## Strange Coincidence

# Programming Snapshot – Python Gambling

Can 10 heads in a row really occur in a coin toss? Or, can the lucky numbers in the lottery be 1, 2, 3, 4, 5, 6? We investigate the law of large numbers.

If you follow the stock market, you are likely to hear people boasting that their buy and sell strategies allegedly only generate profits and never losses. Anyone who has read *A Random Walk down Wall Street* [1] knows that wild stock market speculation only leads to player bankruptcy in the long term. But what about the few fund managers who have managed their deposits for 20 years in such a way that they actually score profits year after year?

According to *A Random Walk down Wall Street*, even in a coin toss competition with enough players, there would be a handful of "masterly" throwers who, to the sheer amazement of all other players, seemingly throw heads 20 times in a row without tails ever showing up. But how likely is that in reality?

#### Heads or Tails?

It is easy to calculate that the probability of the same side of the coin turning up twice in succession is 0.5x0.5 (i.e., 0.25), since the probability of a specific side showing once is 0.5 and the likelihood of two independent events coinciding works out to be the multiplication of the two individual probabilities. The odds on a hat trick (three successes in a row) are therefore 0.5^3=0.125, and a series of 20 is quite unlikely at 0.5^20=0.000000954. But still, if a program only tries often enough, it will happen sometime, and this is what I am testing for today. The short Python `coin_throw()`

generator in Listing 1 [2] simulates coin tosses with a 50 percent probability of heads or tails coming up.

Listing 1

cointhrow

01 #!/usr/bin/env python3 02 import random 03 04 def coin_throw(): 05 sides=('heads', 'tails') 06 07 while True: 08 yield sides[random.randint(0, 1)] 09 10 def experiment(): 11 run = 1 12 max_run = 0 13 prev = "" 14 count = 0 15 16 for side in coin_throw(): 17 count += 1 18 19 if prev == side: 20 run += 1 21 if run > max_run: 22 max_run = run 23 print("max_run: %d %s (%d)" % 24 (max_run, side, count)) 25 else: 26 prev = side 27 run = 1 28 29 experiment()

Listing 1 is divided into the `coin_throw()`

coin toss generator and the experiment evaluation in the `experiment()`

function. In a `for`

loop as embedded in line 16, it looks like `coin_throw()`

would return a long list of result strings consisting of `heads`

or `tails`

, but in reality, the function is a dynamic generator that returns a new value whenever an iteration pump like the `for`

loop asks if there is more coming.

#### In the Generator House

How does the generator, whose output is shown in Figure 1, work? The key is the `yield`

command in line 8 of Listing 1. Python interrupts the execution of the current function for a `yield`

, remembers its internal state, and returns the string (`heads`

or `tails`

) to the calling program in line 8. The `randint(0,1)`

method returns the integer value ` `

or `1`

with 50 percent probability, and the script picks one of the two entries in the `sides`

tuple.

The next time the `coin_throw()`

function is called, the program flow returns to the previous location within the function to continue where `yield`

left off. In this case, this is the endless loop with `while True`

in line 7, whose condition is still true, whereupon another `yield`

command again returns a random value. Thus the generator produces an endless sequence of values in `coin_throw()`

and always performs a new coin toss if the `for`

loop in the main program wants more.

The `experiment()`

function's remaining code stores the number of identical coin toss outcomes in sequence so far in the variable `run`

, the maximum longest sequence so far in `max_run`

, the previous coin toss in `prev`

(`heads`

or `tails`

), and the total number of tosses so far in `count`

.

#### Fatal Double Down

The output from the script in Figure 2 reveals that after 21 million passes, a sequence of 23 tails in succession actually occurred. If a player had doubled the bet for the next game each time they lost, a strategy called "doubling down," they would have needed a total of 2^23 – 1=$8,388,607 of betting capital, assuming an initial bet of $1, not to go bankrupt if they had initially bet on heads, only to see 23 tails in a row with growing frustration.

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