Constructing an Edge in Progressive Betting Systems for Coin Toss Events

[scalpmaster]

I came across a post here which is puzzling:
http://quant.stackexchange.com/questions/4442/coin-toss-system

2 events:
(1) prob(H or T)of a toss = 0.5.
(2) prob(HH or TT) in a cluster of 6 tosses > 0.85.
If you are betting on event(2) .i.e.A group of 6 instead of the very next toss
Can an edge be actually constructed with some kind of progressive betting system?

 

[mathman]

I came across a post here which is puzzling:
http://quant.stackexchange.com/questions/4442/coin-toss-system

2 events:
(1) prob(H or T)of a toss = 0.5.
(2) prob(HH or TT) in a cluster of 6 tosses > 0.85.
If you are betting on event(2) .i.e.A group of 6 instead of the very next toss
Can an edge be actually constructed with some kind of progressive betting system?

What sort of edge do you mean?

Note: I got .96875 (=31/32) for (2).

 

[scalpmaster]

What sort of edge do you mean?

Note: I got .96875 (=31/32) for (2).

Thanks for the reply.Just wondering if we place a bet for consecutive appearance of T or H only after appearance of HTHT or THTH, would there be an advantage since probability is quite high if we count every cluster of 6 tosses as an event instead of every single toss?

 

[Norwegian]

Note: I got .96875 (=31/32) for (2).

Then what did you get for (1)?

Just wondering if we place a bet for consecutive appearance of T or H only after appearance of HTHT or THTH, would there be an advantage since probability is quite high if we count every cluster of 6 tosses as an event instead of every single toss?

Is your coin endowed with some sort of memory that makes future outcomes depend on previous ones?

 

[mathman]

Note to Norwegian - for (1) I assumed it was a fair coin, so that the probability of heads = probability of tails = 1/2.

Note to scalpmaster - Norwegians comment is correct. Past events have no effect on the outcomes of future events for coin tossing.

 

[Norwegian]

Note to Norwegian - for (1) I assumed it was a fair coin, so that the probability of heads = probability of tails = 1/2.

In that case, you may want to reconsider your previous note

Note: I got .96875 (=31/32) for (2).

as you seem to have inconsistent interpretations of "H or T" and "HH or TT".

 

[mathman]

In that case, you may want to reconsider your previous note

as you seem to have inconsistent interpretations of "H or T" and "HH or TT".

My calculation of (2): First toss anything. To avoid HH or TT, each subsequent toss must be opposite of previous toss. For 5 tosses in a row the probability is 1/32. Therefore the probability of getting HH or TT = 31/32.

 

[Norwegian]

My calculation of (2): First toss anything. To avoid HH or TT, each subsequent toss must be opposite of previous toss. For 5 tosses in a row the probability is 1/32. Therefore the probability of getting HH or TT = 31/32.

Sure, but then the probability of getting H or T in (1) must be 100%.

 

[ImaLooser]

I came across a post here which is puzzling:
http://quant.stackexchange.com/questions/4442/coin-toss-system

2 events:
(1) prob(H or T)of a toss = 0.5.
(2) prob(HH or TT) in a cluster of 6 tosses > 0.85.
If you are betting on event(2) .i.e.A group of 6 instead of the very next toss
Can an edge be actually constructed with some kind of progressive betting system?

If you are betting on independent groups of six then results from a previous group will have no effect on the next group.

 

[scalpmaster]

If you are betting on independent groups of six then results from a previous group will have no effect on the next group.

That is not what (2) meant or whether the coin has memory or not.

If you win $1 on 9.6 times and lose $3 every 0.4 times out of 10 groups/counts of 6 tosses (applying martingale if necessary only once on 6th toss), it is a positive expectation game conceptually.

However, the assumption was that you can only start betting on the 5th toss(or 4th) which becomes unclear whether the advantage still stands(which is my question).

Basically, out of 10 groups of 6 tosses, you will find at least 9 groups with a HH or TT in it.
However, this HH or TT can occur on the 2nd, 3rd, 4th or 5th toss, and conventional martingale(limited to once) only gives positive expectation from the 4th toss onwards...unless someone can devise another scheme which still gives positive expectation(maybe smaller) but with more coverage.

 

[Norwegian]

If you win $1 on 9.6 times and lose $3 every 0.4 times out of 10 groups/counts of 6 tosses (applying martingale if necessary only once on 6th toss), it is a positive expectation game conceptually.

Hi there, would you mind informing us of your proposed betting procedure, an in particular how you arrive at your numbers?

And again, since your coin is fair, and I suppose you are given fair odds on each bet, all bets, both individually and collectively, will have zero EV. That is a fact. Any scheme suggesting otherwise does not work, and that includes your attempt above. If you maintain your claim, you are probably breaking the forum rules, and the strict forum police may strike at any moment.

 

[ImaLooser]

Hi there, would you mind informing us of your proposed betting procedure, an in particular how you arrive at your numbers?

And again, since your coin is fair, and I suppose you are given fair odds on each bet, all bets, both individually and collectively, will have zero EV. That is a fact. Any scheme suggesting otherwise does not work, and that includes your attempt above. If you maintain your claim, you are probably breaking the forum rules, and the strict forum police may strike at any moment.

Actually martingales have infinite expectation. But it only works if you have infinite money.

 

[mathman]

Sure, but then the probability of getting H or T in (1) must be 100%.

I believe that he simply mispoke in the statement for (1). P(H) = P(T) = 0.5 is what I assumed he meant.