Let’s play a game.
Here are the rules:
You get to choose.
You can either take ONE fun size candy bar, OR, you can risk it all and role the dice.
If you roll an even number, you get that number of fun size bars from my bucket.
If you roll an odd, I get that number of bars from your bag.
Do you want to play?
Lemme guess, you don’t get a lot of Trick or Treaters in Halloween, do you? 🙂
Even or odd is 50/50, so you’d come out ahead if YOU took evens (2, 4, 6) and let them have odds (1, 3, 5).
No charge.
Funny thing about that. The kids will recognize the value of the guaranteed payout, one fun sized bar, no risk. It is the adults that will start trying to game the system…
I would be curious to see if anyone actually takes you up on that, and if so, what is there (approx.) age.
Yes and thanks to my skillfully deployed set of goblin bookies I am all but guaranteed a favorable outcome,
http://magiccards.info/query?q=Goblin+bookie&v=card&s=cname
I’ll roll the dice. 50% odds of getting a minimum of 2 candy bars vs. 50% odds of losing a maximum of 11 candy bars, and sure there’s a 1 in 6 chance of losing 7, but there’s better odds than that of getting either 6 or 8.
Is it two six-sided dice? The rules are unclear here.
I was visualizing one six-sided die. That puts the odds of winning at 50%, but the pay-out is weighted in favor of the player; on average, you win one more than you’d otherwise lose (you could lose 1, 3, or 5 [avg. 3], or win 2, 4, or 6 [avg. 4]).
Alternatively, if the die is a standard 10-sided, the odds are the same but the pay-out is weighted against the player; they’re numbered 0-9 instead of 1-10. While still technically a “win”, a zero roll gets you nothing; there’s only a 40% chance of a paying win, but still a 50% chance of losing (avg. win is 4, avg. loss is 5).
But now we’re comically over-thinking it, aren’t we? LOL! 😀