Probability question

katodog

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Nov 2, 2006
Location
Rochester,NY
Not sure if this is the right place for this question but here it goes.

Lets say you have 5 "bonus" symbols and 47 blank symbols. No weighting and every symbol has an equal chance of coming in. Lets say you randomly draw five "cards", what are the odds you will get 3 bonus symbols?? what are the odds of drawing 4 bonus symbols and what are the odds of drawing five?
I am struggling with this and any help is appreciated.
 

KasinoKing

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Well... I think this is right:

3 cards = 5/52 x 4/51 x 3/50 = 1 in 2,210
4 cards = 5/52 x 4/51 x 3/50 x 2/49 = 1 in 54,145
5 cards = 5/52 x 4/51 x 3/50 x 2/49 x 1/48 = 1 in 2,589,960

Pretty sure that's right...

KK
 

lnspin

Senior Member
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Well... I think this is right:

3 cards = 5/52 x 4/51 x 3/50 = 1 in 2,210
4 cards = 5/52 x 4/51 x 3/50 x 2/49 = 1 in 54,145
5 cards = 5/52 x 4/51 x 3/50 x 2/49 x 1/48 = 1 in 2,589,960

Pretty sure that's right...

KK

Thats 100% correct if it was being drawn from a deck of cards, without the drawn card being put back into the deck, think he was referring to a slot with say 5 independent reels.

So would be 5/52 x 5/52 x 5/52 and so on for 4th reel and 5th reel.. Im too lazy to mutiply it.
 

Embalu

Dormant account
Joined
Mar 29, 2004
Location
Hungary
Well... I think this is right:

3 cards = 5/52 x 4/51 x 3/50 = 1 in 2,210

This calculation would be right if we drew only 3 card, but in fact we draw 5 cards and need 3 winning cards from this five.

If we take this into consideration and also put back the drawn cards into the deck as lnspin suggested, here is the formula for the probability of getting exactly n winning cards from the 5:

Probability for n winning cards = combin(5,n) x (5/52)[SUP]n[/SUP] x (47/52)[SUP]5-n[/SUP]

where combin(5,n) = 5! / (n! x (5-n)!)

So:

combin (5,0) = 1
combin (5,1) = 5
combin (5,2) = 10
combin (5,3) = 10
combin (5,4) = 5
combin (5,5) = 1

For example, combin (5,3) = 10 means that if we have 3 winning cards of 5, there are 10 possible combinations, here they are ('x' means winning and '.' means losing cards):

xxx..
xx.x.
xx..x
x.xx.
x.x.x
x..xx
.xxx.
.xx.x
.x.xx
..xxx

So, here are the formulas for the given number of winning cards:

Probability for 0 winning cards = 1 x (47/52)[SUP]5[/SUP]
Probability for 1 winning cards = 5 x (5/52) x (47/52)[SUP]4[/SUP]
Probability for 2 winning cards = 10 x (5/52)[SUP]2[/SUP] x (47/52)[SUP]3[/SUP]
Probability for 3 winning cards = 10 x (5/52)[SUP]3[/SUP] x (47/52)[SUP]2[/SUP]
Probability for 4 winning cards = 5 x (5/52)[SUP]4[/SUP] x (47/52)
Probability for 5 winning cards = 1 x (5/52)[SUP]5[/SUP]

And the results:

Probability for 0 winning cards = 60.32156% (or 1 in 1.65778)
Probability for 1 winning cards = 32.08594% (or 1 in 3.11663)
Probability for 2 winning cards = 6.826795% (or 1 in 14.6482)
Probability for 3 winning cards = 0.726255% (or 1 in 137.693)
Probability for 4 winning cards = 0.038631% (or 1 in 2588.62)
Probability for 5 winning cards = 0.000822% (or 1 in 121665)
-----------------------------------------------------------
Total: 100%
 
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