How much happiness does money buy?

livemint.com:

Imagine a game in which you roll a six-sided dice and win a number of dollars equal to the score. So a one gets you $1, a two gets $2 and so on. How much should you pay to play?

Those who paid attention in probability class already know the answer. The potential outcomes are $1, $2, $3, $4, $5 or $6, each of which has a one-in-six chance of coming up. Your expected winnings are the sum of each result multiplied by its odds of occurring. Add it all up and this comes to $3.50. And so you have your price: if you can play for anything less than $3.50, you should. If you can pay $3 to play, for example, then on average you will make a net gain of $0.50.

Nice, but hardly enough to set anyone’s pulse racing. After all, there is a 50% chance of making either nothing or a loss and you can’t do much with a few dollars anyway. But suppose you are instead offered 10,000 rolls. This is a lot more interesting. Your intuition says you are now bound to average close to $3.50 per throw, since repetition will smooth away the effects of luck. A mathematician would confirm your instincts: with that many throws, the odds of the average falling far from $3.50 are near zero. If each roll costs $3, in other words, you are virtually guaranteed a profit of around $5,000 (or 10,000 lots of $0.50). You would be a fool to turn down such a deal—so much so that, if you lack the $30,000 needed to play, you should borrow it.

Now consider a final variation: instead of 10,000 rolls, you are back to just one. This time, though, your winnings will be $10,000 times the score on the dice and the cost to play is $30,000. How keen are you now? The expected profit is still $5,000, but the risk of losing at least $10,000 (by rolling a one or two) has risen from