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Thank you for that - I can see the logic - and also a simpler way to get to the same result - if the first four cards dealt are clubs , the odds are 9 out of 39 against the other card being a club - but because the odd card could be the first, second etc card dealt that becomes 9 out of 39 x 5 , which is 9 out of 195 , which is 21.8 to 1.
Which still seems weird , because when you have four of the suit there are 9 more of that suit in the pack, and 39 others , so you would think the ratio would be 9 to 39 , or 1 in 4.3
But then I thought if you applied that to just six cards, five of which were clubs , then applying that logic you would get a flush 1 out of two goes, whereas in fact you would only get it 1 in 6 goes as the odd card would appear in the first five cards dealt five times out of six !
Thanks for your help on here now, just glad to have got it resolved in my head !!
Sorry - I am not making myself clear - I am not saying you are being dealt ( for instance ) four clubs, what is the odds of the last card being a club.
I am saying you could be dealt a non-club, then four clubs - or a club, then a non-club, then three clubs - etc - So how many times will you get five clubs off the deal , rather than four plus a card of another suit ( that could be the first, 2nd,3rd,4th, or 5th card dealt.)
maybe this is a better way of putting it -
For every one occasion that you 'flop' five clubs on the deal , on how many occasions will you flop four clubs plus one card of a different suit (in any order ) on the deal ?
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