Lots of good stuff on predicting using increasingly larger and larger sets of data.

Here’s a brain teaser, as phrased by wikipedia (spoilers):

*> You have two indistinguishable envelopes that each contain money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. You pick at random, but before you open the envelope, you are offered the chance to take the other envelope instead.*

This should remind you of the Monty Hall problem, where you are asked if you would switch after making a decision. So, I ran a simple Monte Carlo simulation to see what would happen:

def run(simulations=1000): env1, env2 = 0 env2 = 0 for i in range(simulations): envelopes = [random.randint(2,10)*2] select = random.randint(0,1) if select==1: envelopes.append(envelopes[0]/2) elif select==0: envelopes.append(envelopes[0]*2) env1 += envelopes[0] env2 += envelopes[1] return env1, env2, env2*1.0/env1

# plotting the simulation results sim1, sim2 = [], [] for i in range(1000): a, b, c = run(1000) sim1.append(a) sim2.append(b) hist(sim1) hist(sim2)

In this case, which envelope you choose first doesn’t matter. You aren’t sure which one has more money, and there’s no distinguishable way to tell, therefore, the odds of you picking the envelope with more money is 50/50. Let’s call that the *first* envelope, meaning the one you choose first, just to have a point of reference.

Because one envelope always has twice the amount of the other with 50/50 odds, I flip a coin. If the coin is heads (1), the second envelope has 1/2 of the amount in envelope one; if it’s tails (0), then the second envelope has 2x the amount. The odds of the second envelope (the one you switch to) either have twice the money, or half the money. I run 1000 simulations and keep the tally for the first and second envelope. *The end result is the sum of the contents for 1000 results of envelope A and envelope B. *

Here are some results of the simulations:

In [44]: run(10000) Out[44]: (119290, 150884, 1.2648503646575573) In [45]: run(10000) Out[45]: (119894, 148105, 1.2352995145712045) In [46]: run(10000) Out[46]: (120490, 152672, 1.2670927047887792)

Let’s run 1000 simulations of the 1000 results and plot each resulting tally.

Blue is first envelope, green is second. If you did this experiment 1000 times, you would probably have ~$11,000 if you stick with the first envelope, and over ~$13500 if you decided to switch envelopes. This went completely against my initial thoughts, so I checked the code again:

if select==1: envelopes.append(envelopes[0]/2) elif select==0: envelopes.append(envelopes[0]*2)

*Ah ha!* I thought (mistakenly). I’m multiplying the second one and dividing the first. What if I switched the signs?

if select==1: envelopes.append(envelopes[0]*2) elif select==0: envelopes.append(envelopes[0]/2)

I should have expected that. Because the choice to multiply or divide is random, it didn’t matter, as roughly the same amount of enveloped get either doubled or halved regardless of order.

So, here’s what we know so far:

- Switching seems to increase expected payout
- The expected payout increases by a factor of about 25%
- There is enough of a difference in the distributions to accept that switching has a significant effect

It helps to think about the expected payouts on a vector. Assuming the first envelope has $10 in it, the situation would look like this:

0---------10---------20------>

If you switch, the envelope will either contain $20 or $5:

0----5----10---------20------>

It’s plain to see that the potential gain on switching is significantly larger than the loss. Gaining 100% is much more significant in absolute terms than losing 50%. Therefore, expected value of switching – over time, on average – comes to $12.50 or +2.50 (marked by X) which is splitting the difference between -50% ($5 envelope) or +100% ($20)

0----5------{X}------20------>

It pays to switch. But here’s the paradox:* if the act of switching increases your expected payout by 25%, then couldn’t it make sense to constantly switch, therefore perpetually increasing the payout? *

No! When you switch a second time, there are two possible cases:

Case 1, Switching back to the larger denomination: If you switch to the higher payout, and you switch back, you go from $10 to $20 to $10, but since you (unknowingly) had $10 to begin with, the expected additional payout of switching back is always 0.

Case 2, Switching back to the smaller denomination: If you switch to the smaller payout, and you switch back, you go from $10 to $5 to $10, but since you (unknowingly) had $10 to begin with, the expected additional payout of switching back is always 0.

Therefore, the act of switching again has no additional effect on the payout. Which implies that **it is not the act of switching that increases value.** Full stop.

There is certainly an opportunity to increase your payout, but the probability of switching to it is about 50/50 – the same odds of picking it in the first place. It is completely random and therefore no way to foolproof method to increase your payout. The validity of the choice can only be demonstrated after the envelope is opened. While the expected value may increase, the actual contents of the envelopes do not change.

