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I do not think so, as the number of red balls, vs the number of blue balls, is less, since you have already taken one red ball out (unless you simply put the red ball back in the bag, in which case yes, the chance is still 50/50)

NOTE: I no longer frequent this site. If you really need help, PM/DM me and my e.mail will alert me. 

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1 minute ago, comander said:

If it's sampling WITH replacement, the odds are unchanged.
If it's sampling withOUT replacement then the odds shift. 

 

first run WITH replacement: 5/10

second run with replacement: 5/10

 

first run withOUT replacement: 5/10

second run without replacement: 4/9 = (5-1)/(10-1)

@Radium_Angel

But then in that case shouldn't the odds of Mounty Hall problem be the same?

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Ok, say you got super lucky, and pulled out 9 red balls in a row. Unbelievable odds, but for the sake of argument let's say you did.

So what's left is 1 red ball, and 10 blue ones. THe odds are NOT 50/50 you will pick another red ball.

@comander is correct, the more red balls you remove, the smaller your chances of picking another red ball, becomes.

NOTE: I no longer frequent this site. If you really need help, PM/DM me and my e.mail will alert me. 

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  • 2 weeks later...
On 6/2/2021 at 10:08 PM, Wictorian said:

There are 10 blue balls and 10 red balls in a bag. We can calcukate that one has a chance of 1/2 of picking a red ball in each turn, correct? So, if I pick a red ball, is the chance of picking a red balk still same?

the answer is no. there will less chance of picking out a red ball compared to blue balls. for a thing not to happen one would either need zero chances/ probability or probability which is very close to zero. like that one whole Dream Speed Run controversy.

 

yes i have forgotten the math part....

 

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