Pop
Member
Registered: 8th May 03
Location: Reading
User status: Offline
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Ok so I've got a maths question, should be relatively simple but I can't get my brain into gear and want to check what others think.
It's to do with roulette so I'm not after the it won't work etc speech as a few of us are doing this for a laugh/challenge with minimal outlay to see who can get largest % increase on bank before going bust.
Anyway, there are 37 numbers on the table (including 0) which return at odds of 36/1. If you put 1% of your total bank on 35 numbers and get a winner you will be 1% up. Now if you were to back 34 numbers and won you would be 2% up.
Which would be the least riskiest option to get to 2%, betting twice with a 2/37 chance of loss or once with a 3/37 chance? 
Answers on a postcard.
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Balling
Premium Member
Registered: 7th Apr 04
Location: Denmark
User status: Offline
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4 / 74 = 0,06
3 / 37 = 0,09
If my calculations have anything to do with how this stuff works, it would seem that your risk is higher if betting on 34 numbers.
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Seany
Member
Registered: 13th Dec 06
Location: Dunfermline, Fife : Drives Astra cdti Sri
User status: Offline
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The less numbers you bet on the more risk there is,
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Pop
Member
Registered: 8th May 03
Location: Reading
User status: Offline
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Here is my take on it which is a bit different to the replies; I just don't know which is correct.
1% return = 35/37 numbers = 94.6% chance of winning
2% return = 34/37 numbers = 91.9% chance of winning
Now if you do the 1% return bet twice in a row in my mind you are technically risking 4 chances of a loss against 37 numbers = 33/37 = 89.2% chance of winning.
Therefore, my guess would be a single bet to win 2% return caries less risk.
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