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Heads and Tails Game Question

Posted on 2012-09-10
Medium Priority
Last Modified: 2012-09-10
Hey experts,

 Let's say I'm playing a simple head & tails online game in one of the casino betting sites (like willhill.com ), and their winning ratio is 1.9 (i.e. for every 1$ I bet on a head/tail, I get 1.9$, so profit = 0.9$). And let's say, I'm following the strategy below:

Summary of strategy: I always bet on head with amt. x (1$ for example); if I lose I repeat the bet on head by doubling the amount (or increasing it far enough to cover my betting expenses and make a small profit), and if I win I repeat the bet with amt. x...etc

Strategy example:
Step1: 1$ on head - win: back to Step1 (profit: 0.9$) - lose: continue to the next step
Step2: 2$ on head - win: back to Step1 (profit: 0.8$) - lose: continue to the next step
Step3: 4$ on head - win: back to Step1 (profit: 0.6$) - lose: continue to the next step
Step4: 8$ on head - win: back to Step1 (profit: 0.2$) - lose: continue to the next step
Step5: 17$ on head - win: back to Step1 (profit: 0.3$) - lose: continue to the next step
Step6: 36$ on head - win: back to Step1 (profit: 0.4$) - lose: continue to the next step
Step7: 76$ on head - win: back to Step1 (profit: 0.4$) - lose: continue to the next step
Step8: 161$ on head - win: back to Step1 (profit: 0.9$) - lose: continue to the next step
Step9: 339$ on head - win: back to Step1 (profit: 0.1$) - lose: continue to the next step
Step10: 714$ on head - win: back to Step1 (profit: 0.6$) - lose: continue to the next step
with the chance of reaching Step10 less than 0.1%

My question simply would be that isn't this a winning strategy that someone surely could've tried to follow and try to win as much as possible by repeating it continuously? :) or am I surely missing something here?
Question by:mte01
LVL 85

Expert Comment

ID: 38383264
chance of reaching Step10 less than 0.1%
loss on reaching step 10 more than $1000
average loss per dollar bet, $0.05
LVL 10

Accepted Solution

abbright earned 1000 total points
ID: 38383279
Your strategy looks similar to this: http://en.wikipedia.org/wiki/Martingale_(betting_system)

According to this link:
This, combined with the fact that strings of consecutive losses actually occur more often than common intuition suggests, can bankrupt a gambler quickly.
Casino betting limits eliminate the effectiveness of using the martingale strategy.
LVL 27

Expert Comment

ID: 38383301
Also see:   http://en.wikipedia.org/wiki/Saint_Petersburg_paradox

Where you see that even a game with a positive, infinite expected return is guaranteed to be a losing proposition.
LVL 16

Assisted Solution

HooKooDooKu earned 1000 total points
ID: 38383450
Assuming a fair coin (50% head, 50% tails), a payoff of 2:1 is REQUIRED just to break even in the long run.  There is no way to "beat the house" in any game in which the odds favor the house.  Hundreds of books have been written about the subject, and if any of them had any merrit, Las Vagas would have been shut down from losses decades ago.

When it comes to gambling, the only game ever shown that it was POSSIBLE to "beat the house" is Black-Jack.  Black-Jack is different because the odds are not only dependant on the cards on the table, they are also dependant on what is left in the deck (i.e. what cards have already been played).  

When starting from a new deck, the odds of winning/loosing by a skilled Black-Jack player are on the order of 48/52 +/- (he will win 48 percent of the time and loose 52 percent of the time... with a 2:1 payout, the house wins in the long run).  But as the cards are played, as more low cards get played than 10 point cards, the odds shift in favor of the player (as as more 10 point cards get played than low cards, the odds shift more in favor of the house).  The net result is that for the average person pulling up to a Black-Jack table at some random point in time, the house makes money.  But for very skilled players (card counters) they track what cards have and have not been played.  When the odds are against them, they bet small.  When the odds are for they, they bet high.

So if the odds are say 48/52 in favor of the house, the player will bet $10 and loose more often than they win.  But when the odds are say 52/48 in favor of the player, the player will bet $40 and win more often than they lose.  Their wins when they bet high more than offset their loses when they bet low.

Of course this is all just a gross simplication of something that is much more complex.  But the point is that there is no such thing as a betting pattern that can ever beat the house.

As for abbright's point about losing streaks... I once played black-jack where I lost 34 out of 36 hands.  Keeping in mind that when you "correctly" play your Black-Jack hands, your win/loss odds should be in the neiborhood of 48/52.  My win-loss record for that particular round of Black-Jack was 6/94 in favor of the house.  No betting scheme can ever overcome that sort of "bad luck".  It also supports ozo point of how the Martingale scheme can cause you to quickly find yourself having to place a $1,000 bet in hopes of covering an initial $1 loss (and it's a bet you STILL have about a 50% chance of loosing and then finding yourself in the place of needed to now place a $2,000 bet).

Again, to support abbright's point of how frequent a string of losses can occur:  If you flip a coin (head you win, tails you loose) something on the order of about 200 times, the odds are that you will find at least one point in that sequence of 200 "plays" where you will find 10 tails in-a-row.  Yes, there is also some point where you will also get 10 heads in-a-row.  But for those 10 heads, you will have gained $10.  For thos 10 tails, you will have lost a total of $1,000 finding youself needing to place a $1,000 bet in HOPES that the next 50/50 flip will clear all your losses.  (In otherwords, in hopes of winning $1 bets on a coin toss, even when the payout is 2:1, giving the house no advantage, if you "play" 200 times, you've got to come to the table with $1,000 or at some point during those 200 plays, you will likely find a situation where you have no more money with nothing left to bet with to cover your losses.

FYI: A gambling lesson to take away from this "question" is that if you sit down at a $5 Black-Jack table, you are likely to hit a big enough losing streak that you will loose all your money if you don't come to the table with at least $200.

Author Comment

ID: 38383709
Thx a lot for your usefull feedback guys, and especially abbright & HooKooDooKu. And so you made it more clear for me on how much the odds are high of getting something that there is a 0.1% chance of it happening.

I made an excel compilation of the odds & amounts, and it turned out that with a starting bet of 0.1$ (available at a place like william hill), and a maximum betting limit of 500,000$ (which is in reality much lower than that in all betting places), you can play your strategy 50 times a day (for an average profit of 2.5$ a day), and there is a chance once in 115 years to put approx. 265,000$ after getting 20 tails in a row (which is still lower than my 500K theoretical limit). So you can do it this way, but the payout would be very small.

For now, I decided to continue playing it this way only once a day with the chance of getting 10 tails in a row and having to put in 1507$ in the 11th step approx. once every 5.6 years :) Thx again!

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