Thread: probability gambling

. In the XVII century Lord de Mere, a vivid gambler, asked Pascal, one of the great mathematicians of his time: If a player rolls a die 4 times what should I bet on: a) he rolls at least one 6; or b) he does not roll a 6 at all? What would you bet on?
so for this question I would say a. because it would a 4 in 6 chance or a 0.667 chance
anybody have another way of showing this because I think i might me wrong

Hello, mr.toronto!

If a player rolls a die 4 times, what should I bet on:
a) he rolls at least one 6, or
b) he does not roll a 6 at all?
What would you bet on?

Your reasoning is incorrect . . .

One any roll, $\displaystyle P(6) \:=\:\frac{1}{6},\;P(\text{other}) \:=\:\frac{5}{6}$

b) The probability that he rolls no 6's is: .$\displaystyle \left(\frac{5}{6}\right)^4 \:=\:\frac{625}{1296} \:\approx\:0.482 $

a) Hence: .$\displaystyle P(\text{at least one 6}) \:=\:1 - 0.482 \:=\:0.518$

k so he should bet on at least one six since there is a 0.518% he will roll at least one six

Originally Posted by mr.toronto

. In the XVII century Lord de Mere, a vivid gambler, asked Pascal, one of the great mathematicians of his time: If a player rolls a die 4 times what should I bet on: a) he rolls at least one 6; or b) he does not roll a 6 at all? What would you bet on?
so for this question I would say a. because it would a 4 in 6 chance or a 0.667 chance
anybody have another way of showing this because I think i might me wrong

Work from the case where no sixs are rolled.

Probability of not rolling a six on a single roll is 5/6

Probability of this occuring on all four rolls (5/6)^4 ~= 0.4823

Probability of rolling at least one six = 1- prob of rolling no sixes = 1-(5/6)^4 ~= 0.5177

RonL

Originally Posted by mr.toronto

k so he should bet on at least one six since there is a 0.518% he will roll at least one six

If the bet is being offered at evens, otherwise it depends on the odds he is being offered.

RonL