Q: You are in a game with a bankroll of 13 gold coins. The game involves a 13 pots lined up. There is a 96% chance that one of the pots has 22 gold coins in it. You get to inspect a pot by paying 1 gold coin. The game organizer tells you that there is a 90% chance that the very first pot has the gold coins in it. You pay 1 gold coin to inspect the first pot and you find there is no gold in it. Should you continue to play?

Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)

A: This is good example of a puzzle where at first blush it appears that there is no merit in continuing. It almost gives the impression that "most" of your probability of winning is lost right from the get go as the first pot, which was known to have a 90% chance of having the 13 gold coins does not contain gold.

Let us assume that the probability of winning is \(x\) downstream of the first pot (that is pots 2 through 13). Next, lets estimate the probability of not winning at all in this game. The probability of not winning at the first pot is \(1 - 0.9 = 0.1\) and that of not winning on any of the remaining pots is \( 1 - x\). The net probability we know to be \( 1 - 0.96 = 0.04\). Thus we can state this as

$$ 0.1 \times ( 1 - x ) = 0.04 $$

Solving for \( x \) yields

$$x = 60\%$$

You are left with 12 gold coins. Your expected pay off is \( 0.6 \times 22 = 13.2\) coins, so you must play.

As an aside, notice that the probability of a win is

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A: This is good example of a puzzle where at first blush it appears that there is no merit in continuing. It almost gives the impression that "most" of your probability of winning is lost right from the get go as the first pot, which was known to have a 90% chance of having the 13 gold coins does not contain gold.

Let us assume that the probability of winning is \(x\) downstream of the first pot (that is pots 2 through 13). Next, lets estimate the probability of not winning at all in this game. The probability of not winning at the first pot is \(1 - 0.9 = 0.1\) and that of not winning on any of the remaining pots is \( 1 - x\). The net probability we know to be \( 1 - 0.96 = 0.04\). Thus we can state this as

$$ 0.1 \times ( 1 - x ) = 0.04 $$

Solving for \( x \) yields

$$x = 60\%$$

You are left with 12 gold coins. Your expected pay off is \( 0.6 \times 22 = 13.2\) coins, so you must play.

As an aside, notice that the probability of a win is

__independent__of the number of pots that are lined up. But the expected pay off will vary.Some must buy books on Probability

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Nice puzzle. BTW you should make it clear that 90% refers to the *unconditional* chance that the first pot has the 22 coins, at first I thought you meant the chance the first pot has the gold given that one of the pots has it.

ReplyDeleteAlso, a nice, intuitive and quick way to see the answer is to draw a box and partition it into gold exists, gold doesn't exist, and gold is in first pot.

Cheers,

Matt

Thanks Matt.

DeleteYour point, noted.

Interesting. I'm with id that I didn't realize the 90% was independent of the 96% number.

ReplyDeleteAlso, I always find the wording "Your expected pay off is 0.6×22=13.2 coins, so you must play." interesting. The percent return is 10% (1.2 coins above the 12 started with) as of the second round. So the risk/reward is 40% for 10% (40% chance of losing everything for a 10% return). I don't think that falls in the "must play" category. You should play, from a game theory perspective, but from a risk perspective the risk/reward ratio is pretty bad.

The first pot makes much more sense, of course. A 10% risk for a 2200% (you only risk 1 coin) payout... that's closer to a "must play" scenario.

Thanks for the positive feedback. Agree that the risk/reward isn't that great from a utility perspective. Just crafting math fun here :)

ReplyDelete

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