So back to the philosophical debate of choice. How do you know which is the right one to pick? Flip a coin. You’ll be 100% right 50% of the time.

Half the money I spend on advertising is wasted; the trouble is I don't know which half -- John Wanamaker

I’ve often been told that the best odds in Vegas are at the craps table, with only about a 1% edge going to the house. Yet, I don’t know of any player that went to Vegas with $5000 to come home with $4950. They’re either up much more, or down much less. Ironically, expected value should rarely be expected. The healthiest, and most sensible, thing to do is to think about as my friend JVS did, “Just give me one of the envelopes and I’ll be happy. Either way, it’s more money than I have now.” And he’s right. The actual payout in relation to what you had before will always be +X, because no matter what envelope you choose, you will always have +X more dollars than you’ve had before. Trying to optimize for something that is completely random will just drive you insane…

I saw a pretty perplexing question having to do with probability the other day

*> How do you randomly generate a number from 1:7 with a six-sided die?*

It’s a pretty devious question. Each face number 1:6 has exactly 1/6 probability to show up. How do you convert this to a 1/7 probability? I thought about this for a while. I wrestled with common factors, common multiples and really long combinatorial examples drawn by hand. I even googled it. The answer I found proved to be lengthy:

There are a bunch of options. Here's an easy one: roll the die twice, keeping track of the first and second roll. There are 36 outcomes: (1,1), (1,2), (1,3), ..., (6,6) (The first number is what you got on the first roll; the second is what you got on the second.) If you get a (6,6), just re-roll the die twice again until you get a non-(6,6). Now there are 35 equally-likely outcomes, so divide them into 7 groups of 5 corresponding to the 7 choices among which you want to choose. Here's an easy calculation that will do it. Suppose you roll a (x, y) with x = number on first roll and y = number on second roll. First calculate the following number: N = 6(x-1) + y.

I hate complicated solutions. Here’s another method (in python, of course):

import random # import this to use the random feature dicerolls = [ [] for i in range(7) ] for i in range(len(dicerolls)): dicerolls[i] = [ int(random.random()*6) for number in range(500)] # print out winners and identify my random number from 1:7 winner = [sum(i) for i in dicerolls ] print winner print max(winner), "- random number:", winner.index(max(winner))+1

**How it works:** Roll N*7 times. Each result gets put into one of 7 lists, like so (when N = 13):

1: [1,3,3,1,4,1,4,3,4,5,5,6,6]

2: [5,3,4,4,2,3,4,2,3,3,2,4,3]

3: [5,2,4,6,2,3,4,2,3,2,4,5,4]

**…
**7: [2,4,2,3,4,5,1,6,2,4,6,2,4]

Because each individual roll is independent and random, it doesn’t matter how many times you roll, as long as each list is given the same amount of rolls. What’s important is that there are a sufficient number of rolls to prevent ties (in my code, I go with 2^7). After N rolls, you can sum the contents of the list, and the largest position is the outcome of your randomly generated number:

Run with N=500, here are the sums of each list:

[1236, 1207, 1246, 1233, 1273, 1248, 1298] 1298 - random number: 7

In this case, since the 7th list has the highest sum, your randomly generated number between 1:7 is 7. Easy!

I discovered an interesting property about the Monty Hall problem.

When the answer to the Monty Hall problem is explained, often people say “the probability of the car being behind the door if you switch **doubles.**” Not necessarily true! Let’s look at what happens when there are multiple doors using my handy dandy Monty Hall Monte Carlo class:

When there are three doors (blue), yes, the probability of winning goes from 1/3 to 2/3 (double) by always switching. However, if you use more doors, the probability of winning the car by switching shrinks! At 7 doors, there’s hardly any difference in win percentage whether you switching, hold or randomly switch.

Here’s another view of it.

The bottom dot on each line represents never switching, the center represents random switching. You can see that no matter the number of doors, the odds of winning the car are are always higher if you always switch the door you choose. However, the advantage seems to decay quite quickly, almost to the point of exhibiting no advantage near 7 doors.

**UPDATE: Fellow Zipfian Alum Alex Mentch Explains:**

*With 3 doors, your initial pick has a 1/3 chance of having the car and the other two doors have 2/3. **Remove one of those other two doors and the remaining door has the 2/3 chance.*

*At 4 doors, your initial pick has 1/4 and the other 3 have 3/4. **Remove one door and the 3/4 chance is shared amongst 2 doors, so 3/8 each. At 5 doors, your initial pick has 1/5 and the other 4 have 4/5.*

*Remove one door and the remaining doors have 4/5 * 1/3 = 4/15. *

Statistician Lisa Goldberg explains the Monty Hall problem using Bayesian Methods.