This Week's 538 Riddle (3/25/16)

[silverscreenselect]

Again, it's a math expected value problem:

Suppose a casino invents a new game that you must pay $250 to play. The game works like this: The casino draws random numbers between 0 and 1, from a uniform distribution. It adds them together until their sum is greater than 1, at which time it stops drawing new numbers. You get a payout of $100 each time a new number is drawn.

For example, suppose the casino draws 0.4 and then 0.7. Since the sum is greater than 1, it will stop after these two draws, and you receive $200. If instead it draws 0.2, 0.3, 0.3, and then 0.6, it will stop after the fourth draw and you will receive $400. Given the $250 entrance fee, should you play the game?

Specifically, what is the expected value of your winnings?

At first glance, it would seem you should play, since you'll always "win" at lead $200, and you have the potential to win a lot more. Since the expected value of any draw is .5, it would seem that at least half the time, your first two numbers won't add up to 1 and you win. But I'm sure there's more to it.

Here's the link to the site with last week's answer, which people here already guessed (as did 42% of the people who responded):

http://fivethirtyeight.com/features/sho ... sino-game/

[Bob78164]

Again, it's a math expected value problem:

Suppose a casino invents a new game that you must pay $250 to play. The game works like this: The casino draws random numbers between 0 and 1, from a uniform distribution. It adds them together until their sum is greater than 1, at which time it stops drawing new numbers. You get a payout of $100 each time a new number is drawn.

For example, suppose the casino draws 0.4 and then 0.7. Since the sum is greater than 1, it will stop after these two draws, and you receive $200. If instead it draws 0.2, 0.3, 0.3, and then 0.6, it will stop after the fourth draw and you will receive $400. Given the $250 entrance fee, should you play the game?

Specifically, what is the expected value of your winnings?

At first glance, it would seem you should play, since you'll always "win" at lead $200, and you have the potential to win a lot more. Since the expected value of any draw is .5, it would seem that at least half the time, your first two numbers won't add up to 1 and you win. But I'm sure there's more to it.

Here's the link to the site with last week's answer, which people here already guessed (as did 42% of the people who responded):

http://fivethirtyeight.com/features/sho ... sino-game/

That part is easy. The likelihood of a loss is exactly 50% (integrate (1 - x)dx from 0 to 1), so the likelihood of a win is also exactly 50%. Any loss will be exactly $50, and some wins will be more than $50, so the expectation value is clearly positive.

What will require work is determining the likelihood of n numbers for n >= 4. --Bob

[TheConfessor]

This seems like a poorly worded question. It is not clear to me what they are trying to say. If they draw a random number between 0 and 1, does that mean they might draw 0.54763229? Or is it limited to the one-tenth increments used in their example? And if it is the latter, can the same number repeat, or are they drawing from a finite set of 9 possible numbers, that is, .1, .2, .3, .4, .5, .6, .7, .8, and .9, without replacement? Or could they draw .1 ten times in a row and still not exceed 1? Or if the random numbers could be something like .0000000001, then there would be no limit to the possible number of draws.

[Bob78164]

This seems like a poorly worded question. It is not clear to me what they are trying to say. If they draw a random number between 0 and 1, does that mean they might draw 0.54763229? Or is it limited to the one-tenth increments used in their example? And if it is the latter, can the same number repeat, or are they drawing from a finite set of 9 possible numbers, that is, .1, .2, .3, .4, .5, .6, .7, .8, and .9, without replacement? Or could they draw .1 ten times in a row and still not exceed 1? Or if the random numbers could be something like .0000000001, then there would be no limit to the possible number of draws.

The numbers can be anywhere between 0 and 1. It's a uniform distribution, meaning that the probability of drawing a number less than x in the interval is x. The example given makes clear that repeats are possible, although since any single point is a set of measure zero, that possibility can be safely neglected. --Bob

[BackInTex]

I don't have enough facts. Does the casino include free drinks and a discount to the all-you-can eat prime rib buffet?

[Pastor Fireball]

I don't have enough facts. Does the casino include free drinks and a discount to the all-you-can eat prime rib buffet?

*sarcafont* No handouts, you socialist!