"Whenever you can, share. You never know who all will be able to see far away standing upon your shoulders!"

I write mainly on topics related to **science and technology**.

Sometimes, I create tools and animation.

Sept. 4, 2021

Author - manisar

Gambler's fallacy (aka Monte Carlo fallacy) is one of those counterintuitive phenomena that become so simple and obvious once we look at them the right away. We do not need any math or equations to understand the sorcery.

But then, it appears in a slightly different form, or we look at it after some time, and we are again taken for a surprise.

In its most common instance, it's the tendency to think that a ** tails** is highly likely to show up in a coin toss if the

Jump to the test tool right away.

This reminds me of Schroeder Stairs... something that changes its meaning or sense depending upon how we are looking at it.

If we rotate the picture of the stairs shown here 180 degrees, these become upside down, which is normal... until we blink! Once we do that, the stairs magically become upright.

**Hover over (or click on touch screens) the picture** and see that happening - don't forget to blink if you continue to see the stairs upside down after rotating the image.

Ok, back to the Gambler's Fallacy.

This is a fallacy (error in reasoning) because it is simply not true!

**After 5 successive heads, a heads is equally like to show up as a tails in the next toss!**

And the right way to look at it this is as follows.

We got 5 successive heads and now we think that a tails is more likely than heads in the next toss. The point that we need to keep in mind is that yes, a tails is definitely much more probable to show up than 6 heads in a row. But here, the comparison in our mind is not fair so far. We should not be comparing "a tails" with "6 successive heads".

What we should be comparing is these - a tails after 5 successive heads vs. a heads after 5 successive heads. I mean, if at the start of the game, I had asked you to bet on 5 heads and a tails (in 6 tosses) in that order, would you have done that?

5 heads + 1 tails in order is only as much probable as 5 heads + 1 heads. Each of these two is just one of the 2^{6} ways of arranging heads and tails in 6 tosses.

Another way to understand it is to ponder over the fact that in having 5 successive heads, the improbable has already happened! Yes, 6 heads in a row are even more improbable, but not more improbable than 5 heads in a row followed by a tails!

Ok, forget theory, it's time to test our intuition.

In the small tool given below, you can see how different strategies for betting in such a given condition compare to each other.

- Number of winning options - from 2 to 10. You can have a two-faced coin, or up to a 10-faced dice.
- Minimum length of a run to bet after. By run is meant the same result appearing successively. E.g. you may want to bet after 5 or more successive heads (or tails) in a coin toss, or 3 or more 5's (or any other number) in a dice roll.
- Number of draws. The bigger the number, the less will be stochastic variations in the comparison.

The computer bets according to each of the following strategies.

**Strategy 1**: We bet on the*same*result that showed up last few times successively. So, if the last 5 times it were all heads, this strategy will bet on heads in the next toss.**Strategy 2**: We bet*randomly*, completely ignoring what has happened hitherto.**Strategy 3**: We bet on any option*other than that*which appeared successively in the last few draws. E.g.- after 5 successive heads in coin toss, we bet on tails.
- after 3 successive 5's in a 6-faced dice roll, we bet on any number from [1, 2, 3, 4, 6].

**Strategy 3 is what the intuition suggests** (unless you are a die-hard mathematician or a thinker with trained intuition!).

Note that the coin and the dices are fair.

Number of faces

Number of draws

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There won't be any difference in the number of winnings following any of the three strategies! If you see any significant difference, increase the number of draws and try out a few more times - so that the total number of wins goes above 100 for each of the three strategies.

This is simply because, with a fair coin or dice, any number, be it the one that showed up last few times in a row, or any other that was randomly selected, is as good as any other in terms of the probability of it showing up in the next trial.

If the coin or the dice is not fair, then it's a different story. Similarly, let's say there is a cost involved in changing your betting option. Then, you can save some money by simply sticking to the same option, even if it has appeared 10 times in a row! You'll not be worse off than betting on any other number.

In the spoiler above, I mentioned that it doesn't matter if we stick to the same number in any number of draws or we apply any other strategy of selecting the number to bet on!

Let's test this. Let's say there are $N$ winning options, and $x$ draws. Then, in each of the $x$ draws, we may bet on:

**s1:**the same number**s2:**a random number**s3:**a number we haven't betted on in the past draws, i.e. we bet on $x$ different numbers in $x$ draws - one in each draw.

If $x$ > $N$, we repeat the numbers every time $N$ options are all exhausted in $N$ draws.**s4:**a random number that hasn't won in the past draws.

If all the numbers are exhausted (have won at least once), we start over assuming no number has won yet.

Number of winning options

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Gambler's Fallacy exists because of our tendency to assume memoryless phenomena to have a memory.

Some other examples of this fallacy are:

- Your chances of winning a lottery do not increase simply because you have played it a thousand times in the past! So, "you do not owe a win just by being a veteran player".

Your chances are only as good as a newbie. - The chances of a crash happening of the plane you are on simply do not increase because you have flown a million miles in the past without having an accident!

As much as you did not like the first example above, I bet the second one must have compensated for the dismay to some extent!

We saw in the preceding paragraph how knowing about the Gambler's Fallacy can help you lose less (you don't play again and again assuming that doing this begets you a higher probability of winning).

**But, the cool thing is that knowing about Gambler's Fallacy can actually help you win more!**

And, that is because in some cases, playing just once (with all your bets in one draw) is actually better than playing multiple times.

That is, placing the same number of bets in a single draw can be really better than placing those bets in different draws.

Putting all your eggs in one basket does work sometimes!

E.g. if each bet costs $1, and your plan was to play 100 times in the current year, it is actually **better to place all those bets in one draw!**

But, there are two caveats:

- Using all bets in the same draw is better - speaking from only the probability point of view.

If, for placing all bets in a single draw, you need to take a loan, or you know that you'll be tempted to play again later down the year, then it's better to spread out your bets over the year. - Exactly how much better is the strategy of placing all the bets together depends upon the
*the number of winning options vs. number of your bets. If this ratio is more than 100, the**difference is insignificant**.*

So, for big lotteries like Lotto Max, where the number of winning options is astronomically larger than the number of bits you might want to place,**both the strategies are practically equal.**

Let's say there are $N$ winning options and we are ready to bet on $x$ on them.

Let's define the two strategies .

**ST1:**We bet on one option at a time (i.e. one bet in one draw), and we keep on doing this until we win or we have made $x$ bets.**ST2:**We bet on all the $x$ options in one draw.

Now, ST1 can actually be executed in a number ways - as per s1 to s4 defined earlier on this page. But, as we saw, these s1 to s4 are are all the same in terms of the probability of winning. So, let's pick s3 from there (betting on a different number in each draw), and here we can focus on ST1, ST2.

Let's denote the two probabilities using $P_1 = P(ST1)$, and $P_2 = P(ST2)$.

$$\begin{align}

P_1 & = P(\text{at least one win in all draws}) \\[4pt]

& =1 - P(\text{no win in any of the draws}) \\[4pt]

& = 1 - \left( \frac{N-1}{N} \right)^x \\ \\

P_2 &= \frac{x}{N}

\end{align}$$

Let's feed in some numbers into these probabilities, and see how they look.

Number of winning options

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So, not only does the idea that *more playing = more winning* (an instance of Gambler’s Fallacy) doesn’t work, in fact, depending upon the situation, going against it (and playing just once instead of lot many times) can actually be beneficial!

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