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odds of 3 flushes in one hand
The_Don90
Member Posts: 9,818
Ok i have no idea what happened here anyways whats the odds of 3 flushes all connecting on the river ?
PlayerActionCardsAmountPotBalance Small blind £0.02£0.02£1.50The_Don90Big blind £0.04£0.06£3.96 Your hole cardsJK Fold Call £0.04£0.10£10.76 Fold Call £0.04£0.14£2.51 Call £0.02£0.16£1.48The_Don90Check Flop 2Q7 Check The_Don90Check Check Check Turn Q Check The_Don90Check Check Check River 4 Bet £0.04£0.20£1.44The_Don90Raise £0.28£0.48£3.68 Call £0.28£0.76£10.48 Call £0.28£1.04£2.23 Fold The_Don90ShowJK Muck35 Muck910 The_Don90WinFlush to the King£0.96 £4.64
PlayerActionCardsAmountPotBalance Small blind £0.02£0.02£1.50The_Don90Big blind £0.04£0.06£3.96 Your hole cardsJK Fold Call £0.04£0.10£10.76 Fold Call £0.04£0.14£2.51 Call £0.02£0.16£1.48The_Don90Check Flop 2Q7 Check The_Don90Check Check Check Turn Q Check The_Don90Check Check Check River 4 Bet £0.04£0.20£1.44The_Don90Raise £0.28£0.48£3.68 Call £0.28£0.76£10.48 Call £0.28£1.04£2.23 Fold The_Don90ShowJK Muck35 Muck910 The_Don90WinFlush to the King£0.96 £4.64
Comments
This is equivalent to odds of: 13,063 to 1.
seems i have a looong way to go before i work out HOW to work out the odds.
Maybe this is where im going wrong? At the moment im feeling my way through tournys,
as a side question, where do i pick up this info?
Player A - 1st card could be anything
Player A - 2nd card must be one of 12 cards of the same suit from 51 cards remaining (hence 12/51)
Player B - 1st card must be one of 11 cards of the same suit from 50 cards remaining (hence 11/50)
Player B - 2nd card must be one of 10 cards of the same suit from 49 cards remaining (hence 10/49)
Player C - 1st card must be one of 9 cards of the same suit from 48 cards remaining (hence 9/48)
Player C - 2nd card must be one of 8 cards of the same suit from 47 cards remaining (hence 8/47)
There are 20 ways of choosing 3 hands from 6 players at the table.
Then exactly two of the first four community cards need to be of the same suit:
1st community card can be one of 7 cards of the same suit from 46 cards remaining (hence 7/46)
2nd community card can be one of 6 cards of the same suit from 45 cards remaining (hence 6/45)
3rd community card can be one of 39 cards of a different suit from 44 cards remaining (hence 39/44)
4th community card can be one of 38 cards of a different suit from 43 cards remaining (hence 38/43)
There are 6 ways of choosing 2 cards from the 4 dealt on the flop and turn.
River card must be one of 5 cards of the same suit from 42 cards remaining (hence 5/42).
To get the overall probability you multiply all these numbers together.
I know that you didn't really want to know but I couldn't help myself.
It's not really of any use to anyone, obviously.
i think ill stick to asking