Expected value of Laplace cube?
I solved the following task:
…but my calculation is wrong according to the solution:
The only difference is that I multiplied the probability of the sixth throw by 1/6, which isn't provided for in the solution. What I was thinking is that the probability of throwing a six on the sixth throw is 1/6, rather than (5/6)^5 for throwing a six after having already thrown five times.
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Hello Phoenix1604,
didn't you take two exponents wrong?
In addition, the consideration is whether a six is diceed in the case of the second litter, since after the second litter the game ends, whether you have a six dice or not. This is why you do not multiply (5/6)^5 again (1/6).
LG Moon
In fact, I took an exponent wrong. Thank you for your efforts: You get a star;
Each throw is to be assessed individually. The previous litters have nothing to do with it. Just because you've thrown 5x NO six, the 6. Don't have to be.
keyword "The Gambler's Ruin".