And here's the proof that common sense is not so common.
It's a free bet - the cost to play is zero - they are by definition both excellent bets because the expectancy on both is an infinite multiple of the risk.
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- Thread starter sittingduck
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And here's the proof that common sense is not so common.
It's a free bet - the cost to play is zero - they are by definition both excellent bets because the expectancy on both is an infinite multiple of the risk.
Quote from Random.Capital:
And here's the proof that common sense is not so common.
It's a free bet - the cost to play is zero - they are by definition both excellent bets because the expectancy on both is an infinite multiple of the risk.
Yeah, but the question is, which is <i>better</i>. But good point.
Quote from sittingduck:
1- Four cards are dealt off the top of a well shuffled deck. You have a choice:
(a) To win $1 if the first card is a club, and the second a diamond, and the third is a heart, and the fourth is a spade, in this order.
(b) To win $1 if the four cards are of four different suits, in any order.
Which option is better? Or are they equivalent?
Justify by showing what is the probability in each case.
(a)
Card #1 = 13/52 chance
Card #2 = 13/51 chance
Card #3 = 13/50 chance
Card #4 = 13/49 chance
(b)
Card #1 = 52/52 chance
Card #2 = 39/51 chance
Card #3 = 26/50 chance
Card #4 = 13/49 chance
I'd say B looks easier
Quote from Random.Capital:
And here's the proof that common sense is not so common.
It's a free bet - the cost to play is zero - they are by definition both excellent bets because the expectancy on both is an infinite multiple of the risk.
That's the best answer I have seen!
Commonsensical all the way.
B is the best probability of prevailing.