Re: Monty at it again
Yes , my answer assumes that there is a re-deal if both or all three events occur . In fact , it isn't even necessary because when both events occur , with probability 1 , you will be given a non 1 card on your second deal , when your first card is a 1 .
Just verify that
1/10*9/10 +1/10^2*9.10+1.10^3*9/10 +....+ =1/10 .
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