a sticky bonus. It took me a few simulations, but the numbers always worked the same.

The probability of reaching your goal or busting is exactly the ratio of starting money / goal. So in your case, 100/200 = 50%

The ratio always work if the bets are fair (0 ev), regardless of bet size and even with huge moneylines, provided that you never overshoot the goal (ie, betting all the $100 in a +500 line).

Most possibly this also answer your next question, Malky

[hhsilver]

Of course the probability is 0.5, as justin said.

The game has to end after an even number of tosses. A more interesting question is, What is the prob that you win after 2, 4 , 6, 8, 10 ......... tosses .

I'll give the answers and let the curious figure how to get them

p(win in 2 tosses) = 1/4
p( 4 ) = 1/8
p( 6 ) = 1/16
p( 8 ) = 1/32
...
...
p ( 2n ) = 1/2^(n+1) : ^ = 'to the power'

to infinity .........

this is a geometric series with a sum of 1/2 , the answer to the original question

(thanks for the memories)

[Malky]

I found the general formula a good while ago, when studying the probability of busting a sticky bonus. It took me a few simulations, but the numbers always worked the same.

The probability of reaching your goal or busting is exactly the ratio of starting money / goal. So in your case, 100/200 = 50%

The ratio always work if the bets are fair (0 ev), regardless of bet size and even with huge moneylines, provided that you never overshoot the goal (ie, betting all the $100 in a +500 line).

Most possibly this also answer your next question, Malky

So if we throw a little juice into the mix, say 5% in favour of the book.... am I right in saying that the probability of reaching the $200 would be 45% and the chance of busing 55%?

And if the goal was $400, at fair (0 ev) the chance of reaching the goal would be 25% ? Does this mean that the chance of busting would be 75%?

[Cookie Monster]

So if we throw a little juice into the mix, say 5% in favour of the book.... am I right in saying that the probability of reaching the $200 would be 45% and the chance of busing 55%?

It is much more complex then. As the bets are not "fair", the bet size matters a lot... betting a lot of micro bets would result on the juice eating you alive; betting all-in is much better. I would suggest a simulation with the specific rules.

And if the goal was $400, at fair (0 ev) the chance of reaching the goal would be 25% ? Does this mean that the chance of busting would be 75%?

Yes. $100 has a 75% of busting, 25% of reaching $400 on fair bets of any size and odds (again, provided that you never overshoot the $400).