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A game is made, in the game it all depends on a fair six sided dice. The game cost 10 coins to play.?
You throw the dice 3 times. If they`re all one`s you lose all coins. If you get 2 one`s you`ll get 1 coin back, if you get 1 one´s youll get 3coins back, if you only get 2,3,4,5,6 youll get 20 coins back.
A: What would be the expected winnings/losings from each game played ?
1 réponse
- il y a 3 ans
Chance of all ones is 1/216 (1/6 * 1/6 * 1/6), chance of no one's is 125/216 (5/6 * 5/6 * 5/6), chance of two ones is 15/216 (1/6 * 1/6 * 5/6) *3 because there are three ways for it to happen. Chance of a single one is 75/216.
These probabilities add up to 216/216, covering all eventualities.
This means the chances of no payout in any one game are 1/216.
Average winnings = 20(125/216) + 3(75/216) + 1(15/216) which seems to be 2.68 coins.
The fly in the ointment is where you write "lose all your coins". Do you mean the game will be played a set number of times?
The more times you play it, the more that 1/216 shot of a triple one will loom large.
Do you mean you lose everything?
If so, in the long run your winnings will always be zero because you will get wiped out at some point and then you can't afford to get back in.
This has been the fate of many a gambler with an infallible "system".
